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 edit

The 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. Let  .

  • (i)  
  • (ii)  

Proof. Routine.  

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   be a ring. If every principal ideal in   is prime, then   is a field.
Proof. Let  . Since   is in  , which is prime,  . Thus, we can write  . Since   is prime,   is a domain. Hence,  .  

Lemma. Let  . Then   is prime if and only if


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). Let   be a multiplicative system. If   is disjoint from  , then there exists a prime ideal   that is maximal among ideals disjoint from  .
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 domain 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. Let   be a domain with the field of fractions  , and  . Then   if and only if every nonzero prime ideal of   contains  .
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. Let   be a ring and  . Then   is the intersection of all minimal Goldman ideals of 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. Let  . Then   is a Goldman ideal if and only if it is the contraction of a maximal ideal in  .

Theorem. The following are equivalent.

  1. For any  ,   is the intersection of all maximal ideals containing  .
  2. Every Goldman ideal is maximal.
  3. 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. Let   be domains such that   is algebraic and of finite type over  . Then   is a Goldman domain if and only if   is a Goldman domain.
Proof. Let   be the fields of fractions of   and  , respectively.  

Theorem A.19. Let   be a Hilbert-Jacobson ring. Then   is a Hilbert-Jacobson ring.
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.  


To do:
Explain why   is a field (or point to a location where it can be understood why it is so...).

Theorem A.5 (prime avoidance). Let   be ideals, at most two of which are not prime, and  . If  , then   for some  .
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. Suppose: for every   and  ,   whenever (1) is the only principal ideal containing  . Then every irreducible is a prime.
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). Let  . If  , then


is exact.

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

Proposition A.6.   is in the Jacobson radical if and only if   is a unit for every  .
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. In  , 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.   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. Let   be a nonzero ring. The following are equivalent.

  1.   is local.
  2. For every  , either   or   is a unit.
  3. 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. If   is a prime ideal, then   is a local ring where   is its unique maximal ideal.

Example. If   is maximal, then   is a local ring. In particular,  is local for any maximal ideal  .

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. If   is primary, then   is prime. Conversely, if   is maximal, then   is primary.
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.   prime   primary.

Theorem A.8 (Primary decomposition). Let   be a noetherian ring. If  , then   is a finite intersection of primary ideals.
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. If   is indecomposable, then the set of zero divisors is a union of minimal primes.

Integral extension edit

Let   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. Let  . Then the following are equivalent.

  1.   is integral over  .
  2.   is finite over  .
  3.   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. Let   be an integral extension where   is a domain. Then

  • (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). Let   be a finitely generated  -algebra. Then we can find   such that

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

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

Exercise. A ring morphism   (where   is an algebraically closed field) extends to   (Answer:

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).   is a noetherian ring if and only if   is noetherian.
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 the ring of continuous functions  .  is not noetherian.

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


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

Theorem. Let   be a noetherian ring. Then  .
Proof. By induction, it suffices to prove the case  .  

Theorem. Let   be a finite-dimensional  -algebra. If   is a domain with the field of fractions  , then  .
Proof. By the noether normalization lemma,   is integral over   where   are algebraically independent over  . Thus,  . On the other hand,  .  

Theorem. Let   be a domain with (ACCP). Then   is a UFD if and only if every prime ideal   of height 1 is principal.
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. A regular local ring is a UFD.

Theorem A.10 (Krull's intersection theorem). Let   be a proper ideal. If   is either a noetherian domain or a local ring, then  .

Theorem A.15. Let  . If   is noetherian,

  for some  .

In particular, the nilradical of   is nilpotent.
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. If   is noetherian, then   is noetherian.

Corollary. If   is noetherian, then   is noetherian.

Zariski topology edit

Given  , 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. We have:

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

Proof. We have:  .  

Exercise. Let   be a local ring. Then   is connected.

Corollary.   is a closed surjection.

Theorem A.12. If   is noetherian for every maximal ideal   and if   is finite for each  , then   is noetherian.

Integrally closed domain edit

Lemma A.8. In a GCD domain, if  , then  .

Proposition A.9. In a GCD domain, every irreducible element is prime.
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). Let 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. 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.   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. If   is a UFD, then   is a UFD. If A is a principal ideal domain, then   is a UFD.

Theorem A.13 (Nagata criterion). Let A be a domain, and   a multiplicatively closed subset generated by prime elements. Then   is a UFD if and only if   is a UFD.