{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "\n", "\n", "\n", "\n", "\n", "\n", "\n", "

\n", "Abstract Algebra: An Interactive Approach, 2e\n", "

\n", "\n", "

\n", "©2015 This notebook is provided with the textbook, "Abstract Algebra: An Interactive Approach, 2nd Ed." by William Paulsen. Users of this notebook are encouraged to buy the textbook.\n", "

\n", "\n", "\n", "\n", "

\n", " Chapter 14
The Theory of Fields\n", "

\n", "\n", "\n", "

Initialization: This cell MUST be evaluated first:

" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "%display latex\n", "try:\n", " load('absalgtext.sage')\n", "except IOError:\n", " load('/media/sf_sage/absalgtext.sage')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Vector Spaces
\r\n", "Extension Fields
\r\n", "Splitting Fields
\r\n", "Sage Interactive Problems
\r\n", "\r\n", "\r\n", "

Vector Spaces

\r\n", "\r\n", "
\r\n", "In order to study fields in depth, we will first need a few results from a first year linear algebra course. However, most linear algebra courses work with vectors \r\n", "and matrices with real numbers for entries, whereas we will generalize the notations to allow arbitrary fields. Nonetheless, most of the proofs will follow the same way \r\n", "for arbitrary fields as for real numbers.

\r\n", "\r\n", "DEFINITION 14.1
\r\n", "Let F be a field. We say that V is a vector space over F if V is an abelian group under addition (+), and for which there \r\n", "is defined a multiplication a·v for all a ∈ F and v ∈ V such that:

\r\n", "\r\n", "1. Whenever a ∈ F and v ∈ V, a·v ∈ V.
\r\n", "2. When a ∈ F, and v, w ∈ V, then a·(v + w) = \r\n", "a·v + a·w.
\r\n", "3. When a, b ∈ F, and v ∈ V, then (a + b)·v = \r\n", "a·v + b·v.
\r\n", "4. When a, b ∈ F, and v ∈ V, then (a·b)·v = \r\n", "a·(b·v).
\r\n", "5. If e is the identity of F, then e·v = v for all v ∈ V.

\r\n", "\r\n", "The members of V are called vectors. The best way to get a feel for vector spaces is to give some examples.

\r\n", "\r\n", "EXAMPLE:
\r\n", "Consider the set of 3-tuples ⟨u1, u2, u3⟩ where u1, u2, and \r\n", "u3 ∈ ℝ. Addition of two vectors is done componentwise, and \r\n", "k·⟨u1, u2, u3⟩ = \r\n", "⟨k u1, k u2, k u3⟩ when k ∈ ℝ. This is a vector space over ℝ, and can \r\n", "be denoted by ℝ3.

\r\n", "\r\n", "EXAMPLE:
\r\n", "We can generalize the previous example using any field F in place of ℝ, and \r\n", "consider n-tuples ⟨u1, u2, … un⟩. Addition is still defined componentwise, \r\n", "and k·⟨u1, u2, … un⟩ = \r\n", "⟨k·u1, k·u2,… k·un⟩. This will give us a \r\n", "vector space over F, which we can denote by Fn.

\r\n", "\r\n", "EXAMPLE:
\r\n", "Let K be a field, and F any subfield of K. Then K is a vector space over F, defining a·v as a \r\n", "product in the field K. Property 5 follows from the fact that the identity of F must also be the identity of K. The other properties follow \r\n", "from the distributive and associative properties of K.

\r\n", "\r\n", "This last example demonstrates the usefulness in studying vector spaces over a field F. In fact, this is the example that we will concentrate on for the \r\n", "remainder of the chapter. The next definition is the key to understanding the properties of a vector space.

\r\n", "\r\n", "DEFINITION 14.2
\r\n", "Let V be a vector space over a field F. We say that a finite \r\n", "set B = {x1, x2, x3, … xn} of vectors in V are \r\n", "linearly dependent if there are elements c1, c2, … cn ∈ F, not all zero, for which\r\n", "

c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn = 0.

\r\n", "We say that the vectors are linearly independent if they are not linearly dependent, that is, if the only way for\r\n", "

c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn = 0.

\r\n", "is for c1 = c2 = ⋯ = cn = 0.

\r\n", "\r\n", "EXAMPLE 14.4
\r\n", "The vectors ⟨1, 4, −1⟩, ⟨2, −3, 1⟩, and ⟨4, 5, −1⟩ are linearly dependent, since there is a nonzero solution to \r\n", "c1 ⟨1, 4, −1⟩ + c2 ⟨2, −3, 1⟩ + c3 ⟨4, 5, −1⟩ = 0, namely, \r\n", "c1 = 2, c2 = 1, and c3 = −1. On the other hand, ⟨2, 0, 1⟩, ⟨0, 0, 3⟩, \r\n", "and ⟨1, 4, 0⟩ are linearly independent, since in order to get \r\n", "c1 ⟨2, 0, 1⟩ + c2 ⟨0, 0, 3⟩ + c3 ⟨1, 4, 0⟩ = 0, we need\r\n", "4 c3 = 0, 2 c1 + c3 = 0, and c1 + 3 c2 = 0. This forces \r\n", "c3 = 0, c1 = 0, and c2 = 0, so there are no nonzero solutions.

\r\n", "\r\n", "DEFINITION 14.3
\r\n", "Let V be a vector space over a field F. A finite set of \r\n", "vectors {x1, x2, x3, … xn} in V is called a basis of V \r\n", "over F if the set is linearly independent, and every element of V can be expressed in the form\r\n", "

a1 x1 + a2 x2 + \r\n", "a3 x3 + ⋯ + an xn,

\r\n", "with a1, a2, a3, … an in F.

\r\n", "\r\n", "Here are some examples, all of which are fairly routine to check:

\r\n", "1. The complex numbers ℂ have a basis {1, i} over the real numbers ℝ.

\r\n", "2. The quaternions ℍ have a basis {1, i, j, k} over ℝ

\r\n", "3. The field ℚ[√2] has a basis {1, √2} over the \r\n", "rational numbers ℚ.

\r\n", "4. Since a field is a vector space over a subfield, the set of real numbers ℝ is a vector space over the rationals. However, there can be no finite \r\n", "basis {x1, x2, x3, … xn} in ℝ for which every real number could be expressed as\r\n", "

a1 x1 + a2 x2 + \r\n", "a3 x3 + ⋯ + an xn,

\r\n", "with a1, a2, … an ∈ ℚ, lest the set of reals be countable, which contradicts \r\n", "Cantor's diagonalization theorem (0.8).

\r\n", "\r\n", "There is an easy way to determine if a particular set of vectors is a basis.\r\n", "\r\n", "

\r\n", " \r\n", "LEMMA 14.1
\r\n", "B = {x1, x2, x3, … xn} is a basis of a vector space V over \r\n", "F if, and only if, every element of V can be expressed uniquely in the form\r\n", "

v = c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn.

\r\n", "The ordered n-tuple ⟨c1, c2, c3, … cn⟩ is called \r\n", "the coefficients of v with respect to B.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We can define a basis in Sage with the command ToBasis.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([[1, 4, -1], [2, -3, 1], [4, 5, -1]])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Notice that we entered a list of lists, which Sage interprets as a list of vectors. This failed because, as we saw before, the set of \r\n", "vectors ⟨1, 4, -1⟩, ⟨2, -3, 1⟩, and ⟨4, 5, -1⟩ were linearly dependent. Here is another \r\n", "attempt.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([[2, 0, 1], [0, 0, 3], [1, 4, 0]])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This time, there was no error message, so we know that these are linearly independent. Once we have defined a basis, we can find the \r\n", "coefficients c1, c2, … cn for any element of the vector space.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Coefficients(B, [2, 3, 4])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This shows that\r\n", "

⟨2, 3, 4⟩ = 5⁄8⟨2, 0, 8⟩ + 9⁄8⟨0, 0, 3⟩ + \r\n", "3⁄4⟨1, 4, 0⟩.

\r\n", "\r\n", "

\r\n", " \r\n", "LEMMA 14.2
\r\n", "Suppose that V is a vector space over F, and B = {x1, x2, x3, … \r\n", "xn} is a basis of V over F. Then any set {y1, y2, y3, … \r\n", "yn, yn+1} of n + 1 elements of V is linearly dependent.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We can now use this lemma to show that any two bases of a finite extension must have the same number of elements.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.1
\r\n", "Let V be a vector space over F. If the sets X = {x1, x2, x3, … \r\n", "xn} and Y = {y1, y2, y3, … \r\n", "ym} are both bases of V over F, then n = m.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "This proposition allows us to make the following definition.

\r\n", "\r\n", "DEFINITION 14.4
\r\n", "Let V be a vector space over F. If there is a basis {x1, x2, x3, … \r\n", "xn} of V over F, we define the dimension of V over F to be the size n of the basis. If there \r\n", "does not exist a finite basis, we say the dimension of V over F is infinite.

\r\n", "\r\n", "Here are some examples:

\r\n", "\r\n", "ℝ3 is a 3-dimensional vector space over ℝ.

\r\n", "\r\n", "ℂ is a 2-dimensional vector space over ℝ.

