Book a Demo!
CoCalc Logo Icon
StoreFeaturesDocsShareSupportNewsAboutPoliciesSign UpSign In
Download

πŸ“š The CoCalc Library - books, templates and other resources

132948 views
License: OTHER
Kernel:
%%html <link href="http://mathbook.pugetsound.edu/beta/mathbook-content.css" rel="stylesheet" type="text/css" /> <link href="https://aimath.org/mathbook/mathbook-add-on.css" rel="stylesheet" type="text/css" /> <style>.subtitle {font-size:medium; display:block}</style> <link href="https://fonts.googleapis.com/css?family=Open+Sans:400,400italic,600,600italic" rel="stylesheet" type="text/css" /> <link href="https://fonts.googleapis.com/css?family=Inconsolata:400,700&subset=latin,latin-ext" rel="stylesheet" type="text/css" /><!-- Hide this cell. --> <script> var cell = $(".container .cell").eq(0), ia = cell.find(".input_area") if (cell.find(".toggle-button").length == 0) { ia.after( $('<button class="toggle-button">Toggle hidden code</button>').click( function (){ ia.toggle() } ) ) ia.hide() } </script>

Important: to view this notebook properly you will need to execute the cell above, which assumes you have an Internet connection. It should already be selected, or place your cursor anywhere above to select. Then press the "Run" button in the menu bar above (the right-pointing arrowhead), or press Shift-Enter on your keyboard.

ParseError: KaTeX parse error: \newcommand{\lt} attempting to redefine \lt; use \renewcommand

Section18.1Fields of Fractions

ΒΆ

Every field is also an integral domain; however, there are many integral domains that are not fields. For example, the integers Z{\mathbb Z} form an integral domain but not a field. A question that naturally arises is how we might associate an integral domain with a field. There is a natural way to construct the rationals Q{\mathbb Q} from the integers: the rationals can be represented as formal quotients of two integers. The rational numbers are certainly a field. In fact, it can be shown that the rationals are the smallest field that contains the integers. Given an integral domain D,D\text{,} our question now becomes how to construct a smallest field FF containing D.D\text{.} We will do this in the same way as we constructed the rationals from the integers.

An element p/q∈Qp/q \in {\mathbb Q} is the quotient of two integers pp and q;q\text{;} however, different pairs of integers can represent the same rational number. For instance, 1/2=2/4=3/6.1/2 = 2/4 = 3/6\text{.} We know that

ab=cd\begin{equation*} \frac{a}{b} = \frac{c}{d} \end{equation*}

if and only if ad=bc.ad = bc\text{.} A more formal way of considering this problem is to examine fractions in terms of equivalence relations. We can think of elements in Q{\mathbb Q} as ordered pairs in ZΓ—Z.{\mathbb Z} \times {\mathbb Z}\text{.} A quotient p/qp/q can be written as (p,q).(p, q)\text{.} For instance, (3,7)(3, 7) would represent the fraction 3/7.3/7\text{.} However, there are problems if we consider all possible pairs in ZΓ—Z.{\mathbb Z} \times {\mathbb Z}\text{.} There is no fraction 5/05/0 corresponding to the pair (5,0).(5,0)\text{.} Also, the pairs (3,6)(3,6) and (2,4)(2,4) both represent the fraction 1/2.1/2\text{.} The first problem is easily solved if we require the second coordinate to be nonzero. The second problem is solved by considering two pairs (a,b)(a, b) and (c,d)(c, d) to be equivalent if ad=bc.ad = bc\text{.}

If we use the approach of ordered pairs instead of fractions, then we can study integral domains in general. Let DD be any integral domain and let

S={(a,b):a,b∈DΒ andΒ bβ‰ 0}.\begin{equation*} S = \{ (a, b) : a, b \in D \text{ and } b \neq 0 \}. \end{equation*}

Define a relation on SS by (a,b)∼(c,d)(a, b) \sim (c, d) if ad=bc.ad=bc\text{.}

Lemma18.1

The relation ∼\sim between elements of SS is an equivalence relation.

