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Abstract Algebra: An Interactive Approach, 2e

©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.

Chapter 14
The Theory of Fields

Initialization: This cell MUST be evaluated first:

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    load('absalgtext.sage')
except IOError:
    load('/media/sf_sage/absalgtext.sage')

Vector Spaces
Extension Fields
Splitting Fields
Sage Interactive Problems

Vector Spaces

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 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 for arbitrary fields as for real numbers.

DEFINITION 14.1
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 is defined a multiplication a·v for all a ∈ F and v ∈ V such that:

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

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

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

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

EXAMPLE:
Let K be a field, and F any subfield of K. Then K is a vector space over F, defining a·v as a 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 from the distributive and associative properties of K.

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 remainder of the chapter. The next definition is the key to understanding the properties of a vector space.

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

c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n = 0.

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

c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n = 0.

is for c₁ = c₂ = ⋯ = c_n = 0.

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

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

a₁ x₁ + a₂ x₂ + a₃ x₃ + ⋯ + a_n x_n,

with a₁, a₂, a₃, … a_n in F.

Here are some examples, all of which are fairly routine to check:

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

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

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

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 basis {x₁, x₂, x₃, … x_n} in ℝ for which every real number could be expressed as

a₁ x₁ + a₂ x₂ + a₃ x₃ + ⋯ + a_n x_n,

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

There is an easy way to determine if a particular set of vectors is a basis.

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

v = c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n.

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

Click here for the proof.

We can define a basis in Sage with the command ToBasis.

B = ToBasis([[1, 4, -1], [2, -3, 1], [4, 5, -1]])

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 vectors ⟨1, 4, -1⟩, ⟨2, -3, 1⟩, and ⟨4, 5, -1⟩ were linearly dependent. Here is another attempt.

B = ToBasis([[2, 0, 1], [0, 0, 3], [1, 4, 0]])

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 coefficients c₁, c₂, … c_n for any element of the vector space.

Coefficients(B, [2, 3, 4])

This shows that

⟨2, 3, 4⟩ = ⁵⁄₈⟨2, 0, 8⟩ + ⁹⁄₈⟨0, 0, 3⟩ + ³⁄₄⟨1, 4, 0⟩.

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

Click here for the proof.

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

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

Click here for the proof.

This proposition allows us to make the following definition.

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

Here are some examples:

ℝ³ is a 3-dimensional vector space over ℝ.

ℂ is a 2-dimensional vector space over ℝ.

The quadternions ℍ is a 4-dimensional vector space over ℝ.

ℝ is an infinite-dimensional vector space over ℚ.

EXAMPLE:
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.

The Conway polynomial of degree 2 over Z₃ is:

ConwayPolynomial(3, 2)

so if we let a be a root of this polynomial, we can define GF(9) with the commands:

InitDomain(3)
AddFieldVar("a")
Define(a^2, a + 1)
ListField()

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 in Z₃. Thus, one possible basis would be the set

B = ToBasis([1, a])

We can now express every element of GF(9) in terms of this basis. Thus,

Coefficients(B, a^3)

shows that a³ = 1 + 2 a. We have shown that GF(9) is a 2-dimensional vector space over Z₃.

EXAMPLE:
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)? If so, find a basis.

The Conway polynomial of degree 4 over Z₃ is:

ConwayPolynomial(3, 4)

So we could define GF(81) by telling Sage that the generator q raised to the 4^th power is q³ + 1.

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 of GF(81) will satisfy. Let us factor the above Conway polynomial in the field GF(9).

AddFieldVar("x")

factor(x^4 + 2*x^3 + 2)

This gives us two possible polynomials. But we also have the compatibility condition, which says that the power of the new generator b will be the standard generator a of GF(9). Hence, we can also factor

factor(x^10 - a)

The only factor that is in both factorizations is x² + (a + 2)x + a. So we know that b satisfies this polynomial, so we can define b² to be (1 + 2a)b + 2a.

AddFieldVar("b")

Define(b^2, (1 + 2*a)*b + 2*a)

We can now define a basis of GF(81) over GF(9).

B = ToBasis([1, a],[b, 2])

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 is GF(9). We can now find components for elements in GF(81).

Coefficients(B, b^2)

This tells us that b² = (2a + 1)b + 2a. We can check this in Sage.

b^2

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.

PROPOSITION 14.2
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 dimension n, then V is a vector space of F of dimension m n. Futhermore, if {x₁, x₂, x₃, … x_m} is a basis of E over F, and {y₁, y₂, y₃, … y_n} is a basis of V over E, then the set

S = {	x1 y1, x2 y1, x3 y1, … xm y1,
	x1 y2, x2 y2, x3 y2, … xm y2,
	x1 y3, x2 y3, x3 y3, … xm y3,
	……………
	x1 yn, x2 yn, x3 yn, … xm yn}

is a basis of V over F.