\r\n", "\r\n", "The quadternions ℍ is a 4-dimensional vector space over ℝ.

\r\n", "\r\n", "ℝ is an infinite-dimensional vector space over ℚ.

\r\n", "\r\n", "EXAMPLE:
\r\n", "Since Z3 is a subfield of GF(9), we can view GF(9) as a vector space over Z3. Determine a basis for this vector space.

\r\n", "\r\n", "The Conway polynomial of degree 2 over Z3 is:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "ConwayPolynomial(3, 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "so if we let a be a root of this polynomial, we can define GF(9) with the commands:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(3)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^2, a + 1)\r\n", "ListField()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "To find a basis, we can observe that every element of GF(9) can be written as x + y a, where x and y are \r\n", "in Z3. Thus, one possible basis would be the set {1, a}.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([1, a])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can now express every element of GF(9) in terms of this basis. Thus,
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Coefficients(B, a^3)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "shows that a3 = 1 + 2 a. We have shown that GF(9) is a 2-dimensional vector space over Z3.

\r\n", "\r\n", "EXAMPLE:
\r\n", "By the same logic, GF(81) will be a 4-dimensional vector space over Z3. But can we also consider hence GF(81) is a vector space over GF(9)? \r\n", "If so, find a basis.

\r\n", "\r\n", "The Conway polynomial of degree 4 over Z3 is:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "ConwayPolynomial(3, 4)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "So we could define GF(81) by telling Sage that the generator q raised to the 4th power is q3 + 1.

\r\n", "\r\n", "However, we ought to be able to define GF(81) as an extension of GF(9). This means we need an irreducible polynomial in GF(9) for which the generator\r\n", "of GF(81) will satisfy. Let us factor the above Conway polynomial in the field GF(9).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"x\")" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^4 + 2*x^3 + 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This gives us two possible polynomials. But we also have the compatibility condition, which says that \r\n", "the (34 − 1)/(32 − 1) = 10th power of\r\n", "the new generator b will be the standard generator a of GF(9). Hence, we can also factor
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^10 - a)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "The only factor that is in both factorizations is x2 + (a + 2)x + a. So we know that b satisfies this polynomial, \r\n", "so we can define b2 to be (1 + 2a)b + 2a." ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"b\")" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Define(b^2, (1 + 2*a)*b + 2*a)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can now define a basis of GF(81) over GF(9).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([1, a],[b, 2])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "
\r\n", "Note that we included two lists for arguments of the ToBasis command. The first list gives a basis for the root field F, which in this case \r\n", "is GF(9). We can now find components for elements in GF(81).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Coefficients(B, b^2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This tells us that b2 = (2a + 1)b + 2a. We can check this in Sage.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "b^2" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This last example shows that it is possible to have a vector space over a vector space, if the latter vector space happens to be a field.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.2
\r\n", "If E is a vector space over F of dimension m, which also happens to be a field, and V is a vector space over E of \r\n", "dimension n, then V is a vector space of F of dimension m n. Futhermore, \r\n", "if {x1, x2, x3, … xm} is a basis of E over F, \r\n", "and {y1, y2, y3, … yn} is a basis of V over E, then the set\r\n", "

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
S = {x1 y1, x2 y1, x3 y1, \r\n", " … xm y1,
x1 y2, x2 y2, x3 y2, \r\n", " … xm y2,
x1 y3, x2 y3, x3 y3, \r\n", " … xm y3,
……………
x1 yn, x2 yn, x3 yn, \r\n", " … xm yn}
\r\n", "\r\n", "is a basis of V over F.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "The main use of vector spaces in abstract algebra is in the case where the vector space happens to be a field. We will explain this possibility in the next \r\n", "section.

\r\n", "\r\n", "\r\n", "

Extension Fields

\r\n", "\r\n", "
\r\n", "In the last section, we found that many of the examples of vector spaces turned out to also be fields. We will give a special name to this situation.

\r\n", "\r\n", "DEFINITION 14.5
\r\n", "If F is a nontrivial subfield of the field K, and K is a finite dimensional vector space over F, we say that K is \r\n", "a finite extension of F. We say the degree, or dimension of the extension is the size of a basis\r\n", "

{x1, x2, x3, … xn}

\r\n", "of K over F.

\r\n", "\r\n", "For example, the "complex numbers mod 3", or GF(9), contains Z3 as a subfield. Clearly GF(9) is a 2-dimensional vector space \r\n", "over Z3, so we can say that GF(9) is a 2-dimensional extension of Z3.

\r\n", "\r\n", "Likewise, we can define GF(8) as follows:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(2)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 1 + a)\r\n", "F = ListField()\r\n", "F" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This time, we see that every element can be expressed in the form\r\n", "

x + a y + a2 z

\r\n", "where x, y and z are in Z2. Hence, {1, a, a2} is a basis of GF(8) over \r\n", "Z2, so GF(8) is a 3-dimensional extension of Z2.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([1,a,a^2])" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Coefficients(B, a^4)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Here are some other examples:

\r\n", "\r\n", "ℂ is a 2-dimensional extension of ℝ.

\r\n", "\r\n", "GF(27) is a 3-dimensional extenion of Z3, regardless of which basis we use.

\r\n", "\r\n", "It seems intuitively obvious that isomorphic fields have the same dimension over some field F contained in both of the fields. Yet this is only true if \r\n", "the isomorphism ϕ maps the base field F to itself.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.3
\r\n", "If K and E are two finite extensions of F, and suppose that there is an isomorphism ϕ from K onto E such \r\n", "that ϕ(x) = x for all x in F, then K and E have the same dimension over F.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We have already defined F[x] to be the smallest ring containing both F and the element x. We would like to make a similar \r\n", "definition for the smallest field containing both F and x.

\r\n", "\r\n", "DEFINITION 14.6
\r\n", "Let K be a field, and let E be a field containing the field K. Let S be a set of elements in E. Let L denote \r\n", "the collection of all subfields of E that contain the field K, along with the set S. Then we define\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
K(S) = ∩H.
H ∈ L
\r\n", "
\r\n", "That is, K(S) is the intersection of all subfields of E that contain both K and S. If S = \r\n", "{a1, a2, a3, … an}, we will write K(a1, \r\n", "a2, a3, … an) for K(S). Thus, if S consists of a single element a, \r\n", "we can write K(a) for K(S).

\r\n", "\r\n", "It is not difficult to see that this definition produces exactly what we want.\r\n", "\r\n", "

\r\n", "\r\n", "LEMMA 14.3
\r\n", "Let K be a subfield of E, and let S be a collection of elements of E. Then K(S) is the smallest field that \r\n", "contains both K and the elements S.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "For example, If K is the real numbers, and i = √−1, then ℝ(i) gives us the \r\n", "complex numbers ℂ. The field ℚ(√2) is the smallest field containing ℚ and \r\n", "√2, which happens to be the same thing as the \r\n", "ring ℚ[√2].

\r\n", "\r\n", "We have yet to produce any field extensions in Sage besides the finite fields of the last chapter. Yet the strategy for defining a field extension is \r\n", "very similar to that of defining a finite field. We begin by finding an irreducible polynomial f(x) in the field F, and creating \r\n", "the field K = F[x]/⟨f(x)⟩.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.4
\r\n", "Let F be a field, and let f(x) be an irreducible polynomial in F[x] of degree n. Then the \r\n", "field K = F[x]/⟨f(x)⟩ is a finite extension of F of dimension n.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "\r\n", "EXAMPLE:
\r\n", "Let F be the field of rational numbers, and let f(x) = x3 − 2. Form the field extension \r\n", "ℚ[x]/⟨x3 − 2⟩ in Sage.

\r\n", "\r\n", "Since the characteristic of ℚ is 0, we begin the definition by the command
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Next, we let a new variable a be a root to the equation x3 − 2. That is, we define a3 to be 2.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"a\")\r\n", "Define(a^3, 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "That's all there is to it! The basis of this extension field is {1, a, a2}. We can verify this with Sage.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "B = ToBasis([1, a, a^2])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can also do divisions in this field.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "1/(a + a^2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "EXPERIMENT:
\r\n", "Define the field ℚ(√2) in Sage using similar comands. What is the basis of the field? Then have Sage simplify the expression

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
2 a − 3 where a = √2.
5 a + 2
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Although this demonstrates that any extension field of the form F[x]/⟨f(x)⟩ can be entered into Sage, we would like to show \r\n", "that any finite extension of a field can be entered into Sage in the same way. That is, we must show that any finite extension of F is \r\n", "isomorphic to F[x]/⟨f(x)⟩ for some polynomial f(x).\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.5
\r\n", "Suppose a field K is a finite extension of F of dimension n. Let y be an element of K. Then there is an irreducible \r\n", "polynomial f(x) in F[x] of degree at most n such that f(y) = 0. That is, when f(x) \r\n", "is treated as a polynomial in K[x], y is a root of f(x). Furthermore, there is a unique polynomial of lowest degree \r\n", "that satisifies these conditions and for which the leading coefficient is equal to 1.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "The irreducible polynomial determined by this proposition is important enough to warrent a special notation.

