Topics in Abstract Algebra/Commutative algebra

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The set of all prime ideals in a commutative ring A is called the spectrum of A and denoted by \operatorname{Spec}(A). (The motivation for the term comes from the theory of a commutative Banach algebra.)

Spec A[edit]

The set of all nilpotent elements in A forms an ideal called the nilradical of A. Given any ideal \mathfrak a, the pre-image of the nilradical of A is an ideal called the radical of \mathfrak a and denoted by \sqrt{\mathfrak a}. Explicitly, x \in \sqrt{\mathfrak a} if and only if x^n \in \mathfrak a for some n.

Proposition A.14. Let \mathfrak i, \mathfrak j \triangleleft A.

  • (i) \sqrt{\mathfrak i^n} = \sqrt \mathfrak i
  • (ii) \sqrt{\mathfrak i \mathfrak j} = \sqrt{\mathfrak i \cap \mathfrak j} = \sqrt \mathfrak i \cap \sqrt \mathfrak j

Proof. Routine. \square

Exercise. A ring has only one prime ideal if and only if its nilradical is maximal.

Exercise. Every prime ideal in a finite ring is maximal.

Proposition A.2. Let A \ne 0 be a ring. If every principal ideal in A is prime, then A is a field.
Proof. Let 0 \ne x \in A. Since x^2 is in (x^2), which is prime, x \in (x^2). Thus, we can write x = ax^2. Since (0) is prime, A is a domain. Hence, 1 = ax. \square

Lemma. Let \mathfrak p \triangleleft A. Then \mathfrak p is prime if and only if

\mathfrak p \subsetneq \mathfrak a \triangleleft A, \mathfrak p \subsetneq \mathfrak b \triangleleft A implies \mathfrak a \mathfrak b \not\subset \mathfrak p

Proof. (\Rightarrow) Clear. (\Leftarrow) Let \overline{x} be the image of x \in A in A / \mathfrak p. Suppose \overline{a} is a zero-divisor; that is, \overline{a}\overline{b} = 0 for some b \in A \backslash \mathfrak p. Let \mathfrak a = (a, \mathfrak p), and \mathfrak b = (b, \mathfrak p). Since \mathfrak a \mathfrak b = ab + \mathfrak p \subset \mathfrak p, and \mathfrak b is strictly larger than \mathfrak p, by the hypothesis, \mathfrak a \subset \mathfrak p. That is, \overline{a} = 0. \square

Theorem A.11 (multiplicative avoidance). Let S \subset A be a multiplicative system. If \mathfrak a \triangleleft A is disjoint from S, then there exists a prime ideal \mathfrak p \supset \mathfrak a that is maximal among ideals disjoint from S.
Proof. Let \mathfrak m be a maximal element in the set of all ideals disjoint from S. Let \mathfrak a and \mathfrak b be ideals strictly larger than \mathfrak m. Since \mathfrak m is maximal, we find a \in \mathfrak a \cap S and b \in \mathfrak b \cap S. By the definition of S, ab \in S; thus, \mathfrak a \mathfrak b \not \subset \mathfrak m. By the lemma, \mathfrak m is prime then. \square

Note that the theorem applies in particular when S contains only 1.

Exercise. A domain A is a principal ideal domain if every prime ideal is principal.

A Goldman domain is a domain whose field of fractions K is finitely generated as an algebra. When A is a Goldman domain, K always has the form A[f^{-1}]. Indeed, if K = A[s_1^{-1}, ..., s_n^{-1}], let s = s_1...s_n. Then K = A[s^{-1}].

Lemma. Let A be a domain with the field of fractions K, and 0 \ne f \in A. Then K = A[f^{-1}] if and only if every nonzero prime ideal of A contains f.
Proof. (\Leftarrow) Let 0 \ne x \in A, and S = \{ f^n | n \ge 0 \}. If (x) is disjoint from S, then, by the lemma, there is a prime ideal disjoint from S, contradicting the hypothesis. Thus, (x) contains some power of f, say, yx = f^n. Then yx and so x are invertible in A[f^{-1}]. (\Rightarrow) If \mathfrak p is a nonzero prime ideal, it contains a nonzero element, say, s. Then we can write: 1 / s = a / f^n, or f^n = as \in \mathfrak p; thus, f \in \mathfrak p. \square

A prime ideal \mathfrak{p} \in \operatorname{Spec}(A) is called a Goldman ideal if A/\mathfrak p is a Goldman domain.