Proof

Since DD is commutative, ab=ba;ab = ba\text{;} hence, ∼\sim is reflexive on D.D\text{.} Now suppose that (a,b)∼(c,d).(a,b) \sim (c,d)\text{.} Then ad=bcad=bc or cb=da.cb = da\text{.} Therefore, (c,d)∼(a,b)(c,d) \sim (a, b) and the relation is symmetric. Finally, to show that the relation is transitive, let (a,b)∼(c,d)(a, b) \sim (c, d) and (c,d)∼(e,f).(c, d) \sim (e,f)\text{.} In this case ad=bcad = bc and cf=de.cf = de\text{.} Multiplying both sides of ad=bcad=bc by ff yields

afd=adf=bcf=bde=bed.\begin{equation*} a f d = a d f = b c f = b d e = bed. \end{equation*}

Since DD is an integral domain, we can deduce that af=beaf = be or (a,b)∼(e,f).(a,b ) \sim (e, f)\text{.}

We will denote the set of equivalence classes on SS by FD.F_D\text{.} We now need to define the operations of addition and multiplication on FD.F_D\text{.} Recall how fractions are added and multiplied in Q:{\mathbb Q}\text{:}

ab+cd=ad+bcbd;abβ‹…cd=acbd.\begin{align*} \frac{a}{b} + \frac{c}{d} & = \frac{ad + b c}{b d};\\ \frac{a}{b} \cdot \frac{c}{d} & = \frac{ac}{b d}. \end{align*}

It seems reasonable to define the operations of addition and multiplication on FDF_D in a similar manner. If we denote the equivalence class of (a,b)∈S(a, b) \in S by [a,b],[a, b]\text{,} then we are led to define the operations of addition and multiplication on FDF_D by

[a,b]+[c,d]=[ad+bc,bd]\begin{equation*} [a, b] + [c, d] = [ad + b c,b d] \end{equation*}

and

[a,b]β‹…[c,d]=[ac,bd],\begin{equation*} [a, b] \cdot [c, d] = [ac, b d], \end{equation*}

respectively. The next lemma demonstrates that these operations are independent of the choice of representatives from each equivalence class.

Lemma18.2

The operations of addition and multiplication on FDF_D are well-defined.

Proof

We will prove that the operation of addition is well-defined. The proof that multiplication is well-defined is left as an exercise. Let [a1,b1]=[a2,b2][a_1, b_1] = [a_2, b_2] and [c1,d1]=[c2,d2].[c_1, d_1] =[ c_2, d_2]\text{.} We must show that

[a1d1+b1c1,b1d1]=[a2d2+b2c2,b2d2]\begin{equation*} [a_1 d_1 + b_1 c_1,b_1 d_1] = [a_2 d_2 + b_2 c_2,b_2 d_2] \end{equation*}

or, equivalently, that

(a1d1+b1c1)(b2d2)=(b1d1)(a2d2+b2c2).\begin{equation*} (a_1 d_1 + b_1 c_1)( b_2 d_2) = (b_1 d_1) (a_2 d_2 + b_2 c_2). \end{equation*}

Since [a1,b1]=[a2,b2][a_1, b_1] = [a_2, b_2] and [c1,d1]=[c2,d2],[c_1, d_1] =[ c_2, d_2]\text{,} we know that a1b2=b1a2a_1 b_2 = b_1 a_2 and c1d2=d1c2.c_1 d_2 = d_1 c_2\text{.} Therefore,

(a1d1+b1c1)(b2d2)=a1d1b2d2+b1c1b2d2=a1b2d1d2+b1b2c1d2=b1a2d1d2+b1b2d1c2=(b1d1)(a2d2+b2c2).\begin{align*} (a_1 d_1 + b_1 c_1)( b_2 d_2) & = a_1 d_1 b_2 d_2 + b_1 c_1 b_2 d_2\\ & = a_1 b_2 d_1 d_2 + b_1 b_2 c_1 d_2\\ & = b_1 a_2 d_1 d_2 + b_1 b_2 d_1 c_2\\ & = (b_1 d_1) (a_2 d_2 + b_2 c_2). \end{align*}
Lemma18.3

