This chapter collects three subjects that extend the algebra of the earlier chapters: the arithmetic of commutative rings, leading to unique factorization; Galois theory, which connects field extensions to group theory; and an introduction to Lie groups and Lie algebras, the continuous analogues of finite groups and their representations. These topics go beyond the course on which this book is based, and the chapter is a survey. Some of the deeper theorems are stated without proof, and we say so each time.
In this section all rings are commutative with identity (Definition 4.1.1).
17.1.1 Ideals and Quotient Rings
Recall from Chapter 10 that an ideal of a ring is an additive subgroup such that for all and . For , the principal ideal generated by is . More generally, for , the set is the smallest ideal containing , the ideal generated by them. We say that divides , written , if for some ; equivalently, .
Definition 17.1.1 (Quotient ring). Let be an ideal of . The quotient ring is the quotient group of by , whose elements are the cosets , with the multiplication .
The multiplication is well defined: if and with , then . The ring axioms for follow from those for , and the projection , , is a surjective ring homomorphism with kernel .
Proposition 17.1.2 (Kernels and quotients). Let be a ring homomorphism. Then is an ideal of , is a subring of , and is an isomorphism of rings .
Proof. The kernel is an additive subgroup, and if and , then . The image is closed under addition and multiplication and contains . By the First Isomorphism Theorem (Theorem 3.3.3) for the additive groups, the map is a well-defined isomorphism of additive groups onto . It respects multiplication because , and it sends to . ∎
Definition 17.1.3 (Integral domain). A ring with is an integral domain if it has no zero divisors: implies or .
In an integral domain we may cancel nonzero factors: if and , then , so .
Definition 17.1.4 (Prime and maximal ideals). Let be an ideal of . Then is prime if implies or , and maximal if the only ideals with are and .
Proposition 17.1.5 (Quotients by prime and maximal ideals). Let be an ideal of .
- is an integral domain if and only if is prime.
- is a field if and only if is maximal.
In particular, every maximal ideal is prime.
Proof. Note that , since .
(i) In , we have if and only if , and if and only if .
(ii) Suppose is a field, and let be an ideal. Choose . Then has an inverse , so , and ; hence . Conversely, suppose is maximal and . The ideal strictly contains , so it is , and for some and . Then .
The last statement holds because a field has no zero divisors. ∎
17.1.2 Divisibility in Integral Domains
Example 17.1.6 (Integral domains). Every field is an integral domain, and so is every subring of a field. For instance , the Gaussian integers , and are subrings of , hence integral domains. If is a field, then is an integral domain, since for nonzero . The ring is not an integral domain, since .
Definition 17.1.7 (Units, associates, irreducible and prime elements). Let be an integral domain.
- A unit is an element with a multiplicative inverse: for some . The units form a group under multiplication.
- Two elements are associates if for some unit .
- A nonzero non-unit is irreducible if implies that or is a unit.
- A nonzero non-unit is prime if implies or .
In an integral domain, and are associates if and only if : if and , then , so either or . A nonzero element is prime if and only if is a prime ideal.
Proposition 17.1.8 (Prime implies irreducible). In an integral domain, every prime element is irreducible.
Proof. Suppose is prime and . Then , so or ; say . Then , and canceling gives , so is a unit. ∎
Note. The converse fails in general: in the element is irreducible but not prime (Example 17.1). It holds in principal ideal domains (Theorem 17.1).
17.1.3 Euclidean Domains
Definition 17.1.9 (Euclidean domain). An integral domain is a Euclidean domain if there is a function , a Euclidean function, such that for all with there exist with
Example 17.1.10 (Euclidean domains).
- , with (division with remainder).
- for a field , with . This is long division: if has leading term and has leading term with , then has smaller degree than , and repeating this gives with or .
- The Gaussian integers , with (Exercise 17.1(2)).
- The Eisenstein integers , where , with . The argument is the same as for , since every complex number lies within distance of an element of .
17.1.4 Principal Ideal Domains
Definition 17.1.11 (Principal ideal domain). An integral domain is a principal ideal domain (PID) if every ideal of is principal, that is, of the form for some .
Theorem 17.1.12 (Every Euclidean domain is a PID). If is a Euclidean domain, then is a principal ideal domain.