Theorem A.21. Let A be a ring and \mathfrak a \triangleleft A. Then \sqrt \mathfrak a is the intersection of all minimal Goldman ideals of A containing \mathfrak a
Proof. By the ideal correspondence, it suffices to prove the case \mathfrak a = \sqrt \mathfrak a = 0. Let 0 \ne f \in A. Let S = \{ f^n | n \ge 0 \}. Since f is not nilpotent (or it will be in \sqrt{(0)}), by multiplicative avoidance, there is some prime ideal \mathfrak g not containing f. It remains to show it is a Goldman ideal. But if \mathfrak p \triangleleft A / \mathfrak g is a nonzero prime, then f \in \mathfrak p since \mathfrak p collapses to zero if it is disjoint from S. By Lemma, the field of fractions of A / \mathfrak g is obtained by inverting f and so \mathfrak g is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero. \square

In some rings, Goldman ideals are maximal; this will be discussed in the next section. On the other hand,

Lemma. Let \mathfrak a \triangleleft A. Then \mathfrak a is a Goldman ideal if and only if it is the contraction of a maximal ideal in A[X].

Theorem. The following are equivalent.

  1. For any \mathfrak a \triangleleft A, \mathfrak a is the intersection of all maximal ideals containing \mathfrak a.
  2. Every Goldman ideal is maximal.
  3. Every maximal ideal in A[X] contracts to a maximal ideal in A.

Proof. Clear. \square

A ring satisfying the equivalent conditions in the theorem is called a Hilbert-Jacobson ring.

Lemma. Let A \subset B be domains such that B is algebraic and of finite type over A. Then A is a Goldman domain if and only if B is a Goldman domain.
Proof. Let K \subset L be the fields of fractions of A and B, respectively. \square

Theorem A.19. Let A be a Hilbert-Jacobson ring. Then A[X] is a Hilbert-Jacobson ring.
Proof. Let \mathfrak q \triangleleft A[X] be a Goldman ideal, and \mathfrak p = \mathfrak q \cap A. It follows from Lemma something that A / \mathfrak p is a Goldman domain since it is contained in a A[X] / \mathfrak q, a Goldman domain. Since A is a Hilbert-Jacobson ring, \mathfrak p is maximal and so A / \mathfrak p is a field and so (TODO: why?) A[X] / \mathfrak q is a field; that is, \mathfrak q is maximal. \square

Theorem A.5 (prime avoidance). Let \mathfrak p_1, ..., \mathfrak p_r \triangleleft A be ideals, at most two of which are not prime, and \mathfrak a \triangleleft A. If \mathfrak a \subset \bigcup_1^r \mathfrak p_i, then \mathfrak a \subset \mathfrak p_i for some \mathfrak i.
Proof. We shall induct on r to find a \in \mathfrak a that is in no \mathfrak p_i. The case r=1 being trivial, suppose we find a \in \mathfrak a such that a \not\in \mathfrak{p}_i for i < r. We assume a \in \mathfrak{p}_r; else, we're done. Moreover, if \mathfrak{p}_i \subset \mathfrak{p}_r for some i < r, then the theorem applies without \mathfrak{p}_i and so this case is done by by the inductive hypothesis. We thus assume \mathfrak p_i \not\subset \mathfrak p_r for all i < r. Now, \mathfrak a \mathfrak p_1 ... \mathfrak p_{r-1} \not\subset \mathfrak {p}_r; if not, since \mathfrak p_r is prime, one of the ideals in the left is contained in \mathfrak p_r, contradiction. Hence, there is b in the left that is not in \mathfrak {p}_r. It follows that a + b \not\in \mathfrak p_i for all i \le r. Finally, we remark that the argument works without assuming \mathfrak p_1 and \mathfrak p_2 are prime. (TODO: too sketchy.) The proof is thus complete. \square

An element p of a ring is a prime if (p) is prime, and is an irreducible if p = xy \Rightarrow either x or y is a unit..