Click here for the proof.

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 section.

Extension Fields

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.

DEFINITION 14.5
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 a finite extension of F. We say the degree, or dimension of the extension is the size of a basis

{x₁, x₂, x₃, … x_n}

of K over F.

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

Likewise, we can define GF(8) as follows:

InitDomain(2)
AddFieldVar("a")
Define(a^3, 1 + a)
F = ListField()
F

This time, we see that every element can be expressed in the form

x + a y + a² z

where x, y and z are in Z₂. Hence, {1, a, a²} is a basis of GF(8) over Z₂, so GF(8) is a 3-dimensional extension of Z₂.

B = ToBasis([1,a,a^2])

Coefficients(B, a^4)

Here are some other examples:

ℂ is a 2-dimensional extension of ℝ.

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

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 the isomorphism ϕ maps the base field F to itself.

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

Click here for the proof.

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 definition for the smallest field containing both F and x.

DEFINITION 14.6
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 the collection of all subfields of E that contain the field K, along with the set S. Then we define

K(S) =	∩	H.
	H ∈ L

That is, K(S) is the intersection of all subfields of E that contain both K and S. If S = {a₁, a₂, a₃, … a_n}, we will write K(a₁, a₂, a₃, … a_n) for K(S). Thus, if S consists of a single element a, we can write K(a) for K(S).

It is not difficult to see that this definition produces exactly what we want.

LEMMA 14.3
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 contains both K and the elements S.

Click here for the proof.

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

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 very similar to that of defining a finite field. We begin by finding an irreducible polynomial f(x) in the field F, and creating the field K = F[x]/⟨f(x)⟩.

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

Click here for the proof.

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

Since the characteristic of ℚ is 0, we begin the definition by the command

InitDomain(0)

Next, we let a new variable a be a root to the equation x³ − 2. That is, we define a³ to be 2.

AddFieldVar("a")
Define(a^3, 2)

That's all there is to it! The basis of this extension field is {1, a, a²}. We can verify this with Sage.

B = ToBasis([1, a, a^2])

We can also do divisions in this field.

1/(a + a^2)

EXPERIMENT:
Define the field ℚ(√2) in Sage using similar comands. What is the basis of the field? Then have Sage simplify the expression

2 a − 3	where a = √2.
5 a + 2

Although this demonstrates that any extension field of the form F[x]/⟨f(x)⟩ can be entered into Sage, we would like to show 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 isomorphic to F[x]/⟨f(x)⟩ for some polynomial f(x).

PROPOSITION 14.5
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 polynomial f(x) in F[x] of degree at most n such that f(y) = 0. That is, when f(x) is treated as a polynomial in K[x], y is a root of f(x). Furthermore, there is a unique polynomial of lowest degree that satisifies these conditions and for which the leading coefficient is equal to 1.

Click here for the proof.

The irreducible polynomial determined by this proposition is important enough to warrent a special notation.

DEFINITION 14.7
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 Proposition 14.5 that has a leading coeffient of 1 to be the irreducible polynomial of a over F, denoted

Irr_F(a, x).

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

Irr_ℝ(√2, x) = x − √2.

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

Irr_ℝ(i, x) = x² + 1.

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 − ¹⁄₂. However, we want the leading coefficient of the polynomial to be 1, so we write

Irr_ℚ(cos(^π⁄₉), x) = x³ − ³⁄₄ x − ¹⁄₈.

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 enter the field ℚ(cos(^π⁄₉)) into Sage. If we let a = cos(^π⁄₉), we see that a satisfies x³ − ³⁄₄ x − ¹⁄₈ = 0. If we put all terms besides the leading term on the other side of the equation, we find that a satisfies the equation

x³ = ³⁄₄ x + ¹⁄₈.

That is,

a³ = ³⁄₄ a + ¹⁄₈.

Let us enter this fact into Sage.

InitDomain(0)
AddFieldVar("a")
Define(a^3, 3*a/4 + 1/8)

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 equation. That's all there is to it! We can perform operations of this field such as

(2+a)*(3-2*a)

a^5

1/a

In this field, the a represents the number cos(^π⁄₉) = 0.9396926... . Have we really defined the field ℚ(cos(^π⁄₉))? Actually, we have defined the field ℚ[x]/⟨x³ − ³⁄₄ x − ¹⁄₈⟩ in Sage, but we can prove that these two fields are isomorphic.