Proof. Let be a Euclidean function and an ideal. If , then . Otherwise choose with minimal. Clearly . For , write with or . Since , minimality forces , so . ∎
In particular , , and are principal ideal domains.
Note. The converse fails: the ring is a principal ideal domain that has no Euclidean function (Motzkin, 1949).
Theorem 17.1.13 (In a PID, irreducible prime). Let be a principal ideal domain and a nonzero non-unit. Then is irreducible if and only if is prime.
Proof. () This is Proposition 17.1.8.
() Suppose is irreducible and . Since is a PID, for some . Then , so , and since is irreducible, either is a unit or is a unit. If is a unit, then , so and . If is a unit, then , so for some , and ; both terms are divisible by , so . ∎
Corollary 17.1.14 (Irreducible elements generate maximal ideals). If is a principal ideal domain and is irreducible, then is a maximal ideal, so is a field.
Proof. Since is not a unit, . If for an ideal , then , so is a unit (and ) or an associate of (and ). The last statement is Proposition 17.1.5. ∎
17.1.5 Unique Factorization Domains
Definition 17.1.15 (Unique factorization domain). An integral domain is a unique factorization domain (UFD) if:
- every nonzero non-unit is a product of irreducible elements, ; and
- this factorization is unique up to order and associates: if with all irreducible, then and, after reordering, each is an associate of .
Lemma 17.1.16 (Ascending chains in a PID). In a principal ideal domain, every ascending chain of ideals is eventually constant.
Proof. The union is an ideal: if and , then , so , and for all . So for some , which lies in some . Then for all . ∎
Theorem 17.1.17 (Every PID is a UFD). Every principal ideal domain is a unique factorization domain.
Proof. Existence. Suppose some nonzero non-unit is not a product of irreducibles. Then is not irreducible, so with nonzero non-units, and one of is not a product of irreducibles (otherwise would be); call it . Then : the inclusion holds since , where is the other factor, a non-unit; and if , then with a unit, and canceling would give . Repeating the argument with in place of , and so on, gives a strictly ascending chain , contradicting Lemma 17.1.16.
Uniqueness. We use induction on . Suppose with all factors irreducible. By Theorem 17.1.13, is prime, and since it divides the product , it divides some (by induction on the number of factors). Reorder so that . Then , and since is irreducible and is not a unit, is a unit: and are associates. Canceling gives . If , the left side is , and since irreducibles are not units this forces . Likewise, if then . Otherwise is irreducible and an associate of , and the induction hypothesis, applied to , finishes the proof. ∎
Example 17.1.18 (Failure of unique factorization). In we have The norm is multiplicative: . If is a unit, then , so ; and only for . So the units are , and associates have the same norm. No element has norm 2 or 3, since has no integer solutions.
Now , , and . If one of these four elements were a product of two non-units, then would be proper factors of 4, 9, or 6, so one of them would equal 2 or 3, which is impossible. So all four are irreducible. They are pairwise non-associate, having different norms, except and , whose quotient is not . So the two factorizations of 6 are genuinely different, and is not a UFD. In particular is irreducible but not prime: , but divides neither factor, since .
17.1.6 Noetherian Rings
Definition 17.1.19 (Noetherian ring). A ring is Noetherian if it satisfies the ascending chain condition on ideals: every ascending chain of ideals of is eventually constant.
Proposition 17.1.20 (Noetherian means finitely generated ideals). A ring is Noetherian if and only if every ideal of is finitely generated.
Proof. Suppose every ideal is finitely generated, and let be a chain of ideals. As in the proof of Lemma 17.1.16, the union is an ideal. It is generated by finitely many elements, which all lie in some ; then , and for .
Conversely, suppose some ideal is not finitely generated. Choose ; since , choose ; since , choose ; and so on. Then is a chain that is never constant, so is not Noetherian. ∎
Note. Every PID is Noetherian, since every ideal is generated by one element (or by Lemma 17.1.16). In particular fields, , and are Noetherian. The polynomial ring in infinitely many variables is not: .
Theorem 17.1.21 (Hilbert Basis Theorem). If is a Noetherian ring, then so is the polynomial ring .
Proof. Let be an ideal; by Proposition 17.1.20 we must show that is finitely generated. For , let be the set of coefficients of in the polynomials with or . Since is an ideal, is an ideal of ; its nonzero elements are the leading coefficients of the elements of of degree exactly . If has degree at most , then has degree at most , and its coefficient of is the coefficient of in ; hence . Since is Noetherian, there is an with for all .
