# Topics in Abstract Algebra/Commutative algebra

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

## Spec A

editThe set of all nilpotent elements in forms an ideal called the *nilradical* of . Given any ideal , the pre-image of the nilradical of is an ideal called the radical of and denoted by . Explicitly, if and only if for some .

**Proposition A.14.**

- (i)
- (ii)

*Proof.* Routine.

**Exercise.**

**Exercise.**

**Proposition A.2.**

*Proof.* Let . Since is in , which is prime, . Thus, we can write . Since is prime, is a domain. Hence, .

**Lemma.**

- implies

*Proof.* ( ) Clear. ( ) Let be the image of in . Suppose is a zero-divisor; that is, for some . Let , and . Since , and is strictly larger than , by the hypothesis, . That is, .

**Theorem A.11 (multiplicative avoidance).**

*Proof.* Let be a maximal element in the set of all ideals disjoint from . Let and be ideals strictly larger than . Since is maximal, we find and . By the definition of , ; thus, . By the lemma, is prime then.

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

**Exercise.**

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

A *Goldman domain* is a domain whose field of fractions is finitely generated as an algebra. When is a Goldman domain, *K* always has the form . Indeed, if , let . Then .

**Lemma.**

*Proof.* ( ) Let , and . If is disjoint from , then, by the lemma, there is a prime ideal disjoint from , contradicting the hypothesis. Thus, contains some power of , say, . Then and so are invertible in ( ) If is a nonzero prime ideal, it contains a nonzero element, say, . Then we can write: , or ; thus, .

A prime ideal is called a *Goldman ideal* if is a Goldman domain.

**Theorem A.21.**

*A*containing

*Proof.* By the ideal correspondence, it suffices to prove the case . Let . Let . Since is not nilpotent (or it will be in ), by multiplicative avoidance, there is some prime ideal not containing . It remains to show it is a Goldman ideal. But if is a nonzero prime, then since collapses to zero if it is disjoint from . By Lemma, the field of fractions of is obtained by inverting and so is a Goldman ideal. Hence, the intersection of all Goldman ideals reduces to zero.

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

**Lemma.**

**Theorem.**

- For any , is the intersection of all maximal ideals containing .
- Every Goldman ideal is maximal.
- Every maximal ideal in contracts to a maximal ideal in .

*Proof.* Clear.

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

**Lemma.**

*Proof.* Let be the fields of fractions of and , respectively.

**Theorem A.19.**

*Proof.* Let be a Goldman ideal, and . It follows from Lemma something that is a Goldman domain since it is contained in a , a Goldman domain. Since is a Hilbert-Jacobson ring, is maximal and so is a field and so is a field; that is, is maximal.

**Theorem A.5 (prime avoidance).**

*Proof.* We shall induct on to find that is in no . The case being trivial, suppose we find such that for . We assume ; else, we're done. Moreover, if for some , then the theorem applies without and so this case is done by by the inductive hypothesis. We thus assume for all . Now, ; if not, since is prime, one of the ideals in the left is contained in , contradiction. Hence, there is in the left that is not in . It follows that for all . Finally, we remark that the argument works without assuming and are prime. (TODO: too sketchy.) The proof is thus complete.

An element *p* of a ring is a *prime* if is prime, and is an irreducible if either or is a unit..

We write if , and say divides . In a domain, a prime element is irreducible. (Suppose . Then either or , say, the former. Then , and . Canceling out we see is a unit.) The converse is false in general. We have however:

**Proposition.**

*Proof.* Let be an irreducible, and suppose and . Since implies and , there is a such that . But then and so ( is an irreducible.) Thus, .

**Theorem A.16 (Chinese remainder theorem).**

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

**Proposition A.6.**

*Proof.* Let be in the Jacobson radical. If is not a unit, it is in a maximal ideal . But then we have: , which is a sum of elements in ; thus, in , contradiction. Conversely, suppose is not in the Jacobson radical; that is, it is not in some maximal ideal . Then is an ideal containing but strictly larger. Thus, it contains , and we can write: with and . Then , and would cease to be proper, unless is a non-unit.

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.**

**Theorem A.17 (Hopkins).**

*A*be a ring. Then the following are equivalent.

*A*is artinian*A*is noetherian and every prime ideal is maximal.- is finite and discrete, and is noetherian for all maximal ideal .

*Proof.* (1) (3): Let be prime, and . Since is artinian (consider the short exact sequence), the descending sequence stabilizes eventually; i.e., for some unit *u*. Since is a domain, is a unit then. Hence, is maximal and so is discrete. It remains to show that it is finite. Let be the set of all finite intersections of maximal ideals. Let be its *minimal* element, which we have by (1). We write . Let be an arbitrary maximal ideal. Then and so by minimality. Thus, for some *i*. (3) (2): We only have to show is noetherian.

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

**Proposition A.17.**

- is local.
- For every , either or is a unit.
- The set of non-units is an ideal.

*Proof.* (1) (2): If is a non-unit, then is the Jacobson radical; thus, is a unit by Proposition A.6. (2) (3): Let , and suppose is a non-unit. If is a unit, then so are and . Thus, is a non-unit. Suppose are non-units; we show that is a non-unit by contradiction. If is a unit, then there exists a unit such that . Thus either or is a unit, whence either or is a unit, a contradiction. (3) (1): Let be the set of non-units. If is maximal, it consists of nonunits; thus, where we have the equality by the maximality of .

**Example.**

**Example.**

Let be a local noetherian ring.

A. Lemma

- (i) Let be a proper ideal of . If is a finite generated -module, then .
- (ii) The intersection of all over is trivial.