We write x | y if (x) \ni y, and say x divides y. In a domain, a prime element is irreducible. (Suppose x = yz. Then either x | y or x | z, say, the former. Then sx = y, and sxz = x. Canceling x out we see z is a unit.) The converse is false in general. We have however:

Proposition. Suppose: for every x and y, (x) \cap (y) = (xy) whenever (1) is the only principal ideal containing (x, y). Then every irreducible is a prime.
Proof. Let p be an irreducible, and suppose p | xy and p \not| y. Since (p) \cap (x) = (px) implies px | xy and p | y, there is a d such that (1) \ne (d) \supset (p, x). But then d | p and so p | d (p is an irreducible.) Thus, p | x. \square

Theorem A.16 (Chinese remainder theorem). Let \mathfrak a_1, ..., \mathfrak a_n \triangleleft A. If \mathfrak a_j + \mathfrak a_i = (1), then

\prod \mathfrak a_i \to A \to A / \mathfrak a_1 \times \cdots \times \mathfrak a_n \to 0

is exact.

The Jacobson radical of a ring A is the intersection of all maximal ideals.

Proposition A.6. x \in A is in the Jacobson radical if and only if 1 - xy is a unit for every y \in A.
Proof. Let x be in the Jacobson radical. If 1-xy is not a unit, it is in a maximal ideal \mathfrak m. But then we have: 1 = (1 - xy) + xy, which is a sum of elements in \mathfrak m; thus, in \mathfrak m, contradiction. Conversely, suppose x is not in the Jacobson radical; that is, it is not in some maximal ideal \mathfrak m. Then (x, \mathfrak m) is an ideal containing \mathfrak m but strictly larger. Thus, it contains 1, and we can write: 1 = xy + z with y \in A and z \in \mathfrak m. Then 1 - xy \in \mathfrak m, and \mathfrak m would cease to be proper, unless 1 - xy is a non-unit. \square

Note that the nilradical is contained in the Jacobson radical, and they coincide in particular if prime ideals are maximal (e.g., the ring is a principal ideal domain). Another instance of this is:

Exercise. In A[X], the nilradical and the Jacobson radical coincide.

Theorem A.17 (Hopkins). Let A be a ring. Then the following are equivalent.

  1. A is artinian
  2. A is noetherian and every prime ideal is maximal.
  3. \operatorname{Spec}(A) is finite and discrete, and A_\mathfrak m is noetherian for all maximal ideal \mathfrak m.

Proof. (1) \Rightarrow (3): Let \mathfrak p \triangleleft A be prime, and x \in A / \mathfrak p. Since A / \mathfrak p is artinian (consider the short exact sequence), the descending sequence (x^n) stabilizes eventually; i.e., x^n = u x^{n+1} for some unit u. Since A / \mathfrak p is a domain, x is a unit then. Hence, \mathfrak p is maximal and so \operatorname{Spec}(A) is discrete. It remains to show that it is finite. Let S be the set of all finite intersections of maximal ideals. Let \mathfrak i \in S be its minimal element, which we have by (1). We write \mathfrak i = \mathfrak m_1 \cap ... \cap \mathfrak m_n. Let \mathfrak m be an arbitrary maximal ideal. Then \mathfrak m \cap \mathfrak i \in S and so \mathfrak m \cap \mathfrak i = \mathfrak i by minimality. Thus, \mathfrak m = \mathfrak m_i for some i. (3) \Rightarrow (2): We only have to show A is noetherian. \square

A ring is said to be local if it has only one maximal ideal.

Proposition A.17. Let A be a nonzero ring. The following are equivalent.