\r\n", "\r\n", "DEFINITION 14.7
\r\n", "If a field K is a finite extension of F, and a is an element of K, we define the polynomial f(x) given by \r\n", "Proposition 14.5 that has a leading coeffient of 1 to be the irreducible polynomial of a over F, denoted\r\n", "

IrrF(a, x).

\r\n", "For example, Irrℚ(√2, x) = x2 − 2, since \r\n", "x2 − 2 is the simpliest polynomial with rational coefficients for which √2 is a root. Note \r\n", "that if we were to allow real coefficients, we could come up with a simpler polynomial:\r\n", "

Irrℝ(√2, x) = \r\n", "x − √2.

\r\n", "Here are some other examples. If i = √−1, we can see that the simpliest polynomial over the real \r\n", "numbers containing i as a root is\r\n", "

Irrℝ(i, x) = x2 + 1.

\r\n", "Finally, consider the number cos(π⁄9). We found in §11.4 that this number is a root of the polynomial 4 x3 − 3 x − \r\n", "1⁄2. However, we want the leading coefficient of the polynomial to be 1, so we write\r\n", "

Irrℚ(cos(π⁄9), x) = x3 − 3⁄4 x − 1⁄8.

\r\n", "Once we find the irreducible polynomial for an element a, it is not hard to program Sage to mimick the field ℚ(a). For example, let us \r\n", "enter the field ℚ(cos(π⁄9)) into Sage. If we let a = cos(π⁄9), we see that a satisfies \r\n", "x3 − 3⁄4 x − 1⁄8 = 0. If we put all terms besides the leading term on \r\n", "the other side of the equation, we find that a satisfies the equation\r\n", "

x3 = 3⁄4 x + 1⁄8.

\r\n", "That is,\r\n", "

a3 = 3⁄4 a + 1⁄8.

\r\n", "Let us enter this fact into Sage.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 3*a/4 + 1/8)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "The first command tells Sage that we are working with a field of characteristic 0, and the other commands identify a as one solution to the above \r\n", "equation. That's all there is to it! We can perform operations of this field such as
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "(2+a)*(3-2*a)" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "a^5" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "1/a" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "In this field, the a represents the number cos(π⁄9) = 0.9396926... . Have we really defined the \r\n", "field ℚ(cos(π⁄9))? Actually, we have defined the \r\n", "field ℚ[x]/⟨x3 − 3⁄4 x − \r\n", "1⁄8⟩ in Sage, but we can prove that these two fields are isomorphic.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.6
\r\n", "Let F be a subfield of K, and suppose f(x) is an irreducible polynomial in F[x] that has a root w in the \r\n", "larger field K. Then\r\n", "

F(w) ≈ F[x]/⟨f(x)⟩.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "Notice that the field ℚ(cos(π⁄9)) was three dimensional over ℚ, since the basis could \r\n", "be given as {1, a, a2}. Also, the irreducible polynomial x3 − 3⁄4 x − \r\n", "1⁄8 was a third degree polynomial. This is not a coindidence, as we can easily prove.\r\n", "\r\n", "

\r\n", "\r\n", "COROLLARY 14.1
\r\n", "Let K be a finite extension of a field F, and let u be an element in K. If \r\n", "f(x) = IrrF(u, x) has degree n, then F(u) has dimension n over F.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "Notice that we never had to tell Sage that a = cos(π⁄9) in our \r\n", "definition of ℚ(cos(π⁄9)). Rather, we simply told Sage that \r\n", "a satisfies the equation\r\n", "

a3 = 3⁄4 a + 1⁄8.

\r\n", "But there are two other solutions to this equation: −cos(2π⁄9), and cos(4π⁄9). How does Sage know that the field is \r\n", "not ℚ(−cos(2π⁄9)) or ℚ(cos(4π⁄9))?

The answer is of course that these \r\n", "fields are both isomorphic to ℚ(cos(π⁄9)), so Sage doesn't need to know the exact value \r\n", "of a. In fact, we can prove that if we start with isomorphic fields, and extend both of them by two elements for which the irreducible polynomials \r\n", "correspond, then the two field extensions will be isomorphic.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.7
\r\n", "Let ƒ be an isomorphism between a field K and a field E. Let M be a finite extension of K, and let u be \r\n", "in M. Let\r\n", "

p(x) = c0 + c1 x + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn

\r\n", "be IrrK(u, x). Define\r\n", "

h(x) = ƒ(c0) + ƒ(c1) x + \r\n", "ƒ(c2) x2 + \r\n", "ƒ(c3) x3 + ⋯ + ƒ(cn) xn

\r\n", "which is in E[x]. Suppose there is a finite extension of E for which there is a root of h(x), called v. Then there \r\n", "is an isomorphism μ from K(u) to E(v) for which μ(u) = v, and \r\n", "μ(t) = ƒ(t) for all t in K.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "The usual application of this proposition is when K and E are the same field, as in the case ℚ(cos(π⁄9)) and \r\n", "ℚ(−cos(2π⁄9)), in which \r\n", "case we not only can prove that ℚ(cos(π⁄9)) and \r\n", "ℚ(−cos(2π⁄9)) are isomorphic, but we can impose further \r\n", "conditions on the isomorphism.\r\n", "\r\n", "

\r\n", "\r\n", "COROLLARY 14.2
\r\n", "If K is a finite extension of a field F, and u and v are two elements in K such \r\n", "that IrrF(u, x) = IrrF(v, x), then there is an isomorphism μ between \r\n", "F(u) and F(v) such that μ(u) = v, and μ(t) = t for \r\n", "all t in F.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We discovered in Corollary 13.2 that every finite field could be expressed in the form Zp[x]/⟨f(x)⟩, with \r\n", "f(x) an irreducible polynomial in Zp[x]. \r\n", "It is natural to ask whether\r\n", "any finite extension of a field can be represented in the form F[x]/⟨f(x)⟩ for some polynomial f(x) in \r\n", "F[x]. Although there are some fields that are exceptions, ℚ and ℝ are not among them. \r\n", "Once we have proven this, we will be able to enter\r\n", "any finite extension of ℚ or ℝ into Sage\r\n", "using the same technique that was used for finite fields.

\r\n", "\r\n", "\r\n", "

Splitting Fields

\r\n", "\r\n", "
\r\n", "We have already seen that given an irreducible polynomial f(x) in F[x], we can construct a \r\n", "field F[x]/⟨f(x)⟩ for which f(x) has a root in this new field. This raises an interesting question: Can we \r\n", "construct a field for which f(x) factors completely in the new field? Let us demonstrate with some examples.

\r\n", "\r\n", "Let f(x) = x3 + x2 − 2 x − 1. We begin by showing that this polynomial is irreducible \r\n", "over the rationals.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "factor(x^3 + x^2 - 2*x - 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "The output is unchanged, indicating that the polynomial is irreducible.

\r\n", "\r\n", "If a is one root of this polynomial, we can define ℚ(a) in Sage as follows:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"a\")\r\n", "Define(a^3, - a^2 + 2*a + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This defines a ring with characteristic 0, and containing an element a with a3 + a2 − 2 a − 1 = \r\n", "0. We can now see if the polynomial factors over the new field.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^3 + x^2 - 2*x - 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This shows that the polynomial factors completely in the field ℚ(a). Recall that this is very similar to the case of finite fields.

\r\n", "\r\n", "EXPERIMENT:
\r\n", "Note that the polynomial x2 + 4 x + 2 does not factor over the rationals.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "factor(x^2 + 4*x + 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "See if this polynomial factors in the field ℝ(√2). First define the \r\n", "field ℝ(√2) in Sage with the commands:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"a\")\r\n", "Define(a^2, 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Now use the factor command to see if x2 + 4 x + 2 factors in this new field.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "EXAMPLE:
\r\n", "Find a field for which x3 − 2 factors completely.

\r\n", "One root is clearly a = ∛2, so we can define the field ℚ(a) by the commands:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Does the polynomial x3 − 2 factor completely in this new field? Let's try it.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^3 - 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This polynomial of course factors, since a is defined to be a root. But it did not factor completely, as the \r\n", "polynomial x3 + x2 − 2 x − 1 did. Rather, there is a quadratic term \r\n", "x2 + a x + a2. How can we get the polynomial x3 − 2 to factor completely into \r\n", "linear terms? We can define a new element, b, to be a root of x2 + a x + a2. That \r\n", "is, b2 must be −a2 − a b.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"b\")\r\n", "Define(b^2, -a^2 - a*b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can now factor x3 − 2 in the field ℚ(a, b).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^3 - 2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Notice that ℚ(a) is a 3-dimensional extension of ℚ, and ℚ(a, b) is a 2-dimensional extension of ℚ(a). Thus, \r\n", "by proposition 14.2, ℚ(a, b) is a 6-dimensional extension of ℚ, with \r\n", "basis \r\n", "

{1, a, a2, b, a b, a2 b}.

\r\n", "\r\n", "EXPERIMENT:
\r\n", "Prove that the polynomial x2 + a x + a2 = x2 + \r\n", "∛2 x + ∛4 is irreducible in the \r\n", "field ℚ(∛2) by showing that the roots are complex.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "EXAMPLE:\r\n", "A longer example of this process is the polynomial x4 − x + 1. Find a field extension for which this factors.