For each , the ideal is finitely generated, so we may choose , each zero or of degree at most , whose coefficients of generate . Let be the ideal generated by all the with . We claim . If not, choose of minimal degree (note ), with leading coefficient .
- If , then , so , where is the coefficient of in . The polynomial lies in , has degree less than , and is not in (as ).
- If , then , so , and lies in and has degree less than .
Either way the minimality of is contradicted. So is finitely generated. ∎
Corollary 17.1.22 (Polynomial rings in several variables). For any Noetherian ring and , the ring is Noetherian. In particular is Noetherian for every field , and so is .
17.1.7 Polynomials over a Unique Factorization Domain
Gauss’s lemma compares factorizations in and in . A nonzero polynomial is primitive if no prime number divides all of its coefficients. Every nonzero can be written as with and primitive: multiply by a common denominator of its coefficients, and divide by the greatest common divisor of its coefficients, so that .
Lemma 17.1.23 (Gauss’s lemma). The product of two primitive polynomials in is primitive.
Proof. Let be primitive, and suppose a prime divides every coefficient of . Reducing coefficients mod is a ring homomorphism , , and . Since is a field, is an integral domain, so or . That is, divides every coefficient of or every coefficient of , a contradiction. ∎
Proposition 17.1.24 (Factorizations over and over ). If is a product of nonconstant polynomials , then with , , and .
Proof. Write and with and primitive. Then , and is primitive by Gauss’s lemma. Write in lowest terms with , so that . If a prime divides , then divides every coefficient of ; as , it divides every coefficient of , which is impossible. So , and . ∎
Proposition 17.1.25 (Eisenstein’s criterion). Let with , and let be a prime such that , for all , and . Then is irreducible in .
Proof. The units of are the nonzero constants, so if were reducible in it would be a product of two nonconstant polynomials, and by Proposition 17.1.24 we could write with of degrees . The product of the leading coefficients of and is , so divides neither of them, and the reductions mod satisfy , . Now in . This ring is a Euclidean domain, hence a UFD (Theorems 17.1.12 and 17.1.17), and is irreducible in it; so by unique factorization and for some . Since , the constant terms of and vanish, that is, divides the constant terms and . Then divides , a contradiction. ∎
For instance, , , and are irreducible in (take ).
The same ideas give a general theorem. For an integral domain , the fraction field is built from as is built from (and as is built from in Chapter 4): its elements are the fractions with , , where if and only if .
Theorem 17.1.26 (Polynomial rings over a UFD). If is a unique factorization domain, then so is . In particular, and (for a field ) are unique factorization domains.
Proof (sketch). In a UFD, greatest common divisors exist (up to units), so one can define primitive polynomials in as above. Every prime element of gives an integral domain (Proposition 17.1.5), so the proof of Gauss’s lemma works with in place of , and so does Proposition 17.1.24, with in place of . It follows that the irreducible elements of are the irreducible elements of and the primitive polynomials that are irreducible in . Factorizations in are then obtained, and shown to be unique, from those in and in , which is a UFD because it is a Euclidean domain. The second statement follows by induction on , since . ∎
Note. We have shown the implications and none of them reverses. The ring is a PID but not Euclidean. The ring is a UFD by Theorem 17.1.26 but not a PID: if , then forces or , and forces ; but , since every element of has an even constant term. Finally is an integral domain but not a UFD (Example 17.1.18).
Exercise 17.1.27 (Exercises).
- Show that is not a PID by finding an ideal that is not principal. (Hint: consider .)
- Prove that is a Euclidean domain with . (The key step: for any with , the quotient lies within distance of some Gaussian integer.)
Show solution
(1) Let and . First, . Indeed, the right side is an additive subgroup containing and , and it is closed under multiplication by , since and ; so it is an ideal containing . Conversely, if , then with , so it lies in . In particular , and .
Suppose . Then and , so by multiplicativity of the norm (Example 17.1.18), divides and , hence divides 2. No element has norm 2, so , , and , a contradiction. So is not principal.
(2) is an integral domain, being a subring of . Given with , write with . Choose integers with and , and set and . Since is multiplicative on , So with or .