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

- where are in ,

and thus

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

An ideal is said to be *primary* if every zero-divisor in is nilpotent. Explicitly, this means that, whenever and , . In particular, a prime ideal is primary.

**Proposition.**

*Proof.* The first part is clear. Conversely, if is maximal, then is a maximal ideal in . It must be unique and so is local. In particular, a zero-divisor in is nonunit and so is contained in ; hence, nilpotent.

**Exercise.**

**Theorem A.8 (Primary decomposition).**

*Proof.* Let be the set of all ideals that is not a finite intersection of primary ideals. We want to show is empty. Suppose not, and let be its maximal element. We can write as an intersection of two ideals strictly larger than . Indeed, since is not prime by definition in particular, choose and such that . As in the proof of Theorem A.3, we can write: where is the set of all such that . By maximality, . Thus, they are finite intersections of primary ideals, but then so is , contradiction.

**Proposition.**

## Integral extension

editLet be rings. If is a root of a monic polynomial , then is said to be *integral over* . If every element of is integral over , then we say is *integral over* or is an *integral extension* of . More generally, we say a ring morphism is *integral* if the image of is integral over . By replacing with , it suffices to study the case , and that's what we will below do.

**Lemma A.9.**

- is integral over .
- is finite over .
- is contained in an -submodule of that is finite over .

*Proof.* (1) means that we can write:

Thus, spans . Hence, (1) (2). Since (2) (3) vacuously, it remains to show (3) (1). Let be generated over by . Since , we can write

where . Denoting by the matrix , this means that annihilates . Hence, by (3). Noting is a monic polynomial in we get (1).

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 containing . (Proof: if and are integral elements, then and are contained in , finite over .) It is also clear that integrability is transitive; that is, if is integral over and is integral over , then is integral over .

**Proposition.**

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

*Proof.* (i) Suppose is a field, and let . Since and is integral over , we can write:

Multiplying both sides by we see . For the rest, let . We have an integral equation:

- .

Since is a domain, if is the minimal degree of a monic polynomial that annihilates , then it must be that . This shows that , giving us (ii). Also, if is a field, then is invertible and so is .

**Theorem (Noether normalization).**

- is integral over .
- are algebraically independent over .
- are a separating transcendence basis of the field of fractions of if is separable over .

**Exercise A.10 (Artin-Tate).**

**Exercise.**

## Noetherian rings

edit**Exercise.**

The next theorem furnishes many examples of a noetherian ring.

**Theorem A.7 (Hilbert basis).**

*Proof.* By induction it suffices to prove is noetherian. Let . Let be the set of all coefficients of polynomials of degree in . Since , there exists such that

- .

For each , choose finitely many elements of whose coefficients generate . Let be an ideal generated by for all . We claim . It is clear that . We prove the opposite inclusion by induction on the degree of polynomials in . Let , the leading coefficient of and the degree of . Then . If , then

In particular, if , then has degree strictly less than that of and so by the inductive hypothesis . Since , then. If , then and the same argument shows .

**Exercise.**

Let be a noetherian local ring with . Let . Then is called an *ideal of definition* if is artinian.

**Theorem.**

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

**Theorem.**

*Proof.* By induction, it suffices to prove the case .

**Theorem.**

*Proof.* By the noether normalization lemma, is integral over where are algebraically independent over . Thus, . On the other hand, .

**Theorem.**

*Proof.* ( ) By Theorem A.10, contains a prime element . Then

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

**Theorem.**

**Theorem A.10 (Krull's intersection theorem).**

**Theorem A.15.**

- for some .

*Proof.* It suffices to prove this when . Thus, the proof reduces to proving that the nilradical of *A* is nilpotent. Since is nilpotent, we have finitely many nilpotent elements that spans . The power of any linear combination of them is then a sum of terms that contain the high power of some if we take the sufficiently high power. Thus, is nilpotent.

**Proposition A.8.**

**Corollary.**

## Zariski topology

editGiven , let . (Note that .) It is easy to see

- , and .

It follows that the collection of the sets of the form includes the empty set and and is closed under intersection and finite union. In other words, we can define a topology for by declaring to be closed sets. The resulting topology is called the *Zariski topology*. Let , and write .

**Proposition A.16.**

- (i) is quasi-compact.
- (ii) is canonically isomorphic to .

*Proof.* We have: .

**Exercise.**

**Corollary.**

**Theorem A.12.**

## Integrally closed domain

edit**Lemma A.8.**

**Proposition A.9.**

*Proof.* Let be an irreducible, and suppose . Then . If , is a unit, the case we tacitly ignore. Thus, by the lemma, , say, is a nonunit. Since is irreducible, and so .

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

**Theorem (undefined: ACC).**

*A*be a ring that satisfies the ascending chain conditions on principal ideals (example: noetherian ring). Then every

*x*in is a finite product of irreducibles.

**Theorem A.10.**

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

*Proof.* (3) (2): Let . If is nonzero, it then contains a nonzero element *x*, which we factor into irreducibles: . Then for some . Finally, irreducibles are prime since is a GCD domain. (2) (1): Let be the set of all products of prime elements. Clearly, satisfies the hypothesis of Theorem A.11 (i.e., closed under multiplication). Suppose, on the contrary, there is a nonzero nonunit . It is easy to see that since , and are disjoint. Thus, by Theorem A.11, there is a prime ideal containing and disjoint from . But, by (2), contains a prime element ; that is, intersects , contradiction. (1) (3): By uniqueness of factorization, it is clear that is a GCD domain.

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

**Corollary.**

**Theorem A.13 (Nagata criterion).**

*A*be a domain, and a multiplicatively closed subset generated by prime elements. Then is a UFD if and only if is a UFD.