\r\n", "\r\n", "We can verify that this is irreducible over the rational numbers with Sage.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "factor(x^4 - x + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Let us force a to be a root of this polynomial
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"a\")\r\n", "Define(a^4, a - 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can now factor x4 − x + 1 in the field ℚ(a).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^4 - x + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This leaves a third degree polynomial still unfactored, so we will define b to be a root of this third degree polynomial.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"b\")\r\n", "Define(b^3, 1 - a^3 - a^2*b - a*b^2)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can factor x4 − x + 1 over the field ℚ(a, b).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^4 - x + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This still leaves an unfactored quadratic. Let c be a root of this quadratic.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"c\")\r\n", "Define(c^2, -a^2 - a*b - b^2 - a*c - b*c)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We now can factor x4 − x + 1 over the field ℚ(a, b, c).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^4 - x + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We finally did it! Notice that the dimension of the field ℚ(a, b, c) is 4·3·2 = 24. It is easy to see from this example \r\n", "that this procedure can be carried out over any polynomial.

\r\n", "\r\n", "EXPERIMENT:
\r\n", "Try the polynomial x5 + 20 x + 16. How many extensions does it take before this polynomial finally factors completely?
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "

\r\n", "\r\n", "LEMMA 14.4
\r\n", "Let F be a field, and let f(x) be a polynomial in F[x] of degree n whose leading coeffient \r\n", "is cn. Then there is a finite extension K of F such that\r\n", "

f(x) = cn·(x − u1)·(x − \r\n", "u2)·(x − u3) ⋯ (x − un),

\r\n", "where u1, u2, u3, … un are elements in K. Furthermore, the dimension \r\n", "of K over F is at most n!.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "DEFINITION 14.8
\r\n", "If K is a field for which the polynomial f(x) in F[x] factors as\r\n", "

f(x) = cn·(x − u1)·(x − \r\n", "u2)·(x − u3) ⋯ (x − un),

\r\n", "then the field F(u1, u2, u3, … un) is called the splitting field for \r\n", "the polynomial f(x).

\r\n", "\r\n", "For example, the splitting field of x3 + x2 − 2 x − 1 was found to be ℚ(a), where a \r\n", "is one root of the polynomial. Thus, the splitting field is a 3-dimensional extension of ℚ. The splitting field of x3 − 2 turned out to \r\n", "be a 6-dimensional extension of ℚ. The splitting field of x4 − x − 1 turned out to be a 24-dimensional extension \r\n", "of ℚ, which Lemma 14.4 points out is the largest possible dimension for a fourth degree polynomial. Finally, the experiment shows that the splitting field of x5 + 20 x + 16 is a 60 dimensional extension of ℚ.

\r\n", "\r\n", "EXAMPLE:
\r\n", "Let us look at one more example which is rather interesting.\r\n", "Find the splitting field for the polynomial x5 − 5 x + 12.

\r\n", "\r\n", "We first show that this is irreducible over the rationals:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "factor(x^5 - 5*x + 12)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Let us define a to be one root of this polynomial
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"a\")\r\n", "Define(a^5, 5*a - 12)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Can we factor x5 − 5 x + 12 in the field ℚ(a)? Let's see.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^5 - 5*x + 12)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Rather than having an irreducible forth degree equation as we did in the polynomial x5 + 20 x + 16 the polynomial factored into \r\n", "two irreducible quadratic equations. Let us define b to be a root of the second of the two quadratics, since this one is simplier.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"b\")\r\n", "Define(b^2, 1 + a/2 + a^3/2 + b - (a + a^2 + a^3 + a^4)*b/4)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Can we factor x5 − 5 x + 12 in the field ℚ(a,b)?
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^5 - 5*x + 12)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This time the polynomial factored completely. Thus, we have found a splitting field for the polynomial x5 − 5 x + 12. Note \r\n", "that ℚ(a) is a 5-dimensional extension of ℚ, and ℚ(a, b) is a 2-dimensional extension of ℚ(a). Thus, \r\n", "by Proposition 14.2, ℚ(a, b) is only a 10-dimensional extension of ℚ, far less than the maximum dimension of 120.

\r\n", "\r\n", "This example raises the obvious question: What if we let b be a root of the first irreducible polynomial instead of the second? Would we get the \r\n", "same splitting field? We would like to prove that the splitting fields are in fact uniquely determined up to isomorphism. It is actually easier to prove a stonger \r\n", "statement for which we can use the principle of induction.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.8
\r\n", "Let ϕ be an isomorphism from the field F to a field E. Let\r\n", "

f(x) = c0 + c1 x + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn

\r\n", "be a polynomial in F[x]. Then\r\n", "

g(x) = ϕ(c0) + ϕ(c1) x + \r\n", "ϕ(c2) x2 + \r\n", "ϕ(c3) x3 + ⋯ + ϕ(cn) xn

\r\n", "is a polynomial in E[x]. Suppose that K is a splitting field of f(x) over F, and L is a splitting field \r\n", "of g(x) over E. Then there is an isomorphism μ from K to L, such \r\n", "that μ(t) = ϕ(t) for all t in F.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We can use this proposition to show that indeed the splitting field of a polynomial is uniquely determined up to isomorphism.\r\n", "\r\n", "

\r\n", "\r\n", "COROLLARY 14.3
\r\n", "If f(x) is a polynomial in F[x], then all splitting fields of f(x) are isomorphic.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "In §13.3, we studied the properties of cyclotomic polynomials. It will be important later on to determine the splitting fields of these polynomials. Let us look \r\n", "at a few examples to see if we can determine a pattern.

\r\n", "\r\n", "The ninth cyclotomic polynomial Φ9(x) is given as
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "Cyclotomic(9,\"x\")" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "If we can let a be a root of x6 + x3 + 1 by the commands
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^6, - 1 - a^3)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We finally can see if the cyclotomic polynomial factors in ℚ(a).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^6 + x^3 + 1)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This shows us that the splitting field of the ninth cyclotomic polynomial is ℚ(a), which is a 6-dimensional extension of ℚ.

\r\n", "\r\n", "EXPERIMENT:
\r\n", "Find the splitting field of Φ8(x) = x4 + 1.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can generalize these observations in the following proposition.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.9
\r\n", "The splitting field of the nth cyclotomic polynomial Φn(x) has dimension ϕ(n) over ℚ, \r\n", "where ϕ(n) is Euler's totient function. In fact, the splitting field is given as ℚ(ωn), where \r\n", "ωn is a primitive nth root of unity.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "\r\n", "In all of the cases we have looked at, we found the splitting fields of irreducible polynomials. However, the polynomial does not have to be irreducible for it to have \r\n", "an interesting splitting field. Consider the polynomial x4 − 5 x2 + 6. If we factor this over the rationals,
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "var(\"x\")\r\n", "factor(x^4 - 5*x^2 + 6)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We find that this factors! In fact, we can easily see all 4 roots: ±√2 and \r\n", "±√3. If we define a to be the solution to x2 − 2 = 0, (that \r\n", "is, a = √2), we have the following \r\n", "factorization in ℚ(√2).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^2, 2)\r\n", "factor(x^4 - 5*x^2 + 6)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "The x2 − 3 remains unfactored, as we would expect. Thus, we need to define b to solve x2 − 3 = 0,
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "AddFieldVar(\"b\")\r\n", "Define(b^2, 3)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "and factor the polynomial over the field ℚ(√2, √3).
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^4 - 5*x^2 + 6)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Thus, we see that the splitting field of x4 − 5 x2 + 6 is \r\n", "ℚ(√2, √3). This field is a four dimensional extension \r\n", "of ℚ.

\r\n", "\r\n", "EXPERIMENT:
Find the splitting field of the polynomial x4 − 10 x2 + 1. What is the dimension of the splitting field \r\n", "over ℚ?
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "These examples show that the splitting fields can vary dramatically from polynomial to polynomial. In the next chapter we will see how the splitting field can be used \r\n", "to actually solve a polynomial in terms of square roots and cube roots, whenever this is possible. But first, we need to explore some interesting properties of \r\n", "splitting fields.

\r\n", "\r\n", "Let us begin by defining the splitting field of x3 − 2 as follows:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0, \"x\")\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 2)\r\n", "AddFieldVar(\"b\")\r\n", "Define(b^2, - a^2 - a*b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "We can see whether the polynomial x2 + 3 factors in the splitting field:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^2 + 3)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Thus, we see that √−3 is in the field ℚ(a, b). Let us see if another polynomial factors \r\n", "in ℚ(a, b)—x6 + 108.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "factor(x^6 + 108)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "ℚ(a, b) turns out to contain the splitting field of x6 + 108!