  1. A is local.
  2. For every x \in A, either x or 1 - x is a unit.
  3. The set of non-units is an ideal.

Proof. (1) \Rightarrow (2): If x is a non-unit, then x is the Jacobson radical; thus, 1 - x is a unit by Proposition A.6. (2) \Rightarrow (3): Let x, y \in A, and suppose x is a non-unit. If xy is a unit, then so are x and y. Thus, xy is a non-unit. (TODO: remains to show x+y is a nonunit.) (3) \Rightarrow (1): Let \mathfrak i be the set of non-units. If \mathfrak m \triangleleft A is maximal, it consists of nonunits; thus, \mathfrak m \subset \mathfrak i where we have the equality by the maximality of \mathfrak m. \square

Example. If p is a prime ideal, then A_p is a local ring where p is its unique maximal ideal.

Example. If \sqrt \mathfrak i is maximal, then A/\mathfrak i is a local ring. In particular, A/\mathfrak m^n, (n \ge 1)is local for any maximal ideal \mathfrak m.

Let (A, \mathfrak m) be a local noetherian ring.

A. Lemma

  • (i) Let \mathfrak i be a proper ideal of A. If M is a finite generated \mathfrak i-module, then M = 0.
  • (ii) The intersection of all \mathfrak{m}^k over k \ge 1 is trivial.

Proof: We prove (i) by the induction on the number of generators. Suppose M cannot be generated by strictly less than n generators, and suppose we have x_1, ... x_n that generates M. Then, in particular,

x_1 = a_1 x_1 + a_2 x_2 + ... + a_n x_n where a_i are in \mathfrak i,

and thus

(1 - a_1) x_1 = a_2 x_2 + ... + a_n x_n

Since a_1 is not a unit, 1 - a_1 is a unit; in fact, if 1 - a_1 is not a unit, it belongs to a unique maximal ideal \mathfrak{m}, which contains every non-units, in particular, a_1, and thus 1 \in \mathfrak{m}, which is nonsense. Thus we find that actually x_2, ..., x_n generates M; this contradicts the inductive hypothesis.\square

An ideal \mathfrak q \triangleleft A is said to be primary if every zero-divisor in A/\mathfrak q is nilpotent. Explicitly, this means that, whenever xy \in \mathfrak q and y \not\in \mathfrak q, x \in \sqrt{\mathfrak q}. In particular, a prime ideal is primary.

Proposition. If \mathfrak q is primary, then \sqrt{\mathfrak q} is prime. Conversely, if \sqrt{\mathfrak q} is maximal, then \mathfrak q is primary.
Proof. The first part is clear. Conversely, if \sqrt \mathfrak q is maximal, then \mathfrak m = \sqrt \mathfrak q / \mathfrak q is a maximal ideal in A/\mathfrak q. It must be unique and so A/\mathfrak q is local. In particular, a zero-divisor in A/\mathfrak q is nonunit and so is contained in \mathfrak m; hence, nilpotent. \square

Exercise. \sqrt{\mathfrak q} prime \not\Rightarrow \mathfrak q primary.

Theorem A.8 (Primary decomposition). Let A be a noetherian ring. If \mathfrak i \triangleleft A, then \mathfrak i is a finite intersection of primary ideals.
Proof. Let S be the set of all ideals that is not a finite intersection of primary ideals. We want to show S is empty. Suppose not, and let \mathfrak i be its maximal element. We can write \mathfrak i as an intersection of two ideals strictly larger than \mathfrak i. Indeed, since \mathfrak i is not prime by definition in particular, choose x \not\in \mathfrak i and y \not\in \mathfrak i such that xy \in \mathfrak i. As in the proof of Theorem A.3, we can write: \mathfrak i = \mathfrak j (\mathfrak i + x) where \mathfrak j is the set of all a \in A such that ax \in \mathfrak i. By maximality, \mathfrak j, \mathfrak i + x \not\in S. Thus, they are finite intersections of primary ideals, but then so is  \mathfrak i, contradiction. \square

Proposition. If (0) is indecomposable, then the set of zero divisors is a union of minimal primes.

