# Commutative Algebra/Jacobson rings and Jacobson spaces

## Definition and elementary characterisations[edit]

**Definition 14.1**:

A **Jacobson ring** is a ring such that every prime ideal is the intersection of some maximal ideals.

Before we strive for a characterisation of Jacobson rings, we shall prove a lemma first which will be of great use in one of the proofs in that characterisation.

**Lemma 14.2**:

Let be a Jacobson ring and let be an ideal. Then is a Jacobson ring.

**Proof**:

Let be prime. Then is prime. Hence, according to the hypothesis, we may write

- ,

where the are all maximal. As is surjective, we have . Hence, we have

- ,

where the latter equality follows from implying that for all , , where and and thus . Since the ideals are maximal, the claim follows.

**Theorem 14.3**:

Let be a ring. The following are equivalent:

- is a Jacobson ring.
- Every radical ideal (see def. 13.1) is an intersection of maximal ideals.
- For every prime the Jacobson radical of equals the zero ideal.
- For every ideal , the Jacobson radical of is equal to the nilradical of .

**Proof 1**: We prove 1. 2. 3. 4. 1.

1. 2.: Let be a radical ideal. Due to theorem 13.3,

- .

Now we may write each prime ideal containing as the intersection of maximal ideals (we are in a Jacobson ring) and hence obtain 1. 2.

2. 3.: Let be prime. In particular, is radical. Hence, we may write

- ,

where the are maximal. Now suppose that is contained within the Jacobson radical of . According to theorem 13.7, is a unit within , where is arbitrary. We want to prove . Let thus be such that . Then and thus with and , that is . Let be the inverse of , that is . This means for all , and in particular, . Hence , contradiction.

3. 4.: Let . Assume there exists and a prime ideal such that , but for all maximal . Let be the canonical projection. Since preimages of prime ideals under homomorphism are prime, is prime.

Let be a maximal ideal within . Assume . Let be the canonical projection. As in the first proof of theorem 12.2, is maximal.

We claim that is maximal. Assume , that is for a suitable . Since , , contradiction. Assume is strictly contained within . Let . Then . If , then , contradiction. Hence and thus , that is .

Furthermore, if , then . Now since . Hence, , that is, , a contradiction to .

Thus, is contained within the Jacobson radical of .

4. 1.: Assume is prime not the intersection of maximal ideals. Then

- .

Hence, there exists an such that for every maximal ideal of .

The set is multiplicatively closed. Thus, theorem 12.3 gives us a prime ideal such that .

Let be a maximal ideal of that does not contain . Let be the canonical projection. We claim that is a maximal ideal containing . Indeed, the proof runs as in the first proof of theorem 12.2. Furthermore, does not contain , for if it did, then . Thus we obtained a contradiction, which is why every maximal ideal of contains .

Since within , the Jacobson radical equals the Nilradical, is also contained within all prime ideals of , in particular within . Thus we have obtained a contradiction.

**Proof 2**: We prove 1. 4. 3. 2. 1.

1. 4.: Due to lemma 3.10, is a Jacobson ring. Hence, it follows from the representations of theorem 13.3 and def. 13.6, that Nilradical and Jacobson radical of are equal.

4. 3.: Since is a radical ideal (since it is even a prime ideal), has no nilpotent elements and thus it's nilradical vanishes. Since the Jacobson radical of that ring equals the Nilradical due to the hypothesis, we obtain that the Jacobson radical vanishes as well.

3. 2.: I found no shorter path than to combine 3. 1. with 1. 2.

2. 1.: Every prime ideal is radical.

**Remaining arrows**:

1. 3.: Let be a prime ideal of . Now suppose that is contained within the Jacobson radical of . According to theorem 13.7, is a unit within , where is arbitrary. Write

- ,

where the are maximal. We want to prove . Let thus be such that . Then and thus with and , that is . Let be the inverse of , that is . This means for all , and in particular, . Hence , contradiction.

3. 1.: Let be prime. If is maximal, there is nothing to show. If is not maximal, is not a field. In this case, there exists a non-unit within , and hence, by theorem 12.1 or 12.2 (applied to where is a non-unit), contains at least one maximal ideal. Furthermore, the Jacobson radical of is trivial, which is why there are some maximal ideals of such that

- .

As in the first proof of theorem 12.2, are maximal ideals of . Furthermore,

- .