PROPOSITION 14.6
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 larger field K. Then

F(w) ≈ F[x]/⟨f(x)⟩.

Click here for the proof.

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

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

Click here for the proof.

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

a³ = ³⁄₄ a + ¹⁄₈.

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

The answer is of course that these fields are both isomorphic to ℚ(cos(^π⁄₉)), so Sage doesn't need to know the exact value 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 correspond, then the two field extensions will be isomorphic.

PROPOSITION 14.7
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 in M. Let

p(x) = c₀ + c₁ x + c₂ x² + c₃ x³ + ⋯ + c_n xⁿ

be Irr_K(u, x). Define

h(x) = ƒ(c₀) + ƒ(c₁) x + ƒ(c₂) x² + ƒ(c₃) x³ + ⋯ + ƒ(c_n) xⁿ

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 is an isomorphism μ from K(u) to E(v) for which μ(u) = v, and μ(t) = ƒ(t) for all t in K.

Click here for the proof.

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

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

Click here for the proof.

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

Splitting Fields

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

Let f(x) = x³ + x² − 2 x − 1. We begin by showing that this polynomial is irreducible over the rationals.

InitDomain(0, "x")
factor(x^3 + x^2 - 2*x - 1)

The output is unchanged, indicating that the polynomial is irreducible.

If a is one root of this polynomial, we can define ℚ(a) in Sage as follows:

AddFieldVar("a")
Define(a^3, - a^2 + 2*a + 1)

This defines a ring with characteristic 0, and containing an element a with a³ + a² − 2 a − 1 = 0. We can now see if the polynomial factors over the new field.

factor(x^3 + x^2 - 2*x - 1)

This shows that the polynomial factors completely in the field ℚ(a). Recall that this is very similar to the case of finite fields.

EXPERIMENT:
Note that the polynomial x² + 4 x + 2 does not factor over the rationals.

InitDomain(0, "x")
factor(x^2 + 4*x + 2)

See if this polynomial factors in the field ℝ(√2). First define the field ℝ(√2) in Sage with the commands:

AddFieldVar("a")
Define(a^2, 2)

Now use the factor command to see if x² + 4 x + 2 factors in this new field.

EXAMPLE:
Find a field for which x³ − 2 factors completely.

One root is clearly a = ∛2, so we can define the field ℚ(a) by the commands:

InitDomain(0, "x")
AddFieldVar("a")
Define(a^3, 2)

Does the polynomial x³ − 2 factor completely in this new field? Let's try it.

factor(x^3 - 2)

This polynomial of course factors, since a is defined to be a root. But it did not factor completely, as the polynomial x³ + x² − 2 x − 1 did. Rather, there is a quadratic term x² + a x + a². How can we get the polynomial x³ − 2 to factor completely into linear terms? We can define a new element, b, to be a root of x² + a x + a². That is, b² must be −a² − a b.

AddFieldVar("b")
Define(b^2, -a^2 - a*b)

We can now factor x³ − 2 in the field ℚ(a, b).

factor(x^3 - 2)

Notice that ℚ(a) is a 3-dimensional extension of ℚ, and ℚ(a, b) is a 2-dimensional extension of ℚ(a). Thus, by proposition 14.2, ℚ(a, b) is a 6-dimensional extension of ℚ, with basis

{1, a, a², b, a b, a² b}.

EXPERIMENT:
Prove that the polynomial x² + a x + a² = x² + ∛2 x + ∛4 is irreducible in the field ℚ(∛2) by showing that the roots are complex.

EXAMPLE: A longer example of this process is the polynomial x⁴ − x + 1. Find a field extension for which this factors.

We can verify that this is irreducible over the rational numbers with Sage.

InitDomain(0, "x")
factor(x^4 - x + 1)

Let us force a to be a root of this polynomial

AddFieldVar("a")
Define(a^4, a - 1)

We can now factor x⁴ − x + 1 in the field ℚ(a).

factor(x^4 - x + 1)

This leaves a third degree polynomial still unfactored, so we will define b to be a root of this third degree polynomial.

AddFieldVar("b")
Define(b^3, 1 - a^3 - a^2*b - a*b^2)

We can factor x⁴ − x + 1 over the field ℚ(a, b).

factor(x^4 - x + 1)

This still leaves an unfactored quadratic. Let c be a root of this quadratic.

AddFieldVar("c")
Define(c^2, -a^2 - a*b - b^2 - a*c - b*c)

We now can factor x⁴ − x + 1 over the field ℚ(a, b, c).

factor(x^4 - x + 1)

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 that this procedure can be carried out over any polynomial.