\r\n", " \r\n", "EXPERIMENT:
\r\n", "\r\n", "Try factoring the polynomial\r\n", "

x6 + 6 x4 − 12 x3 + 36 x2 − 36 x + 36

\r\n", "over the field ℚ(a, b), which is the splitting field of x3 − 2. This polynomial was choosen to have a \r\n", "root in ℚ(a, b). Does this polynomial factor completely in ℚ(a, b)?
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "These examples reveal a startling fact: Whenever an irreducible polynomial in ℚ[x] has just one root in a splitting field, then the polynomial \r\n", "factors completely in that splitting field. This property characterizes splitting fields from other extensions of ℚ.\r\n", "\r\n", "

\r\n", "\r\n", "LEMMA 14.5
\r\n", "Let K be the splitting field of a polynomial f(x) in F[x]. Then if p(x) is an irreducible polynomial \r\n", "in F[x] for which there is one root in K, then p(x) factors completely in K.

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "EXPERIMENT:
\r\n", "Try factoring the polynomial x6 + x3 − 6 over the field ℚ(a, b), the splitting field \r\n", "of x3 − 2. Notice that this polynomial factors, but does not factor completely in this field. Why doesn't this example \r\n", "contradict Lemma 14.5?
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "The fact that the splitting field of x6 + 108 is the same as the splitting field of x3 − 2 reveals another curious property \r\n", "of splitting fields. Rather than having to make an "extension of an extension" to define the splitting field ℚ(a, b), we could have \r\n", "defined the same field using a single extension of the element w, where \r\n", "w = 6√−108. This is done with the commands:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)\r\n", "AddFieldVar(\"w\")\r\n", "Define(w^6, -108)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "DEFINITION 14.9
\r\n", "We say that a finite extension of a field K is called a simple extension if it can be expressed as K(a) for some \r\n", "element a.

\r\n", "\r\n", "What we have observed is that the splitting field of x3 − 2, even though it was originally described as an extension of an \r\n", "extension, is in fact a simple extension of ℚ of dimension 6.

\r\n", "\r\n", "Let us show, using the splitting fields, that an extension of an extension will usually form a simple extension.\r\n", "\r\n", "

\r\n", "\r\n", "PROPOSITION 14.10
\r\n", "Let F be a field, and let K be a finite-dimensional extension of F. Suppose that K = F(u, v) with \r\n", "u, v in K. Let L be the splitting field of the polynomial g(x) = IrrF(v, x), \r\n", "and suppose that there are no multiple roots of g(x) in the field L. Then there is an element w of K such \r\n", "that K = F(w).

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "We can easily extend this result by induction to any number of extensions of ℚ.\r\n", "\r\n", "

\r\n", "\r\n", "COROLLARY 14.4
\r\n", "Let K be a finite-dimensional extension of F, with K = F(u1, u2, u3, \r\n", "… un), and suppose that none of the polynomials IrrF(ui, x) have multiple roots in each of \r\n", "their splitting fields. Then there exists an element w in K such that K = F(w).

\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "

\r\n", "Sage has a function SimpleExtension that finds one of the many elements w for which the \r\n", "field ℚ(a, b, …) = ℚ(w).

\r\n", "\r\n", "EXAMPLE:
\r\n", "\r\n", "We have seen the splitting field of x3 − 2 is \r\n", "ℚ(∛2, ω3∛2). Find a single \r\n", "element w such that the splitting field is ℚ(w).

\r\n", "\r\n", "We first set up the splitting field as\r\n", "ℚ(a, b), where a3 = 2, and b2 = −a2 − a b.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 2)\r\n", "AddFieldVar(\"b\")\r\n", "Define(b^2, -a^2 - a*b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "To find an element in this field which forms a simple extension, we compute
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "SimpleExtension(a, b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This shows that ℚ(a, b) = ℚ(a + 2 b), which is a simple extension. What element is a + 2 b? Let us \r\n", "compute its sixth power.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "expand((a + 2*b)^6)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Thus, we see that a + 2 b is one root of x6 + 108.

\r\n", "\r\n", "How does this command work? The key is in the proof of Proposition 14.10. Within the proof, we found that F(u, v) = F(u + y v), \r\n", "where y is any number such that

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
y ≠ uₕ − u
v − vₖ
\r\n", "whenever uh is a root of IrrF(u, x), and vk is a root \r\n", "of IrrF(v, x).

\r\n", "\r\n", "EXAMPLE:
\r\n", "Consider the splitting field of x4 − 5 x2 + 6, which is \r\n", "ℚ(√2, √3). Since this field is the splitting field of \r\n", "a polynomial without multiple roots, we can apply Proposition 14.10 to show \r\n", "that ℚ(√2, √3) = ℚ(w) for some \r\n", "element w. But what is that element?

\r\n", "\r\n", "Note that Irrℚ(√2, x) = x2 − 2, which has roots \r\n", "of ±√2. Likewise, Irrℚ(√3, x) = \r\n", "x2 − 3 which has roots of ±√3. Hence, we must pick a rational value of y \r\n", "that is not equal to\r\n", "

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
±√2 − √2 .
√3 ± √3
\r\n", "
\r\n", "That is, y cannot equal 0 or −√2/3. Any other rational value of y will do, so for convenience \r\n", "we can take y = 1. Then w = u + y v = √2 + \r\n", "√3.

\r\n", "\r\n", "We can also have Sage find an element w for us.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^2, 2)\r\n", "AddFieldVar(\"b\")\r\n", "Define(b^2, 3)\r\n", "SimpleExtension(a, b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "Thus Sage verifies that ℚ(√2, √3) = \r\n", "ℚ(√2 + √3).

\r\n", "\r\n", "It is curious to note that √2 + √3 is one of the roots of \r\n", "the polynomial x4 − 10 x2 + 1. Thus, x4 − 10 x2 + 1 and \r\n", "x4 − 5 x2 + 6 have the same splitting field, even though one of the polynomials factors over the rationals while the \r\n", "other doesn't.

\r\n", "\r\n", "EXAMPLE:
\r\n", "Consider ℚ(∛2, √2). This is not \r\n", "a splitting field, but it is contained in the splitting field of f(x) = (x3 − 2)(x2 − 2), which does \r\n", "not have multiple roots, so we can still apply Proposition 14.10 to show \r\n", "that ℚ(∛2, √2) = ℚ(w) for some \r\n", "element w. But what is that element?

\r\n", "\r\n", "First we define a = ∛2 and b = √2.
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "InitDomain(0)\r\n", "AddFieldVar(\"a\")\r\n", "Define(a^3, 2)\r\n", "AddFieldVar(\"b\")\r\n", "Define(b^2, 2)\r\n", "SimpleExtension(a, b)" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "This shows that ℚ(∛2, √2) = \r\n", "ℚ(∛2 + √2).

\r\n", "\r\n", "There is in fact an easier way to find a simple extension in this case. Merely not that ⁶√2 ∈ \r\n", "ℚ(∛2, √2), since \r\n", "⁶√2 = \r\n", "√2⁄∛2. Yet \r\n", "√2 = \r\n", "(⁶√2)3, and ∛2 = \r\n", "(⁶√2)2. So \r\n", "ℚ(∛2, √2) = \r\n", "ℚ(⁶√2).

\r\n", "\r\n", "The fact that we can convert an extension of an extension to a simple extension will simplify many of the proofs involving splitting fields. In particular, it will \r\n", "allow us to explore the automorphisms of the splitting fields. In the next chapter we will discover that the automorphisms of the splitting fields determine much \r\n", "of the information about the roots of the polynomial, and whether they can be expressed in terms of square roots and cube roots. This beautiful correlation is \r\n", "referred to as Galois theory.\r\n", "\r\n", "









































\r\n", "\r\n", "

Proofs:

\r\n", "\r\n", "\r\n", "\r\n", "Proof of Lemma 14.1:

\r\n", "\r\n", "If B is a basis, then every element v ∈ V can be expressed in the form c1 x1 + \r\n", "c2 x2 + c3 x3 + ⋯ + cn xn. Suppose that\r\n", "

v = a1 x1 + a2 x2 + \r\n", "a3 x3 + ⋯ + an xn

\r\n", "is another such expression. Then\r\n", "

(a1 − c1) x1 + \r\n", "(a2 − c2) x2 + (a3 − c3) x3 + ⋯ \r\n", "+ (an − cn) xn = v − v = 0.

\r\n", "But the vectors in B are linearly independent, so the only way that the combination of vectors could be 0 is \r\n", "for ai − ci = 0 for all 1 ≤ i ≤ n. Hence, ai = ci for all i, \r\n", "and the representation is unique.

\r\n", "\r\n", "On the other hand, if every v ∈ V can be uniquely represented as c1 x1 + \r\n", "c2 x2 + c3 x3 + ⋯ + cn xn, then in \r\n", "particular 0 has only one representation, namely 0 = 0 x1 + 0 x2 + 0 x3 + ⋯ + \r\n", "0 xn. Thus, the vectors in B are linearly independent, and so B is a basis.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Lemma 14.2:

\r\n", "\r\n", "Suppose that Y = {y1, y2, y3, … \r\n", "yn, yn+1} are linearly independent, so that all of these vectors are nonzero.

\r\n", "\r\n", "Our goal is to show, with a suitable rearrangement of the vectors in B, that \r\n", "

{y1, y2, … \r\n", "yk−1, yk, xk+1, … xk}

\r\n", "is a basis for every 0 ≤ k ≤ n.