Integral extension[edit]

Let A \subset B be rings. If b \in B is a root of a monic polynomial f \in A[X], then b is said to be integral over A. If every element of B is integral over A, then we say B is integral over A or B is an integral extension of A. More generally, we say a ring morphism f: A \to B is integral if the image of A is integral over B. By replacing A with f(A), it suffices to study the case A \subset B, and that's what we will below do.

Lemma A.9. Let b \in B. Then the following are equivalent.

  1. b is integral over A.
  2. A[b] is finite over A.
  3. A[b] is contained in an A-submodule of B that is finite over A.

Proof. (1) means that we can write:

b^{n+r} = -(b^{r+n-1} a_{n-1} + ... b^{r+1} a_1 + b^r a_0)

Thus, 1, b, ..., b^{n-1} spans A[b]. Hence, (1) \Rightarrow (2). Since (2) \Rightarrow (3) vacuously, it remains to show (3) \Rightarrow (1). Let M_{/A[b]} be generated over A by x_1, ..., x_n. Since b x_i \in M, we can write

b x_i = \sum_{j=1}^n c_{ij} x_j

where c_{kj} \in A. Denoting by C the matrix c_{ij}, this means that \det(bI-C) annihilates M. Hence, \det(bI-C) = 0 by (3). Noting \det(bI-C) is a monic polynomial in b we get (1). \square

The set of all elements in B that are integral over A is called the integral closure of A in B. By the lemma, the integral closure is a subring of B containing A. (Proof: if x and y are integral elements, then A[xy] and A[x-y] are contained in A[x, y], finite over A.) It is also clear that integrability is transitive; that is, if C is integral over B and B is integral over A, then C is integral over A.

Proposition. Let f:A \to B be an integral extension where B is a domain. Then

  • (i) A is a field if and only if B is a field.
  • (ii) Every nonzero ideal of B has nonzero intersection with A.

Proof. (i) Suppose B is a field, and let x \in A. Since x^{-1} \in B and is integral over A, we can write:

x^{-n} = -(a_{n-1}x^{-(n-1)} + ... + a_1 x^{-1} + a_0)

Multiplying both sides by x^{n-1} we see x^{-1} \in A. For the rest, let 0 \ne b \in B. We have an integral equation:

-a_0 = b^n + a_{n-1}b^{n-1} + ... + a_1 b = b(b^{n-1} + a_{n-1}b^{n-2} + ... + a_1).

Since B is a domain, if n is the minimal degree of a monic polynomial that annihilates b, then it must be that a_0 \ne 0. This shows that bB \cap A \ne 0, giving us (ii). Also, if A is a field, then a_0 is invertible and so is b. \square

Theorem (Noether normalization). Let A be a finitely generated k-algebra. Then we can find z_1, ..., z_d such that

  1. A is integral over k[z_1, ..., z_d].
  2. z_1, ..., z_d are algebraically independent over k.
  3. z_1, ..., z_d are a separating transcendence basis of the field of fractions K of A if K is separable over k.

Exercise A.10 (Artin-Tate). Let A \subset B \subset C be rings. Suppose A is noetherian. If C is finitely generated as an A-algebra and integral over B, then B is finitely generated as an A-algebra.

Exercise. A ring morphism f:A \to \Omega (where \Omega is an algebraically closed field) extends to F: A[b] \to \Omega (Answer: http://www.math.uiuc.edu/~r-ash/ComAlg/)

Noetherian rings[edit]

Exercise. A ring is noetherian if and only if every prime ideal is finitely generated. (See T. Y. Lam and Manuel L. Reyes, A Prime Ideal Principle in Commutative Algebra for a systematic study of results of this type.)

The next theorem furnishes many examples of a noetherian ring.

