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(b) Show that .f / D fKŒx is a maximal ideal in KŒx if, and only if, f is irreducible.
(c) Conclude that Zn D Z=nZ is a field if, and only if, ˙n is prime, and that KŒx=.f / is a field if, and only if, f is irreducible.
6.3.8. If J is an ideal of the ring R, show that J Œx is an ideal in RŒx and furthermore RŒx=J Œx Š .R=J /Œx. Hint: Find a natural homomorphism from RŒx onto .R=J /Œx with kernel J Œx.
6.3.9. For any ring R, and any natural number n, we can define the matrix ring Matn.R/ consisting of n-by-n matrices with entries in R. If J is an ideal of R, show that Matn.J / is an ideal in Matn.R/ and furthermore Matn.R/=Matn.J / Š Matn.R=J /. Hint: Find a natural homomorphism from Matn.R/ onto Matn.R=J / with kernel Matn.J /.
6.3.10. Let R be a commutative ring. Show that RŒx=xRŒx Š R.
6.3.11. This exercise gives a version of the Chinese remainder theorem.
(a) Let R be a ring, P and Q ideals in R, and suppose that P \ Q D f0g, and P C Q D R. Show that the map x 7! .x C P; x C Q/ is an isomorphism of R onto R=P ˚ R=Q. Hint: Injectivity is clear. For surjectivity, show that for each a; b 2 R, there exist x 2 R, p 2 P , and q 2 Q such that x C p D a, and x C q D b.
(b) More generally, if P CQ D R, show that R=.P \Q/ Š R=P ˚ R=Q.
6.3.12.
(a) Show that integers m and n are relatively prime if, and only if, mZ C nZ D Z if, and only if, mZ \ nZ D mnZ. Conclude that if m and n are relatively prime, then Zmn Š Zm ˚ Zn as rings.
(b) State and prove a generalization of this result for the ring of polynomials KŒx over a field K.
6.4. Integral Domains
The product of nonzero elements of a ring can be zero. Here are some familiar examples: Let R be the ring of real–valued functions on a set X and let A be a proper subset of X . Let f be the characteristic function of A, that is, the function satisfying f .a/ D 1 if a 2 A and f .x/ D 0 if x 2 X n A. Then f and 1 f are nonzero elements of R whose product is zero.
In Z6, Œ3Œ2 D Œ0.
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0 1
Let x be the 2-by-2 matrix . Compute that x ¤ 0 but 0 0 x2 D 0.
Definition 6.4.1. An integral domain is a commutative ring with identity element 1 in which the product of any two nonzero elements is nonzero.
You are asked to verify the following examples in the Exercises.
Example 6.4.2.
(a) The ring of integers Z is an integral domain.
(b) Any field is an integral domain.
(c) If R is an integral domain, then RŒx is an integral domain. In particular, KŒx is an integral domain for any field K.
(d) If R is an integral domain and R0 is a subring containing the identity, then R0 is an integral domain.
(e) The ring of formal power series RŒŒx with coefficients in an integral domain is an integral domain.
There are two common constructions of fields from integral domains.
One construction is treated in Corollary 6.3.14 and Exercise 6.3.7. Namely, if R is a commutative ring with 1 and M is a maximal ideal, then the quotient ring R=M is a field.
Another construction is that of the field of fractions1 of an integral do main. This construction is known to you from the formation of the rational numbers as fractions of integers. Given an integral domain R, we wish to construct a field from symbols of the form a=b, where a; b 2 R and b ¤ 0. You know that in the rational numbers, 2=3 D 8=12; the rule is a=b D a0=b0 if ab0 D a0b. It seems prudent to adopt the same rule for any integral domain R.
Lemma 6.4.3. Let R be an integral domain. Let S be the set of symbols of the form a=b, where a; b 2 R and b ¤ 0. The relation a=b a0=b0 if ab0 D a0b is an equivalence relation on S.
Proof. Exercise 6.4.9.