\r\n", "\r\n", "If k = 0, then this set is the original set B, which is a basis. So let us use induction to assume that it is true for the previous case, that is, \r\n", "that {y1, y2, … \r\n", "yk−1, xk, xk+1, … xk} is a basis.

\r\n", "\r\n", "We then can express yk = a1 y1 + a2 y2 + ⋯ \r\n", "+ ak−1 yk−1 + ak xk + ⋯ \r\n", "+ an xn. Since the vectors in Y are linearly independent, we see that at least one \r\n", "of ak, ak+1, … an is nonzero. By rearranging the remaining elements of B, we can \r\n", "suppose that ak ≠ 0.

\r\n", "\r\n", "Then xk = ak-1(yk − a1 y1 − ⋯ \r\n", "− ak−1 yk−1 − ak+1 xk+1 \r\n", "− ⋯ − an xn).

\r\n", "\r\n", "Any element v ∈ V can be expressed as v = c1 y1 + c2 y2 \r\n", "+ ⋯ + ck−1 yk−1 + ck xk + ⋯ \r\n", "+ cn xn. By substituting for the value of xk, we see that v can be expressed as a \r\n", "linear combination of \r\n", "

{y1, y2, … \r\n", "yk−1, yk, xk+1, … xn}.

\r\n", "If this set were linearly dependent, there would be a nonzero solution to\r\n", "

c1 y1 + c2 y2 \r\n", "+ ⋯ + ck−1 yk−1 + ck yk + ⋯ \r\n", "+ cn xn = 0.

\r\n", "Then ck ≠ 0 lest there also be a nonzero solution to c1 y1 + c2 y2 \r\n", "+ ⋯ + ck−1 yk−1 + ck xk + ⋯ \r\n", "+ cn xn = 0, but we are assuming that {y1, y2, … \r\n", "yk−1, xk, xk+1, … xn} is a basis.

\r\n", "\r\n", "But substituting the value for yk gives\r\n", "

ck(a1 y1 + a2 y2 + ⋯ \r\n", "+ ak−1 yk−1 + ak xk + ⋯ \r\n", "+ an xn) + \r\n", "c1 y1 + c2 y2 \r\n", "+ ⋯ + ck−1 yk−1 + ck+1 yk+1 \r\n", "+ ⋯ + cn xn = 0.

\r\n", "This is a nonzero solution to b1 y1 + b2 y2 \r\n", "+ ⋯ + bk−1 yk−1 + bk xk + ⋯ \r\n", "+ bn xn = 0, since bk = ck ak ≠ 0. Thus, the \r\n", "set {y1, y2, … \r\n", "yk−1, yk, xk+1, … xn} is linearly independent, and hence \r\n", "is a basis of V.

\r\n", "\r\n", "Now we can use the induction to say that {y1, y2, … yn} is a basis of V, but \r\n", "then yn+1 can be expressed in terms of {y1, y2, … yn}, which shows \r\n", "that Y is in fact linearly dependent.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.1:

\r\n", "\r\n", "Suppose that n is not equal to m. By exchanging the roles of X and Y if necessary, we can assume \r\n", "that n < m. Then we can use Lemma 14.2 to show that {y1, y2, y3, … \r\n", "yn+1} is linearly dependent, hence Y is not a basis of V. So we must have n = m.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.2:

\r\n", "\r\n", "Since {y1, y2, y3, … yn} is a basis for V over E, we can write \r\n", "any element of V in the form\r\n", "

c1 y1 + c2 y2 + \r\n", "c3 y3 + ⋯ + cn yn.

\r\n", "where c1, c2, c3, … cn are in E.

\r\n", "\r\n", "Since {x1, x2, x3, … xm} is a basis of E over F, we can in turn \r\n", "write\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
c1 = a1,1 x1 + a2,1 x2 + \r\n", " a3,1 x3 + ⋯ + am,1xm,
c2 = a1,2 x1 + a2,2 x2 + \r\n", " a3,2 x3 + ⋯ + am,2xm,
c3 = a1,3 x1 + a2,3 x2 + \r\n", " a3,3 x3 + ⋯ + am,3xm,
……………
cn = a1,n x1 + a2,n x2 + \r\n", " a3,n x3 + ⋯ + am,nxm.
\r\n", "
\r\n", "where each ai,j is in F. Combining these, we see that every element of V can be expressed in the form\r\n", "

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
 a1,1 x1 y1 + a2,1 x2 y1 + \r\n", " a3,1 x3 y1 + ⋯ + am,1xm y1 +
 a1,2 x1 y2 + a2,2 x2 y2 + \r\n", " a3,2 x3 y2 + ⋯ + am,2xm y2 +
 a1,3 x1 y3 + a2,3 x2 y3 + \r\n", " a3,3 x3 y3 + ⋯ + am,3xm y3 +
  ……………
 a1,n x1 yn + \r\n", " a2,n x2 yn + \r\n", " a3,n x3 yn + ⋯ + \r\n", " am,nxm yn.
\r\n", "
\r\n", "Thus, to show that the set S is a basis of V over F, we merely have to show that these vectors are linearly independent. Let us switch to \r\n", "a summation notation for the remainder of the proof. Suppose that there is a nonzero linear combination of these vectors that produces 0, that is\r\n", "

\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
mn
∑∑ai,j xi yj = 0
 i=1  j=1 
\r\n", "\r\n", "for ai,j in F. Then we have\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
mnn⎛m⎞
0 =∑∑ai,j xi yj =∑⎜∑ai,j xi ⎟yj.
⎜⎟
⎜⎟
 i=1  j=1  j=1 ⎝ i=1 ⎠
\r\n", "
\r\n", "Since {y1, y2, y3, … yn} is a basis of V over E, the only way \r\n", "that the right hand expression could be zero is if\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
m
∑ai,j xi = 0
i=1
\r\n", "
\r\n", "for all j = 1, 2, 3, … n. Now {x1, x2, x3, … xm} is a \r\n", "basis of E over F, so the only way that each of these sums could be 0 is if ai,j = 0 for all values of i \r\n", "and j. Since all of the coefficients must be 0, the vectors in S are linearly independent, and therefore S is a basis of V over \r\n", "F of dimension m n.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.3:

\r\n", "\r\n", "Suppose that {x1, x2, x3, … xn} is a basis of K over F. We want \r\n", "to show that {ϕ(x1), ϕ(x2), ϕ(x3), … \r\n", "ϕ(xn)} is a basis of E over F. If v is in E, then ϕ(u) = v for \r\n", "some u in K. Since K is generated by the elements in the basis, we have\r\n", "

u = c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn

\r\n", "for some c1, c2, c3, … cn in F. Then\r\n", "

v = ϕ(u) = ϕ(c1) ϕ(x1) + \r\n", "ϕ(c2) ϕ(x2) + ϕ(c3) ϕ(x3) + \r\n", "⋯ + ϕ(cn) ϕ(xn) = c1 ϕ(x1) + \r\n", "c2 ϕ(x2) + c3 ϕ(x3) + ⋯ + \r\n", "cn ϕ(xn).

\r\n", "Thus, {ϕ(x1), ϕ(x2), ϕ(x3), … \r\n", "ϕ(xn)} generates the field E. Also, if\r\n", "

c1 ϕ(x1) + \r\n", "c2 ϕ(x2) + c3 ϕ(x3) + ⋯ + \r\n", "cn ϕ(xn) = 0,

\r\n", "then ϕ(c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn) = 0, which implies that\r\n", "

c1 x1 + c2 x2 + \r\n", "c3 x3 + ⋯ + cn xn = 0.

\r\n", "since K and E are isomorphic. But since {x1, x2, x3, … xn} is \r\n", "a basis for K, this can only happen if\r\n", "

c1 = c2 = c3 = ⋯ = cn = 0.

\r\n", "So {ϕ(x1), ϕ(x2), ϕ(x3), … \r\n", "ϕ(xn)} is a basis for E over F, and hence K and E have the same dimension over the \r\n", "field F.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Lemma 14.3:

\r\n", "\r\n", "First, we must show that K(S) is a subfield of E. If x and y are in K(S), y ≠ 0, then \r\n", "x and y are in each of the subfields in the collection L. Then x − y and x·y-1 are \r\n", "also in each of the subfields in this collection. Thus, x − y and x·y-1 are in K(S), and \r\n", "so K(S) is a subfield of E.

\r\n", "\r\n", "To show that K(S) is the smallest field containing both K and the elements S, note that K(S) is one of the subfields \r\n", "in the collection L. Thus, any subfield containing K and the elements of S must also contain K(S).