The set of equivalence classes of S,S\text{,} FD,F_D\text{,} under the equivalence relation ∼,\sim\text{,} together with the operations of addition and multiplication defined by

[a,b]+[c,d]=[ad+bc,bd][a,b]β‹…[c,d]=[ac,bd],\begin{align*} [a, b] + [c, d] & = [ad + b c, b d]\\ [ a, b] \cdot [c, d] & = [ac, b d], \end{align*}

is a field.

Proof

The additive and multiplicative identities are [0,1][0,1] and [1,1],[1,1]\text{,} respectively. To show that [0,1][0,1] is the additive identity, observe that

[a,b]+[0,1]=[a1+b0,b1]=[a,b].\begin{equation*} [a, b] + [0, 1] = [ a 1 + b 0, b 1] = [a,b]. \end{equation*}

It is easy to show that [1,1][1, 1] is the multiplicative identity. Let [a,b]∈FD[a, b] \in F_D such that aβ‰ 0.a \neq 0\text{.} Then [b,a][b, a] is also in FDF_D and [a,b]β‹…[b,a]=[1,1];[a,b] \cdot [b, a] = [1,1]\text{;} hence, [b,a][b, a] is the multiplicative inverse for [a,b].[a, b]\text{.} Similarly, [βˆ’a,b][-a,b] is the additive inverse of [a,b].[a, b]\text{.} We leave as exercises the verification of the associative and commutative properties of multiplication in FD.F_D\text{.} We also leave it to the reader to show that FDF_D is an abelian group under addition.

It remains to show that the distributive property holds in FD;F_D\text{;} however,

[a,b][e,f]+[c,d][e,f]=[ae,bf]+[ce,df]=[aedf+bfce,bdf2]=[aed+bce,bdf]=[ade+bce,bdf]=([a,b]+[c,d])[e,f]\begin{align*} [a, b] [e, f] + [c, d][ e, f ] & = [a e, b f ] + [c e, d f]\\ & = [a e d f + b f c e, b d f^2 ]\\ & = [a e d + b c e, b d f ]\\ & = [a d e + b c e, b d f ]\\ & = ( [a, b] + [c, d] ) [ e, f ] \end{align*}

and the lemma is proved.

The field FDF_D in LemmaΒ 18.3 is called the or of the integral domain D.D\text{.}

Theorem18.4

Let DD be an integral domain. Then DD can be embedded in a field of fractions FD,F_D\text{,} where any element in FDF_D can be expressed as the quotient of two elements in D.D\text{.} Furthermore, the field of fractions FDF_D is unique in the sense that if EE is any field containing D,D\text{,} then there exists a map ψ:FDβ†’E\psi : F_D \rightarrow E giving an isomorphism with a subfield of EE such that ψ(a)=a\psi(a) = a for all elements a∈D,a \in D\text{,} where we identify aa with its image in FD.F_D\text{.}

Proof

We will first demonstrate that DD can be embedded in the field FD.F_D\text{.} Define a map ϕ:D→FD\phi : D \rightarrow F_D by ϕ(a)=[a,1].\phi(a) = [a, 1]\text{.} Then for aa and bb in D,D\text{,}

Ο•(a+b)=[a+b,1]=[a,1]+[b,1]=Ο•(a)+Ο•(b)\begin{equation*} \phi( a + b ) = [a+b, 1] = [a, 1] + [b, 1] = \phi(a ) + \phi(b) \end{equation*}

and

Ο•(ab)=[ab,1]=[a,1][b,1]=Ο•(a)Ο•(b);\begin{equation*} \phi( a b ) = [a b, 1] = [a, 1] [b, 1] = \phi(a ) \phi(b); \end{equation*}

hence, Ο•\phi is a homomorphism. To show that Ο•\phi is one-to-one, suppose that Ο•(a)=Ο•(b).\phi(a) = \phi( b)\text{.} Then [a,1]=[b,1],[a, 1] = [b, 1]\text{,} or a=a1=1b=b.a = a1 = 1b = b\text{.} Finally, any element of FDF_D can be expressed as the quotient of two elements in D,D\text{,} since