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Let us denote the quotient of S by the equivalence relation by Q.R/.
For a=b 2 S, write Œa=b for the equivalence class of a=b, an element of Q.R/.
1Do not confuse the terms field of fractions and quotient ring or quotient field.
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Now, we attempt to define operations on Q.R/, following the guiding example of the rational numbers. The sum Œa=b C Œc=d will have to be defined as Œ.ad C bc/=bd and the product Œa=bŒc=d as Œac=bd .
As always, when we define operations on equivalence classes in terms of representatives of those classes, the next thing that has to be done is to check that the operations are well defined. You are asked to check that addition and multiplication are well defined in Exercise 6.4.10.
It is now straightforward, if slightly tedious, to check that Q.R/ is a field, and that the map a 7! Œa=1 is an injective ring homomorphism of R into Q.R/; thus R can be considered as a subring of Q.R/. Moreover, any injective homomorphism of R into a field F extends to an injective homomorphism of Q.R/ into F : '
R
qqqqq
F
q qqqq
Q'
qqqq
Q.R/
The main steps in establishing these facts are the following:
1. Q.R/ is a ring with zero element 0 D Œ0=1 and multiplicative identity 1 D Œ1=1. In Q.R/, 0 ¤ 1.
2. Œa=b D 0 if, and only if, a D 0. If Œa=b ¤ 0, then Œa=bŒb=a D 1. Thus Q.R/ is a field.
3. a 7! Œa=1 is an injective homomorphism of R into Q.R/.
4. If ' W R ! F is an injective homomorphism of R into a field F , then Q ' W Œa=b 7! '.a/='.b/ defines an injective homomorphism of Q.R/ into F , which extends '.
See Exercises 6.4.11 and 6.4.12.
Proposition 6.4.4. If R is an integral domain, then Q.R/ is a field containing R as a subring. Moreover, any injective homomorphism of R into a field F extends to an injective homomorphism of Q.R/ into F .
Example 6.4.5. Q.Z/ D Q.
Example 6.4.6. Q.KŒx/ is the field of rational functions in one variable.
This field is denoted K.x/.
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Example 6.4.7. Q.KŒx1; : : : ; xn/ is the field of rational functions in n variables. This field is denoted K.x1; : : : ; xn/.
We have observed in Example 6.2.3 that in a ring R with multiplicative identity 1, the additive subgroup h1i generated by 1 is a subring and the map k 7! k 1 is a ring homomorphism from Z onto h1i R. If this ring homomorphism is injective, then h1i Š Z as rings. Otherwise, the kernel of the homomorphism is nZ for some n 2 N and h1i Š Z=nZ D Zn as rings.
If R is an integral domain, then any subring containing the identity is also an integral domain, so, in particular, h1i is an integral domain. If h1i is finite, so ring isomorphic to Zn for some n, then n must be prime. (Zn is an integral domain if, and only if, n is prime.)
Definition 6.4.8. If the subring h1i of an integral domain R generated by the identity is isomorphic to Z, the integral domain is said to have characteristic 0. If the subring h1i is isomorphic to Zp for a prime p, then R is said to have characteristic p.
A quotient of an integral domain need not be an integral domain; for example, Z12 is a quotient of Z. On the other hand, a quotient of a ring with nontrivial zero divisors can be an integral domain; Z3 is a quotient of Z12.
Those ideals J in a ring R such that R=J has no nontrivial zero divi sors can be easily characterized. An ideal J is said to be prime if for all a; b 2 R, if ab 2 J , then a 2 J or b 2 J .
Proposition 6.4.9. Let J be an ideal in a ring R. J is prime if, and only if, R=J has no nontrivial zero divisors. In particular, if R is commutative with identity, then J is prime if, and only if, R=J is an integral domain.
Proof. Exercise 6.4.14.
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Corollary 6.4.10. In a commutative ring with identity element, every maximal ideal is prime.