EXPERIMENT:
Try the polynomial x⁵ + 20 x + 16. How many extensions does it take before this polynomial finally factors completely?

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

f(x) = c_n·(x − u₁)·(x − u₂)·(x − u₃) ⋯ (x − u_n),

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

Click here for the proof.

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

f(x) = c_n·(x − u₁)·(x − u₂)·(x − u₃) ⋯ (x − u_n),

then the field F(u₁, u₂, u₃, … u_n) is called the splitting field for the polynomial f(x).

For example, the splitting field of x³ + x² − 2 x − 1 was found to be ℚ(a), where a 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 be a 6-dimensional extension of ℚ. The splitting field of x⁴ − x − 1 turned out to be a 24-dimensional extension 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.

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

We first show that this is irreducible over the rationals:

InitDomain(0, "x")
factor(x^5 - 5*x + 12)

Let us define a to be one root of this polynomial

AddFieldVar("a")
Define(a^5, 5*a - 12)

Can we factor x⁵ − 5 x + 12 in the field ℚ(a)? Let's see.

factor(x^5 - 5*x + 12)

Rather than having an irreducible forth degree equation as we did in the polynomial x⁵ + 20 x + 16 the polynomial factored into two irreducible quadratic equations. Let us define b to be a root of the second of the two quadratics, since this one is simplier.

AddFieldVar("b")
Define(b^2, 1 + a/2 + a^3/2 + b - (a + a^2 + a^3 + a^4)*b/4)

Can we factor x⁵ − 5 x + 12 in the field ℚ(a,b)?

factor(x^5 - 5*x + 12)

This time the polynomial factored completely. Thus, we have found a splitting field for the polynomial x⁵ − 5 x + 12. Note that ℚ(a) is a 5-dimensional extension of ℚ, and ℚ(a, b) is a 2-dimensional extension of ℚ(a). Thus, by Proposition 14.2, ℚ(a, b) is only a 10-dimensional extension of ℚ, far less than the maximum dimension of 120.

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 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 statement for which we can use the principle of induction.

PROPOSITION 14.8
Let ϕ be an isomorphism from the field F to a field E. Let

f(x) = c₀ + c₁ x + c₂ x² + c₃ x³ + ⋯ + c_n xⁿ

be a polynomial in F[x]. Then

g(x) = ϕ(c₀) + ϕ(c₁) x + ϕ(c₂) x² + ϕ(c₃) x³ + ⋯ + ϕ(c_n) xⁿ

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

Click here for the proof.

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

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

Click here for the proof.

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 at a few examples to see if we can determine a pattern.

The ninth cyclotomic polynomial Φ₉(x) is given as

Cyclotomic(9,"x")

If we can let a be a root of x⁶ + x³ + 1 by the commands

InitDomain(0, "x")
AddFieldVar("a")
Define(a^6, - 1 - a^3)

We finally can see if the cyclotomic polynomial factors in ℚ(a).

factor(x^6 + x^3 + 1)

This shows us that the splitting field of the ninth cyclotomic polynomial is ℚ(a), which is a 6-dimensional extension of ℚ.

EXPERIMENT:
Find the splitting field of Φ₈(x) = x⁴ + 1.

We can generalize these observations in the following proposition.

PROPOSITION 14.9
The splitting field of the n^th cyclotomic polynomial Φ_n(x) has dimension ϕ(n) over ℚ, where ϕ(n) is Euler's totient function. In fact, the splitting field is given as ℚ(ω_n), where ω_n is a primitive n^th root of unity.

Click here for the proof.

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 an interesting splitting field. Consider the polynomial x⁴ − 5 x² + 6. If we factor this over the rationals,

var("x")
factor(x^4 - 5*x^2 + 6)

We find that this factors! In fact, we can easily see all 4 roots: ±√2 and ±√3. If we define a to be the solution to x² − 2 = 0, (that is, a = √2), we have the following factorization in ℚ(√2).

InitDomain(0, "x")
AddFieldVar("a")
Define(a^2, 2)
factor(x^4 - 5*x^2 + 6)

The x² − 3 remains unfactored, as we would expect. Thus, we need to define b to solve x² − 3 = 0,

AddFieldVar("b")
Define(b^2, 3)

and factor the polynomial over the field ℚ(√2, √3).

factor(x^4 - 5*x^2 + 6)

Thus, we see that the splitting field of x⁴ − 5 x² + 6 is ℚ(√2, √3). This field is a four dimensional extension of ℚ.

EXPERIMENT:
Find the splitting field of the polynomial x⁴ − 10 x² + 1. What is the dimension of the splitting field over ℚ?

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 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 splitting fields.