Theorem A.7 (Hilbert basis). A is a noetherian ring if and only if A[T_1, ... T_n] is noetherian.
Proof. By induction it suffices to prove A[T] is noetherian. Let I \triangleleft A[T]. Let L_n be the set of all coefficients of polynomials of degree \le n in I. Since L_n \triangleleft A, there exists d such that

L_0 \subset L_1 \subset L_2, ..., \subset L_d = L_{d+1} = ....

For each 0 \le n \le d, choose finitely many elements f_{1n}, f_{2n}, ... f_{m_nn} of I whose coefficients b_{1n}, ... b_{m_nn} generate L_n. Let I' be an ideal generated by f_{jn} for all j, n. We claim I = I'. It is clear that I \subset I'. We prove the opposite inclusion by induction on the degree of polynomials in I. Let f \in I, a the leading coefficient of f and n the degree of f. Then a \in L_n. If n \le d, then

a = a_1 b_{1n} + a_2 b_{2n} + ... + a_{m_n} b_{{m_n}n}

In particular, if g = a_1 f_{1n} + a_2 f_{2n} + ... + a_{m_n} f_{{m_n}n}, then f - g has degree strictly less than that of f and so by the inductive hypothesis f - g \in I'. Since g \in I', f \in I' then. If n \ge d, then a \in L_d and the same argument shows f \in I'. \square

Exercise. Let A be the ring of continuous functions f: [0, 1] \to [0, 1].A is not noetherian.

Let (A, \mathfrak m) be a noetherian local ring with k = A/\mathfrak m. Let \mathfrak i \triangleleft A. Then \mathfrak i is called an ideal of definition if A / \mathfrak i is artinian.

Theorem. \dim_k (\mathfrak m / {\mathfrak m}^2) \ge \dim A

The local ring A is said to be regular if the equality holds in the above.

Theorem. Let A be a noetherian ring. Then \dim A[T_1, ..., T_n] = n + \dim A.
Proof. By induction, it suffices to prove the case n = 1. \square

Theorem. Let A be a finite-dimensional k-algebra. If A is a domain with the field of fractions K, then \dim A = \operatorname{trdeg}_k K.
Proof. By the noether normalization lemma, A is integral over k[x_1, ..., x_n] where x_1, ..., x_n are algebraically independent over k. Thus, \dim A = \dim k[x_1, ..., x_n] = n. On the other hand, \operatorname{trdeg}_k K = n. \square

Theorem. Let A be a domain with (ACCP). Then A is a UFD if and only if every prime ideal \mathfrak p of height 1 is principal.
Proof. (\Rightarrow) By Theorem A.10, \mathfrak p contains a prime element x. Then

0 \subset (x) \subset \mathfrak p

where the second inclusion must be equality since \mathfrak p has height 1. (\Leftarrow) In light of Theorem A.10, it suffices to show that A is a GCD domain. (TODO: complete the proof.) \square

Theorem. A regular local ring is a UFD.

Theorem A.10 (Krull's intersection theorem). Let \mathfrak i \triangleleft A be a proper ideal. If A is either a noetherian domain or a local ring, then \bigcap_{n \ge 1} \mathfrak i^n = 0.

Theorem A.15. Let \mathfrak i \triangleleft A. If A is noetherian,

\sqrt \mathfrak i^n \subset \mathfrak i \subset \sqrt \mathfrak i for some n.

In particular, the nilradical of A is nilpotent.
Proof. It suffices to prove this when \mathfrak i = 0. Thus, the proof reduces to proving that the nilradical of A is nilpotent. Since A is nilpotent, we have finitely many nilpotent elements x_1, ..., x_n that spans \sqrt{(0)}. The power of any linear combination of them is then a sum of terms that contain the high power of some x_j if we take the sufficiently high power. Thus, \sqrt{(0)} is nilpotent. \square

Proposition A.8. If A is noetherian, then \hat A is noetherian.

Corollary. If A is noetherian, then A[[X]] is noetherian.