Proof. If M is a maximal ideal, then R=M is a field by Corollary and therefore an integral domain. By the proposition, M is a prime ideal.
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Example 6.4.11. Every ideal in Z has the form d Z for some d > 0. The ideal d Z is prime if, and only if, d is prime. The proof of this assertion is left as an exercise.
Exercises 6.4 6.4.1. Let R be a commutative ring. An element x 2 R is said to be nilpotent if xk D 0 for some natural number k.
(a) Show that the set N of nilpotent elements of R is an ideal in R.
(b) Show that R=N has no nonzero nilpotent elements.
(c) Show that if S is an integral domain and ' W R ! S is a homomorphism, then N ker.'/.
6.4.2. If x is a nilpotent element in a ring with identity, show that 1 x is
1
invertible. Hint: Think of the power series expansion for , when t is
1
t a real variable.
6.4.3. An element e in a ring is called an idempotent if e2 D e. Show that the only idempotents in an integral domain are 1 and 0. (We say that an idempotent is nontrivial if it is different from 0 or 1. So the result is that an integral domain has no nontrivial idempotents.)
6.4.4. Show that if R is an integral domain, then the ring of polynomials RŒx with coefficients in R is an integral domain.
6.4.5. Generalize the results of the previous problem to rings of polynomials in several variables.
6.4.6. Show that if R is an integral domain, then the ring of formal power series RŒŒx with coefficients in R is an integral domain. What are the units in RŒŒx?
6.4.7.
(a) Show that a subring R0 of an integral domain R is an integral domain, if 1 2 R0.
(b) The Gaussian integers are the complex numbers whose real and imaginary parts are integers. Show that the set of Gaussian integers is an integral domain.
(c) Show that the ring of symmetric polynomials in n variables is an integral domain.
6.4.8. Show that a quotient of an integral domain need not be an integral domain.
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6.4.9. Prove Lemma 6.4.3.
6.4.10. Show that addition and multiplication on Q.R/ is well defined.
This amounts to the following: Suppose a=b a0=b0 and c=d c0=d 0 and show that .ad Cbc/=bd .a0d 0Cc0b0/=b0d 0 and ac=bd a0c0=b0d 0.
6.4.11.
(a) Show that Q.R/ is a ring with identity element Œ1=1 and 0 D Œ0=1.
(b) Show that Œa=b D 0 if, and only if, a D 0.
(c) Show that if Œa=b ¤ 0, then Œb=a is the multiplicative inverse of Œa=b. Thus Q.R/ is a field.
6.4.12.
(a) Show that a 7! Œa=1 is an injective unital ring homomorphism of R into Q.R/. In this sense Q.R/ is a field containing R.
(b) If F is a field such that F R, show that there is an injective unital homomorphism ' W Q.R/ ! F such that '.Œa=1/ D a for a 2 R.
6.4.13. If R is the ring of Gaussian integers, show that Q.R/ is isomorphic to the subfield of C consisting of complex numbers with rational real and imaginary parts.
6.4.14. Let J be an ideal in a ring R. Show that J is prime if, and only if, R=J has no nontrivial zero divisors. (In particular, if R is commutative with identity, then J is prime if, and only if, R=J is an integral domain.)
6.4.15. Every ideal in Z has the form d Z for some d > 0. Show that the ideal d Z is prime if, and only if, d is prime.
6.4.16. Show that a maximal ideal in a commutative ring with identity is prime.
6.5. Euclidean Domains, Principal Ideal Domains, and Unique Factorization
We have seen two examples of integral domains with a good theory of factorization, the ring of integers Z and the ring of polynomials KŒx over a field K. In both of these rings R, every nonzero, noninvertible element has an essentially unique factorization into irreducible factors.
The common feature of these rings, which was used to establish unique factorization is a Euclidean function d W R n f0g ! N [ f0g with the property that d.fg/ maxfd.f /; d.g/g and for each f; g 2 R n f0g there exist q; r 2 R such that f D qg C r and r D 0 or d.r/