Let us begin by defining the splitting field of x³ − 2 as follows:

InitDomain(0, "x")
AddFieldVar("a")
Define(a^3, 2)
AddFieldVar("b")
Define(b^2, - a^2 - a*b)

We can see whether the polynomial x² + 3 factors in the splitting field:

factor(x^2 + 3)

Thus, we see that √−3 is in the field ℚ(a, b). Let us see if another polynomial factors in ℚ(a, b)—x⁶ + 108.

factor(x^6 + 108)

ℚ(a, b) turns out to contain the splitting field of x⁶ + 108!

EXPERIMENT:

Try factoring the polynomial

x⁶ + 6 x⁴ − 12 x³ + 36 x² − 36 x + 36

over the field ℚ(a, b), which is the splitting field of x³ − 2. This polynomial was choosen to have a root in ℚ(a, b). Does this polynomial factor completely in ℚ(a, b)?

These examples reveal a startling fact: Whenever an irreducible polynomial in ℚ[x] has just one root in a splitting field, then the polynomial factors completely in that splitting field. This property characterizes splitting fields from other extensions of ℚ.

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

Click here for the proof.

EXPERIMENT:
Try factoring the polynomial x⁶ + x³ − 6 over the field ℚ(a, b), the splitting field of x³ − 2. Notice that this polynomial factors, but does not factor completely in this field. Why doesn't this example contradict Lemma 14.5?

The fact that the splitting field of x⁶ + 108 is the same as the splitting field of x³ − 2 reveals another curious property of splitting fields. Rather than having to make an "extension of an extension" to define the splitting field ℚ(a, b), we could have defined the same field using a single extension of the element w, where w = ⁶√−108. This is done with the commands:

InitDomain(0)
AddFieldVar("w")
Define(w^6, -108)

DEFINITION 14.9
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 element a.

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

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

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

Click here for the proof.

We can easily extend this result by induction to any number of extensions of ℚ.

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

Click here for the proof.

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

EXAMPLE:

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

We first set up the splitting field as ℚ(a, b), where a³ = 2, and b² = −a² − a b.

InitDomain(0)
AddFieldVar("a")
Define(a^3, 2)
AddFieldVar("b")
Define(b^2, -a^2 - a*b)

To find an element in this field which forms a simple extension, we compute

SimpleExtension(a, b)

This shows that ℚ(a, b) = ℚ(a + 2 b), which is a simple extension. What element is a + 2 b? Let us compute its sixth power.

expand((a + 2*b)^6)

Thus, we see that a + 2 b is one root of x⁶ + 108.

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), where y is any number such that

y ≠	uₕ − u
	v − vₖ

whenever u_h is a root of Irr_F(u, x), and v_k is a root of Irr_F(v, x).

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

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

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 .

---

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

We can also have Sage find an element w for us.

InitDomain(0)
AddFieldVar("a")
Define(a^2, 2)
AddFieldVar("b")
Define(b^2, 3)
SimpleExtension(a, b)

Thus Sage verifies that ℚ(√2, √3) = ℚ(√2 + √3).

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

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

First we define a = ∛2 and b = √2.

InitDomain(0)
AddFieldVar("a")
Define(a^3, 2)
AddFieldVar("b")
Define(b^2, 2)
SimpleExtension(a, b)

This shows that ℚ(∛2, √2) = ℚ(∛2 + √2).

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

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 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 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 referred to as Galois theory.

Proofs:

Proof of Lemma 14.1:

If B is a basis, then every element v ∈ V can be expressed in the form c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n. Suppose that

v = a₁ x₁ + a₂ x₂ + a₃ x₃ + ⋯ + a_n x_n

is another such expression. Then

(a₁ − c₁) x₁ + (a₂ − c₂) x₂ + (a₃ − c₃) x₃ + ⋯ + (a_n − c_n) x_n = v − v = 0.

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

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

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Proof of Lemma 14.2:

Suppose that Y = {y₁, y₂, y₃, … y_n, y_n+1} are linearly independent, so that all of these vectors are nonzero.

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

{y₁, y₂, … y_k−1, y_k, x_k+1, … x_k}

is a basis for every 0 ≤ k ≤ 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, that {y₁, y₂, … y_k−1, x_k, x_k+1, … x_k} is a basis.

We then can express y_k = a₁ y₁ + a₂ y₂ + ⋯

• a_k−1 y_k−1 + a_k x_k + ⋯

• a_n x_n. Since the vectors in Y are linearly independent, we see that at least one of a_k, a_k+1, … a_n is nonzero. By rearranging the remaining elements of B, we can suppose that a_k ≠ 0.
  