Ο•(a)[Ο•(b)]βˆ’1=[a,1][b,1]βˆ’1=[a,1]β‹…[1,b]=[a,b].\begin{equation*} \phi(a) [\phi(b)]^{-1} = [a, 1] [b, 1]^{-1} = [a, 1] \cdot [1, b] = [a, b]. \end{equation*}

Now let EE be a field containing DD and define a map ψ:FDβ†’E\psi :F_D \rightarrow E by ψ([a,b])=abβˆ’1.\psi([a, b]) = a b^{-1}\text{.} To show that ψ\psi is well-defined, let [a1,b1]=[a2,b2].[a_1, b_1] = [a_2, b_2]\text{.} Then a1b2=b1a2.a_1 b_2 = b_1 a_2\text{.} Therefore, a1b1βˆ’1=a2b2βˆ’1a_1 b_1^{-1} = a_2 b_2^{-1} and ψ([a1,b1])=ψ([a2,b2]).\psi( [a_1, b_1]) = \psi( [a_2, b_2])\text{.}

If [a,b][a, b ] and [c,d][c, d] are in FD,F_D\text{,} then

ψ([a,b]+[c,d])=ψ([ad+bc,bd])=(ad+bc)(bd)βˆ’1=abβˆ’1+cdβˆ’1=ψ([a,b])+ψ([c,d])\begin{align*} \psi( [a, b] + [c, d] ) & = \psi( [ad + b c, b d ] )\\ & = (ad +b c)(b d)^{-1}\\ & = a b^{-1} + c d^{-1}\\ & = \psi( [a, b] ) + \psi( [c, d] ) \end{align*}

and

ψ([a,b]β‹…[c,d])=ψ([ac,bd])=(ac)(bd)βˆ’1=abβˆ’1cdβˆ’1=ψ([a,b])ψ([c,d]).\begin{align*} \psi( [a, b] \cdot [c, d] ) & = \psi( [ac, b d ] )\\ & = (ac)(b d)^{-1}\\ & = a b^{-1} c d^{-1}\\ & = \psi( [a, b] ) \psi( [c, d] ). \end{align*}

Therefore, ψ\psi is a homomorphism.

To complete the proof of the theorem, we need to show that ψ\psi is one-to-one. Suppose that ψ([a,b])=abβˆ’1=0.\psi( [a, b] ) = ab^{-1} = 0\text{.} Then a=0b=0a = 0b = 0 and [a,b]=[0,b].[a, b] = [0, b]\text{.} Therefore, the kernel of ψ\psi is the zero element [0,b][ 0, b] in FD,F_D\text{,} and ψ\psi is injective.

Example18.5

Since Q{\mathbb Q} is a field, Q[x]{\mathbb Q}[x] is an integral domain. The field of fractions of Q[x]{\mathbb Q}[x] is the set of all rational expressions p(x)/q(x),p(x)/q(x)\text{,} where p(x)p(x) and q(x)q(x) are polynomials over the rationals and q(x)q(x) is not the zero polynomial. We will denote this field by Q(x).{\mathbb Q}(x)\text{.}

We will leave the proofs of the following corollaries of TheoremΒ 18.4 as exercises.

Corollary18.6

Let FF be a field of characteristic zero. Then FF contains a subfield isomorphic to Q.{\mathbb Q}\text{.}

Corollary18.7

Let FF be a field of characteristic p.p\text{.} Then FF contains a subfield isomorphic to Zp.{\mathbb Z}_p\text{.}