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.4:

\r\n", "\r\n", "From Proposition 13.1, K = F[x]/⟨f(x)⟩ is a field that contains F as a \r\n", "subfield. Let y = x + ⟨f(x)⟩ in K. If we treat f(x) as a polynomial in K[x], we find \r\n", "that f(y)= 0. Consider the set {1, y, y2, y3, … yn−1}. We \r\n", "wish to show that this set is a basis for K. That is, we wish to show that every element of K can be expressed uniquely as\r\n", "

k = a1 1 + a2 y + a3 y2 + \r\n", "⋯ + an yn−1,

\r\n", "where the a1, a2, a3, … an are in F. Any \r\n", "element k ∈ K can be expressed as k = g(x) + ⟨f(x)⟩ for some polynomial g(x) in \r\n", "F[x]. By the division algorithm theorem (12.1), there exist unique polynomials q(x) and r(x) such that\r\n", "

g(x) = f(x)·q(x) + r(x),

\r\n", "where either r(x) = 0, or the degree of r(x) is less than n. Then\r\n", "

r(x) = a1 + a2 x + a3 x2 + \r\n", "⋯ + an xn−1

\r\n", "for some a1, a2, a3, … an in F. Note that we can now write\r\n", "

k = g(x) + ⟨f(x)⟩ = r(x) + ⟨f(x)⟩ = \r\n", "a1 1 + a2 y + a3 y2 + \r\n", "⋯ + an yn−1.

\r\n", "Since r(x) is unique, k is uniquely determined as a linear combination of {1, y, y2, y3, \r\n", "… yn−1}. Thus, by Lemma 14.1, {1, y, y2, y3, \r\n", "… yn−1} is a basis.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.5:

\r\n", "\r\n", "Let y be an element in the field K. Consider the set {1, y, y2, y3, \r\n", "… yn}. Since there are n + 1 elements in this set, and K has dimension n over F, by Lemma 14.2 these \r\n", "are linearly dependent, so there is a nonzero solution to\r\n", "

a0 + a1 y + a2 y2 + \r\n", "a3 y3 + ⋯ + an yn = 0

\r\n", "with a0, a1, a2, … an in F. Thus, there is a nonzero polynomial\r\n", "

a0 + a1 x + a2 x2 + \r\n", "a3 x3 + ⋯ + an xn

\r\n", "in F[x] for which y is a root when treated as a polynomial in K[x].

\r\n", "\r\n", "Let us now show uniqueness. Let f(x) be a polynomial of lowest possible degree in F[x] such that f(y) = 0. Since \r\n", "F is a field, we can divide this polynomial by its leading coefficient to obtain a polynomial with a leading coefficient of 1. Now, if there were two \r\n", "such polynomials, f(x) and g(x), then by the division algorithm theorem (12.1), there exist polynomials q(x) and \r\n", "r(x) such that\r\n", "

f(x) = g(x)·q(x) + r(x),

\r\n", "where either r(x) = 0 or the degree of r(x) is strictly less than the degree of g(x). But note that\r\n", "

0 = f(y) = g(y)·q(y) + r(y) = \r\n", "0 + r(y) = 0.

\r\n", "Thus, y is a root of the polynomial r(x). But the degree of f(x) and g(x) was chosen to be \r\n", "minimal. So r(x) = 0, and f(x) is a multiple of g(x). Finally, since both f(x) and \r\n", "g(x) have the same degree and have the same leading term of 1, we have f(x) = g(x). Therefore, there is a \r\n", "unique polynomial in F[x] of minimal degree and leading coefficient of 1 such that f(y) = 0.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.6:

\r\n", "\r\n", "Let us consider the evaluation homomorphism ϕw that maps polynomials in F[x] to elements in F(w):\r\n", "

ϕw(g(x)) = g(w).

\r\n", "By Proposition 12.1, ϕw is a ring homomorphism. The image of this homomorphism contains both F and w, and since \r\n", "F(w) is the smallest field containing both F and w, the image is all of F(w). The kernel \r\n", "of ϕw is the set of polynomials in F[x] that have w as a root. But f(x) is an irreducible \r\n", "polynomial in F[x] containing w as a root. Thus, any polynomial in the kernel is a multiple of f(x). Thus, the kernel \r\n", "of ϕw is ⟨f(x)⟩. Finally, by the first ring isomorphism theorem (10.2), we have \r\n", "that F(w) ≈ F[x]/⟨f(x)⟩.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Corollary 14.1:

\r\n", "\r\n", "By Proposition 14.5, f(x) = IrrF(u, x) exists. By Proposition 14.6, F(u) is isomorphic to the \r\n", "field F[x]/⟨f(x)⟩. By Proposition 14.4, F[x]/⟨f(x)⟩ has dimension n over \r\n", "F. Finally, by Corollary 14.3, two isomorphic extensions of F must have the same dimension over F provided that the isomorphism fixes the \r\n", "elements of F, which the isomorphism in Proposition 14.6 clearly does. Thus, the dimension of F(u) over F is n.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.7:

\r\n", "\r\n", "By Lemma 13.3, we can extend ƒ to a isomorphism from K[x] to E[x]. By Proposition 12.1, ϕv is \r\n", "a ring homomorphism from E[x] to E(v). We can combine these homomorphisms to produce the homomorphism\r\n", "

ϕv·ƒ : K[x] → E[x] → E(v).

\r\n", "Since the isomorphism in Lemma 13.3 sends x to x, we have that (ϕv·ƒ)(x) = \r\n", "ϕv(ƒ(x)) = ϕv(x) = v. So v is in the image of this combination \r\n", "of homomorphisms, as well as the subfield E. Thus, the image of ϕv·ƒ is E(v). The kernal \r\n", "of ϕv is the set of polynomials in E[x] with v as a root. But h(x) is an irreducible polynomial \r\n", "in E[x] for which h(v) = 0. Thus, the kernel of ϕv is the ideal ⟨h(x)⟩. Since \r\n", "h(x) = ƒ(p(x)), we have that the kernel of ƒ·ϕv \r\n", "is ⟨p(x)⟩. Thus, by the first ring isomorphism theorem (10.2),\r\n", "

K[x]/⟨p(x)⟩ ≈ E(v).

\r\n", "By Proposition 14.6, we also have\r\n", "

K(u) ≈ K[x]/⟨p(x)⟩,

\r\n", "and in this isomorphism, u mapped to the coset x + ⟨p(x)⟩. If we let μ be the combination of these two isomorphisms,\r\n", "

μ : K(u) → K[x]/⟨p(x)⟩ → E(v)

\r\n", "then μ(u) = ϕv(ƒ(x)) = v, and μ(t) = ƒ(t) for \r\n", "all t in K.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Corollary 14.2:

\r\n", "\r\n", "We simply let ƒ be the identity mapping from F to itself, and use Proposition 14.7. Then p(x) and h(x) are \r\n", "both equal to IrrF(u, x). Since v is another root of h(x), the conclusion follows from the \r\n", "conclusion of Proposition 14.7.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Lemma 14.4:

\r\n", "\r\n", "The proof is by induction on n. If n = 1, then f(x) is a linear function, so its only root is \r\n", "in F. Thus K = F, and the degree of K over F is 1 = 1!.

\r\n", " \r\n", "Suppose that this is true for polynomials of degree less than n. Let p(x) be an irreducible factor of f(x), and consider \r\n", "the field E = F[x]/⟨p(x)⟩. By Proposition 14.4, E is a finite extension of F whose dimension over \r\n", "F is the degree of p(x), which is at most n. Then un = x + ⟨p(x)⟩ is a root \r\n", "of p(x) in the field E, and since p(x) is a factor of f(x), (x − un) is \r\n", "a factor of f(x) in the field E. Thus, we can write f(x) = \r\n", "g(x)·(x − un) for some g(x) in E[x]. Note that g(x) \r\n", "has degree (n − 1), and has the same leading coefficient as f(x). Thus, we can use the induction hypothesis to show that \r\n", "there is a \r\n", "field K that is a finite extension of E with dimension at most (n − 1)! such that g(x) factors completely as\r\n", "

g(x) = cn·(x − u1)·(x − \r\n", "u2)·(x − u3) ⋯ (x − un−1).

\r\n", "Thus,\r\n", "

f(x) = cn·(x − u1)·(x − \r\n", "u2)·(x − u3) ⋯ (x − \r\n", "un−1)·(x − un).

\r\n", "By Proposition 14.2, the dimension of K over F is the product of the dimension of E over F times the dimension of K \r\n", "over E. Thus, the dimension of K over F is at most n·(n − 1)! = n!.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.8:

\r\n", "\r\n", "If f(x) has degree 1, then the roots of f(x) are in F, and the roots of g(x) are in E. Thus, \r\n", "K = E, and L = F, and so the function μ(t) = ϕ(t) satisfies the necessary \r\n", "conditions.

\r\n", "\r\n", "Let us use induction on the degree of the polynomial f(x). That is, we will assume that the proposition is true for all polynomials of \r\n", "degree (n − 1). By Lemma 13.3, the isomorphism ϕ extends to an isomorphism from F[x] to E[x] in such a \r\n", "way that ϕ(x) = x. Thus, if p(x) is an irreducible factor of the polynomial f(x), then \r\n", "ϕ(p(x)) is an irreducible factor of the polynomial g(x) = ϕ(f(x)). Note that every root \r\n", "of p(x) is also a root of f(x), so that p(x) factors completely in the field K. Likewise, \r\n", "ϕ(p(x)) factors completely in the field L.

\r\n", "\r\n", "Let u be a root of p(x) in K, and let v be a root of ϕ(p(x)) in L. By \r\n", "Proposition 14.7, there is an isomorphism θ mapping F(u) to E(v), such \r\n", "that θ(u) = v, and θ(t) = ϕ(t) for all t in F.