  

Then x_k = a_k^-1(y_k − a₁ y₁ − ⋯ − a_k−1 y_k−1 − a_k+1 x_k+1 − ⋯ − a_n x_n).

Any element v ∈ V can be expressed as v = c₁ y₁ + c₂ y₂

• ⋯ + c_k−1 y_k−1 + c_k x_k + ⋯

• c_n x_n. By substituting for the value of x_k, we see that v can be expressed as a linear combination of

{y₁, y₂, … y_k−1, y_k, x_k+1, … x_n}.

If this set were linearly dependent, there would be a nonzero solution to

c₁ y₁ + c₂ y₂ + ⋯ + c_k−1 y_k−1 + c_k y_k + ⋯ + c_n x_n = 0.

Then c_k ≠ 0 lest there also be a nonzero solution to c₁ y₁ + c₂ y₂ + ⋯ + c_k−1 y_k−1 + c_k x_k + ⋯ + c_n x_n = 0, but we are assuming that {y₁, y₂, … y_k−1, x_k, x_k+1, … x_n} is a basis.

But substituting the value for y_k gives

c_k(a₁ y₁ + a₂ y₂ + ⋯ + a_k−1 y_k−1 + a_k x_k + ⋯ + a_n x_n) + c₁ y₁ + c₂ y₂ + ⋯ + c_k−1 y_k−1 + c_k+1 y_k+1 + ⋯ + c_n x_n = 0.

This is a nonzero solution to b₁ y₁ + b₂ y₂ + ⋯ + b_k−1 y_k−1 + b_k x_k + ⋯ + b_n x_n = 0, since b_k = c_k a_k ≠ 0. Thus, the set {y₁, y₂, … y_k−1, y_k, x_k+1, … x_n} is linearly independent, and hence is a basis of V.

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

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Proof of Proposition 14.1:

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

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Proof of Proposition 14.2:

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

c₁ y₁ + c₂ y₂ + c₃ y₃ + ⋯ + c_n y_n.

where c₁, c₂, c₃, … c_n are in E.

Since {x₁, x₂, x₃, … x_m} is a basis of E over F, we can in turn write

c1 =	a1,1 x1 + a2,1 x2 + a3,1 x3 + ⋯ + am,1xm,
c2 =	a1,2 x1 + a2,2 x2 + a3,2 x3 + ⋯ + am,2xm,
c3 =	a1,3 x1 + a2,3 x2 + a3,3 x3 + ⋯ + am,3xm,
	……………
cn =	a1,n x1 + a2,n x2 + a3,n x3 + ⋯ + am,nxm.

where each a_i,j is in F. Combining these, we see that every element of V can be expressed in the form

a1,1 x1 y1 + a2,1 x2 y1 + a3,1 x3 y1 + ⋯ + am,1xm y1 +

a1,2 x1 y2 + a2,2 x2 y2 + a3,2 x3 y2 + ⋯ + am,2xm y2 +

a1,3 x1 y3 + a2,3 x2 y3 + a3,3 x3 y3 + ⋯ + am,3xm y3 +

……………

a1,n x1 yn + a2,n x2 yn + a3,n x3 yn + ⋯ + am,nxm yn.

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 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

m	n
∑	∑	ai,j xi yj = 0
i=1	j=1

for a_i,j in F. Then we have

	m	n		n	⎛	m		⎞
0 =	∑	∑	ai,j xi yj =	∑	⎜	∑	ai,j xi	⎟	yj.
					⎜			⎟
					⎜			⎟
	i=1	j=1		j=1	⎝	i=1		⎠

Since {y₁, y₂, y₃, … y_n} is a basis of V over E, the only way that the right hand expression could be zero is if

m
∑	ai,j xi = 0
i=1

for all j = 1, 2, 3, … n. Now {x₁, x₂, x₃, … x_m} is a basis of E over F, so the only way that each of these sums could be 0 is if a_i,j = 0 for all values of i 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 F of dimension m n.

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Proof of Proposition 14.3:

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

u = c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n

for some c₁, c₂, c₃, … c_n in F. Then

v = ϕ(u) = ϕ(c₁) ϕ(x₁) + ϕ(c₂) ϕ(x₂) + ϕ(c₃) ϕ(x₃) + ⋯ + ϕ(c_n) ϕ(x_n) = c₁ ϕ(x₁) + c₂ ϕ(x₂) + c₃ ϕ(x₃) + ⋯ + c_n ϕ(x_n).