Zariski topology[edit]

Given \mathfrak a \triangleleft A, let \operatorname{V}(\mathfrak a) = \{ \mathfrak p \in \operatorname{Spec}(A) | \mathfrak p \supset \mathfrak a \}. (Note that \operatorname{V}(\mathfrak a) = \operatorname{V}(\sqrt \mathfrak a).) It is easy to see

V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a \mathfrak b)  = V(\mathfrak a \cap \mathfrak b), and \cap_\alpha V(\mathfrak a_\alpha) = V((\mathfrak a_\alpha | \alpha)).

It follows that the collection of the sets of the form \operatorname{V}(\mathfrak a) includes the empty set and \operatorname{Spec}(A) and is closed under intersection and finite union. In other words, we can define a topology for \operatorname{Spec}(A) by declaring \operatorname{Z}(\mathfrak i) to be closed sets. The resulting topology is called the Zariski topology. Let X = \operatorname{Spec}(A), and write X_f = X \backslash V ((f)) = \{ P \in X | P \ni f \}.

Proposition A.16. We have:

  • (i) X_f is quasi-compact.
  • (ii) X_{fg} is canonically isomorphic to \operatorname{Spec}(A[f^{-1}])_g.

Proof. We have: X_f \subset \bigcup_\alpha X_{f_\alpha} = X \backslash V((f_\alpha|\alpha)) \Leftrightarrow (f) \subset (f_\alpha|\alpha) \Leftrightarrow f \in (f_{\alpha_1}, ..., f_{\alpha_n}). \square

Exercise. Let A be a local ring. Then \operatorname{Spec}(A) is connected.


Corollary. \operatorname{Spec}(B) \to \operatorname{Spec}(A) is a closed surjection.

Theorem A.12. If A_m is noetherian for every maximal ideal \mathfrak m and if \{ \mathfrak m \in \operatorname{Max}(A) | x \in m \} is finite for each x \in A, then A is noetherian.

Integrally closed domain[edit]

Lemma A.8. In a GCD domain, if (x, y) = 1 = (x, z), then (x, yz) = 1.

Proposition A.9. In a GCD domain, every irreducible element is prime.
Proof. Let x be an irreducible, and suppose x | yz. Then x | (x, yz). If (x, yz) = 1, x is a unit, the case we tacitly ignore. Thus, by the lemma, d = (x, y), say, is a nonunit. Since x is irreducible, x | d and so x | y. \square

In particular, in a polynomial ring that is a GCD domain, every irreducible polynomial is a prime element.

Theorem (undefined: ACC). Let A be a ring that satisfies the ascending chain conditions on principal ideals (example: noetherian ring). Then every x in A is a finite product of irreducibles.

Theorem A.10. Let A be a domain. The following are equivalent.

  1. Every nonzero nonunit element is a finite product of prime elements.
  2. (Kaplansky) Every nonzero prime ideal contains a prime element.
  3. A is a GCD domain and has (ACC) on principal ideals.

Proof. (3) \Rightarrow (2): Let \mathfrak{p} \in \operatorname{Spec}(A). If \mathfrak{p} is nonzero, it then contains a nonzero element x, which we factor into irreducibles: x = p_1 ... p_n. Then p_j \in \mathfrak{p} for some j. Finally, irreducibles are prime since A is a GCD domain. (2) \Rightarrow (1): Let S be the set of all products of prime elements. Clearly, S satisfies the hypothesis of Theorem A.11 (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit x. It is easy to see that since x \not\in S, (x) and S are disjoint. Thus, by Theorem A.11, there is a prime ideal \mathfrak p containing x and disjoint from S. But, by (2), \mathfrak p contains a prime element y; that is, \mathfrak p intersects S, contradiction. (1) \Rightarrow (3): By uniqueness of factorization, it is clear that A is a GCD domain. \square

A domain satisfying the equivalent conditions in the theorem is called a unique factorization domain or a UFD for short.

Corollary. If A is a UFD, then A[X] is a UFD. If A is a principal ideal domain, then A[[X]] is a UFD.

Theorem A.13 (Nagata criterion). Let A be a domain, and S \subset A a multiplicatively closed subset generated by prime elements. Then A is a UFD if and only if S^{-1}A is a UFD.