\r\n", "\r\n", "Since u is a root of f(x), we can write f(x) = (x − u)·h(x), with \r\n", "h(x) in F(u)[x]. Then\r\n", "

g(x) = ϕ(f(x)) = θ(f(x)) = \r\n", "θ(x − u)·θ(h(x)) = \r\n", "(x − v)·θ(h(x)).

\r\n", "Since h(x) has degree (n − 1), we can use the induction hypothesis. Obviously K is the splitting field \r\n", "of h(x) over F(u), and L is the splitting field of θ(h(x)) over E(v). Thus, \r\n", "by the induction hypothesis the proposition is true for the polynomial h(x), so there is an isomorphism μ such \r\n", "that μ(t) = θ(t) for all t in F(u). Since θ(t) = \r\n", "ϕ(t) for all t in F, we have found an isomorphism with the necessary properties.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Corollary 14.3:

\r\n", "\r\n", "Simply let F = E, and let ϕ(t) = t for all t in F. Then by Proposition 14.8, any two splitting fields \r\n", "of f(x) = g(x) will be isomorphic.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.9:

\r\n", "\r\n", "From the definition of the splitting field, the generator\r\n", "

ωn = e(2π i/n) = \r\n", "cos(2π⁄n) + \r\n", "i sin(2π⁄n)

\r\n", "is a root of the nth cyclotomic polynomial\r\n", "

Φn(x) = \r\n", "(x − ωnk¹)·(x − \r\n", "ωnk²)·(x − \r\n", "ωnk³) ⋯ (x − \r\n", "ωnkⁱ),

\r\n", "where k1, k2, k3, … ki are the integers from 1 to n that are coprime \r\n", "to n. Thus, the splitting field contains ℚ(ωn). Note that all powers of ωn are in this field, and so \r\n", "the nth cyclotomic polynomial factors completely in ℚ(ωn).

\r\n", "\r\n", "Since Φn(x) is irreducible by Gauss' theorem on cyclotomic polynomials (13.2), and the dimension \r\n", "of Φn(x) is ϕ(n), the dimension of the splitting field ℚ(ωn) is ϕ(n).

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Lemma 14.5:

\r\n", "\r\n", "Let u1, u2, u3, … un be the roots of f(x) in \r\n", "K. Then\r\n", "

K = F(u1, u2, u3, … un).

\r\n", "Suppose that p(x) has one root v in K. Consider p(x) as a polynomial in K, and let L be the \r\n", "splitting field of p(x) over K. Let w be any other root of p(x) in L besides v. To show \r\n", "that K = L, we need to show that w is in K, which would show that all roots of p(x) are in K.

\r\n", "\r\n", "By Proposition 14.7, there is an isomorphism ϕ from F(v) to F(w) such that ϕ(v) = w, \r\n", "and ϕ(t) = t for all t in F. (We let ƒ(t) = t, the identity map, and let E and \r\n", "K both be the field F.) By Lemma 13.3 we can extend ϕ to an isomorphism from F(v)[x] to \r\n", "F(w)[x], and ϕ(f(x)) = f(x).

\r\n", "\r\n", "We now want to consider the field K(w). We have\r\n", "

K(w) = F(u1, u2, u3, … \r\n", "un, w) = F(w, u1, u2, u3, … \r\n", "un).

\r\n", "Thus, K(w) is the splitting field of f(x) over the field F(w). Since v is in K,\r\n", "

K = K(v) = F(u1, u2, u3, … \r\n", "un, v) = F(v, u1, u2, u3, … \r\n", "un).

\r\n", "so K is the splitting field of f(x) over the field F(v).

\r\n", "\r\n", "Consequently Proposition 14.8 shows us that the isomorphism ϕ from F(v) to F(w) extends to an \r\n", "isomorphism μ from K to K(w), and μ(v) = w. Also, μ(t) = t for \r\n", "all t in F. Thus, we can use Corollary 14.3 to show that K and K(w) have the same dimension over F. By \r\n", "Proposition 14.2, the dimension of K(w) over F equals the dimension of K(w) over K times the dimension \r\n", "of K over F. Therefore, the dimension of K(w) over K must be 1, so w is in K. Therefore, every root \r\n", "of p(x) is in K, so p(x) factors completely in K.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Proposition 14.10:

\r\n", "\r\n", "If F is a finite field, then K will also be a finite field, and we can simply let w be a generator of the multiplicative \r\n", "group K*, using Proposition 13.4. Thus, we will assume that F is an infinite field. Let f(x) = \r\n", "IrrF(u, x) and g(x) = IrrF(v, x). Let E be the splitting field \r\n", "of g(x) over the field F(u). Since g(x) factors completely in L without double \r\n", "roots, g(x) will also factor completely in E without double roots. Let \r\n", "

v = v1, v2, v3, … vn

\r\n", "be the distinct roots of g(x) in E.

\r\n", "\r\n", "Since u is in E, there is at least one root of f(x) in the field E. Even though f(x) may not factor \r\n", "completely in the field E, we can let \r\n", "

u = u1, u2, u3, … um

\r\n", "be the roots of f(x) over E.

\r\n", "\r\n", "Since F is an infinite field, we can pick some element y of F, such that\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
y ≠ uₕ − u for all 1 ≤ h ≤ m, 1 ≤ k ≤ n.
v − vₖ
\r\n", "
\r\n", "Finally, we let w = u + y v. Let us show that K = F(w). To show that v is in F(w), \r\n", "let p(x) = f(w − y x), and note that p(v) = \r\n", "f(u + y v − y v) = f(u) = 0, so v is a root of p(x). If one of the other roots \r\n", "of g(x) is a root of p(x), then\r\n", "

w − y vk = uh for some h and k,

\r\n", "so u + y v − y vk = uh, giving us\r\n", "\r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", " \r\n", "
y = uₕ − u,
v − vₖ
\r\n", "
\r\n", "and we specifically chose y so that it would avoid these values. Thus, there is only one root in common between g(x) and p(x) \r\n", "in the field E.

\r\n", "\r\n", "Let r(x) = IrrF(w)(v, x). Then r(x) divides the polynomials g(x) \r\n", "and p(x), since both polynomials have v as a root. In fact, we have seen that g(x) and p(x) have no \r\n", "other roots in common, so r(x) has only one root in the field E. But g(x) splits completely in E, and has no \r\n", "multiple roots in E. Thus, r(x) has degree 1, and in fact r(x) = x − v. This proves that v is \r\n", "in F(w). To see that u is in F(w), we note that u = y v − w. Thus, \r\n", "F(u, v) is contained in F(w), while F(w) is obviously contained \r\n", "in F(u, v). Therefore, F(u, v) = F(w).

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "Proof of Corollary 14.4:

\r\n", "\r\n", "We will proceed by induction on n. If n = 1, we can let w = u1, and there is nothing to prove. If n = 2, we can \r\n", "use Proposition 14.10 to find w. Suppose that the corollary is true for the previous case, so that we found a u in K such that \r\n", "

F(u) = F(u1, u2, u3, \r\n", "… un−1).

\r\n", "Let v = un, and since g(x) = \r\n", "IrrF(un, x) does not have multiple roots in its splitting field L, we can use Proposition 14.10 to find \r\n", "a w in K such that F(w) = F(u, v). But then \r\n", "

F(w) = F(u1, u2, u3, \r\n", "… un−1, un).

\r\n", "Thus, the corollary is true for all positive values of n.

\r\n", "\r\n", "Return to text\r\n", "\r\n", "









































\r\n", "\r\n", "\r\n", "\r\n", "

Sage Interactive Problems

\r\n", "\r\n", "
\r\n", "§14.1 #20)
\r\n", "Use Sage to find the coefficients of the vector ⟨3,-2,5⟩ in ℝ3 using the \r\n", "basis {⟨2,-1,4⟩,⟨5,2,1⟩,⟨4,-3,2⟩}.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.1 #21)
\r\n", "Use Sage to find the coefficients of the element a5 in GF(27) over Z3 using the basis \r\n", "{1, a, a2}. Here, a is a root of the Conway polynomial of degree 3 over Z3, which is x3 + 2 x + 1.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.2 #19)
\r\n", "Define the field ℚ(√−3) in Sage, then \r\n", "find 1/(5 + √−3).
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.2 #20)
\r\n", "Define the field ℚ(√5) in Sage. Does the \r\n", "polynomial x2 + 4 x − 1 factor in this field?
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "

\r\n", "For Problems 17 through 20: Define the splitting field of the polynomial in Sage. Determine the dimension of the splitting field over ℚ.

\r\n", "\r\n", "§14.3 #17) x3 + x2 − 4 x + 1.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.3 #18) x5 + x4 − 4 x3 − 3 x2 + 3 x + 1.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.3 #20) x5 − 2.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "
\r\n", "§14.3 #19) x5 + 3 x3 + 5 x + 10.
" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "\r\n", "\r\n", "\r\n", "\r\n", "" ] } ], "metadata": { "kernelspec": { "display_name": "SageMath (stable)", "name": "sagemath" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.14" } }, "nbformat": 4, "nbformat_minor": 0 }