Thus, {ϕ(x₁), ϕ(x₂), ϕ(x₃), … ϕ(x_n)} generates the field E. Also, if

c₁ ϕ(x₁) + c₂ ϕ(x₂) + c₃ ϕ(x₃) + ⋯ + c_n ϕ(x_n) = 0,

then ϕ(c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n) = 0, which implies that

c₁ x₁ + c₂ x₂ + c₃ x₃ + ⋯ + c_n x_n = 0.

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

c₁ = c₂ = c₃ = ⋯ = c_n = 0.

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

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Proof of Lemma 14.3:

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

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 in the collection L. Thus, any subfield containing K and the elements of S must also contain K(S).

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Proof of Proposition 14.4:

From Proposition 13.1, K = F[x]/⟨f(x)⟩ is a field that contains F as a subfield. Let y = x + ⟨f(x)⟩ in K. If we treat f(x) as a polynomial in K[x], we find that f(y)= 0. Consider the set {1, y, y², y³, … yⁿ⁻¹}. We 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

k = a₁ 1 + a₂ y + a₃ y² + ⋯ + a_n yⁿ⁻¹,

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

g(x) = f(x)·q(x) + r(x),

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

r(x) = a₁ + a₂ x + a₃ x² + ⋯ + a_n xⁿ⁻¹

for some a₁, a₂, a₃, … a_n in F. Note that we can now write

k = g(x) + ⟨f(x)⟩ = r(x) + ⟨f(x)⟩ = a₁ 1 + a₂ y + a₃ y² + ⋯ + a_n yⁿ⁻¹.

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

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Proof of Proposition 14.5:

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

a₀ + a₁ y + a₂ y² + a₃ y³ + ⋯ + a_n yⁿ = 0

with a₀, a₁, a₂, … a_n in F. Thus, there is a nonzero polynomial

a₀ + a₁ x + a₂ x² + a₃ x³ + ⋯ + a_n xⁿ

in F[x] for which y is a root when treated as a polynomial in K[x].

Let us now show uniqueness. Let f(x) be a polynomial of lowest possible degree in F[x] such that f(y) = 0. Since 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 such polynomials, f(x) and g(x), then by the division algorithm theorem (12.1), there exist polynomials q(x) and r(x) such that

f(x) = g(x)·q(x) + r(x),

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

0 = f(y) = g(y)·q(y) + r(y) = 0 + r(y) = 0.

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

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Proof of Proposition 14.6:

Let us consider the evaluation homomorphism ϕ_w that maps polynomials in F[x] to elements in F(w):

ϕ_w(g(x)) = g(w).

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

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Proof of Corollary 14.1:

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

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Proof of Proposition 14.7:

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

ϕ_v·ƒ : K[x] → E[x] → E(v).

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

K[x]/⟨p(x)⟩ ≈ E(v).

By Proposition 14.6, we also have

K(u) ≈ K[x]/⟨p(x)⟩,

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

μ : K(u) → K[x]/⟨p(x)⟩ → E(v)

then μ(u) = ϕ_v(ƒ(x)) = v, and μ(t) = ƒ(t) for all t in K.

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Proof of Corollary 14.2:

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

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Proof of Lemma 14.4:

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

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

g(x) = c_n·(x − u₁)·(x − u₂)·(x − u₃) ⋯ (x − u_n−1).

Thus,

f(x) = c_n·(x − u₁)·(x − u₂)·(x − u₃) ⋯ (x − u_n−1)·(x − u_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 over E. Thus, the dimension of K over F is at most n·(n − 1)! = n!.

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Proof of Proposition 14.8:

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, K = E, and L = F, and so the function μ(t) = ϕ(t) satisfies the necessary conditions.

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 degree (n − 1). By Lemma 13.3, the isomorphism ϕ extends to an isomorphism from F[x] to E[x] in such a way that ϕ(x) = x. Thus, if p(x) is an irreducible factor of the polynomial f(x), then ϕ(p(x)) is an irreducible factor of the polynomial g(x) = ϕ(f(x)). Note that every root of p(x) is also a root of f(x), so that p(x) factors completely in the field K. Likewise, ϕ(p(x)) factors completely in the field L.

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

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

g(x) = ϕ(f(x)) = θ(f(x)) = θ(x − u)·θ(h(x)) = (x − v)·θ(h(x)).

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

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Proof of Corollary 14.3:

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

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Proof of Proposition 14.9:

From the definition of the splitting field, the generator

ω_n = e^{(2π i/n)} = cos(^2π⁄_n) + i sin(^2π⁄_n)

is a root of the n^th cyclotomic polynomial

Φ_n(x) = (x − ω_n^k¹)·(x − ω_n^k²)·(x − ω_n^k³) ⋯ (x − ω_n^kⁱ),

where k₁, k₂, k₃, … k_i are the integers from 1 to n that are coprime to n. Thus, the splitting field contains ℚ(ω_n). Note that all powers of ω_n are in this field, and so the n^th cyclotomic polynomial factors completely in ℚ(ω_n).

Since Φ_n(x) is irreducible by Gauss' theorem on cyclotomic polynomials (13.2), and the dimension of Φ_n(x) is ϕ(n), the dimension of the splitting field ℚ(ω_n) is ϕ(n).

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Proof of Lemma 14.5:

Let u₁, u₂, u₃, … u_n be the roots of f(x) in K. Then

K = F(u₁, u₂, u₃, … u_n).

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

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

We now want to consider the field K(w). We have

K(w) = F(u₁, u₂, u₃, … u_n, w) = F(w, u₁, u₂, u₃, … u_n).

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

K = K(v) = F(u₁, u₂, u₃, … u_n, v) = F(v, u₁, u₂, u₃, … u_n).

so K is the splitting field of f(x) over the field F(v).

Consequently Proposition 14.8 shows us that the isomorphism ϕ from F(v) to F(w) extends to an isomorphism μ from K to K(w), and μ(v) = w. Also, μ(t) = t for 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 Proposition 14.2, the dimension of K(w) over F equals the dimension of K(w) over K times the dimension of K over F. Therefore, the dimension of K(w) over K must be 1, so w is in K. Therefore, every root of p(x) is in K, so p(x) factors completely in K.

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Proof of Proposition 14.10:

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 group K^*, using Proposition 13.4. Thus, we will assume that F is an infinite field. Let f(x) = Irr_F(u, x) and g(x) = Irr_F(v, x). Let E be the splitting field of g(x) over the field F(u). Since g(x) factors completely in L without double roots, g(x) will also factor completely in E without double roots. Let

v = v₁, v₂, v₃, … v_n

be the distinct roots of g(x) in E.

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 completely in the field E, we can let

u = u₁, u₂, u₃, … u_m

be the roots of f(x) over E.

Since F is an infinite field, we can pick some element y of F, such that

y ≠	uₕ − u	for all 1 ≤ h ≤ m, 1 ≤ k ≤ n.
	v − vₖ

Finally, we let w = u + y v. Let us show that K = F(w). To show that v is in F(w), let p(x) = f(w − y x), and note that p(v) = f(u + y v − y v) = f(u) = 0, so v is a root of p(x). If one of the other roots of g(x) is a root of p(x), then

w − y v_k = u_h for some h and k,

so u + y v − y v_k = u_h, giving us

y =	uₕ − u	,
	v − vₖ

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) in the field E.

Let r(x) = Irr_F(w)(v, x). Then r(x) divides the polynomials g(x) and p(x), since both polynomials have v as a root. In fact, we have seen that g(x) and p(x) have no 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 multiple roots in E. Thus, r(x) has degree 1, and in fact r(x) = x − v. This proves that v is in F(w). To see that u is in F(w), we note that u = y v − w. Thus, F(u, v) is contained in F(w), while F(w) is obviously contained in F(u, v). Therefore, F(u, v) = F(w).

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Proof of Corollary 14.4:

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 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

F(u) = F(u₁, u₂, u₃, … u_n−1).

Let v = u_n, and since g(x) = Irr_F(u_n, x) does not have multiple roots in its splitting field L, we can use Proposition 14.10 to find a w in K such that But then

F(w) = F(u₁, u₂, u₃, … u_n−1, u_n).

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

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Sage Interactive Problems

§14.1 #20)
Use Sage to find the coefficients of the vector ⟨3,-2,5⟩ in ℝ³ using the basis {⟨2,-1,4⟩,⟨5,2,1⟩,⟨4,-3,2⟩}.

§14.1 #21)
Use Sage to find the coefficients of the element a⁵ in GF(27) over Z₃ using the basis {1, a, a²}. Here, a is a root of the Conway polynomial of degree 3 over Z₃, which is x³ + 2 x + 1.

§14.2 #19)
Define the field ℚ(√−3) in Sage, then find 1/(5 + √−3).

§14.2 #20)
Define the field ℚ(√5) in Sage. Does the polynomial x² + 4 x − 1 factor in this field?

For Problems 17 through 20: Define the splitting field of the polynomial in Sage. Determine the dimension of the splitting field over ℚ.

§14.3 #17) x³ + x² − 4 x + 1.

§14.3 #18) x⁵ + x⁴ − 4 x³ − 3 x² + 3 x + 1.

§14.3 #20) x⁵ − 2.

§14.3 #19) x⁵ + 3 x³ + 5 x + 10.