2. 4.: Let be the nilradical of . We claim that

- .

Let first , that is, . Then , that is and . The other inclusion follows similarly, only the order is in reverse (in fact, we just did equivalences).

Due to the assumption, we may write

- ,

where the are maximal ideals of .

Since is surjective, . Hence,

- ,

where the last equality follows from implying that for and and hence for all . Furthermore, the are either maximal or equal to , since any ideal of properly containing contains one element not contained within , which is why , hence and thus .

Thus, is the intersection of some maximal ideals of , and thus the Jacobson radical of is contained within it. Since the other inclusion holds in general, we are done.

4. 2.: As before, we have

- .

Let now be the Jacobson radical of , that is,

- ,

where the are the maximal ideals of . Then we have by the assumption:

- .

Furthermore, as in the first proof of theorem 12.2, are maximal.

## Goldman's criteria[edit]

Now we shall prove two more characterisations of being a Jacobson ring. These were established by Oscar Goldman.

**Theorem 14.4 (Goldman's first criterion)**:

Let be a ring. is Jacobson if and only if is.

This is the hard one, and we do it right away so that we have it done.

**Proof**:

One direction () isn't too horrible. Let be a Jacobson ring, and let be a prime ideal of . (We shall denote ideals of with a small zero as opposed to ideals of to avoid confusion.)

We now define

- .

This ideal contains exactly the polynomials whose constant term is in . It is prime since

as can be seen by comparing the constant coefficients. Since is Jacobson, for a given that is not contained within , and hence not in , there exists a maximal ideal containing , but not containing . Set . We claim that is maximal. Indeed, we have an isomorphism

via

- .

Therefore, is a field if and only if is. Hence, is maximal, and it does not contain . Since thus every element outside can be separated from by a maximal ideal, is a Jacobson ring.

The other direction is a bit longer.

We have given a Jacobson ring and want to prove Jacobson. Hence, let be a prime ideal, and we want to show it to be the intersection of maximal ideals.

We first treat the case where and is an integral domain.

Assume first that does contain a nonzero element (i.e. is not equal the zero ideal).

Assume is contained within all maximal ideals containing , but not within . Let such that is of lowest degree among all nonzero polynomials in . Since , . Since is an integral domain, we can form the quotient field . Then .

Assume that is not irreducible in . Then , , where , are not associated to . Let such that . Then . As is prime, wlog. . Hence . Thus, and are associated, contradiction.

is Euclidean with the degree as absolute value. Uniqueness of prime factorisation gives a definition of the greatest common divisor. Since is irreducible in and , . Applying the Euclidean algorithm, , . Multiplication by an appropriate constant yields , . Thus, . Hence, is contained within every maximal ideal containing . Further, .

Let be any maximal ideal of not containing . Set

- .

Assume . Then , . We divide by by applying a polynomial long division algorithm working for elements of a general polynomial ring: We successively eliminate the first coefficient of by subtracting an appropriate multiple of . Should that not be possible, we multiply by the leading coefficient of , that shall be denoted by . Then we cannot eliminate the desired coefficient of , but we can eliminate the desired coefficient of . Repeating this process gives us

- ,

for . Furthermore, since this equation implies , we must have since the degree of was minimal among polynomials in . Then

with . By moving such coefficients to , we may assume that no coefficient of is in . Further, is nonzero since otherwise . Denote the highest coefficient of by , and the highest coefficient of by . Since the highest coefficients of and must cancel out (as ),

- .

Thus, and , but , which is absurd as every maximal ideal is prime. Hence, .

According to theorem 12.2, there exists a maximal ideal containing . Now does not equal all of , since otherwise . Hence, and the maximality of imply . Further, is a maximal ideal containing and thus contains . Hence, .

Thus, every maximal ideal that does not contain contains ; that is, for all maximal ideals of . But according to theorem 12.3, we may choose a prime ideal of not intersecting the (multiplicatively closed) set , and since is a Jacobson ring, there exists a maximal ideal containing and not containing . This is a contradiction.

Let now be the zero ideal (which is prime within an integral domain). Assume that there are only finitely many elements in which are irreducible in , and call them . The element

factors into irreducible elements, but at the same time is not divisible by any of , since otherwise wlog.

- ,

which is absurd. Thus, there exists at least one further irreducible element not listed in , and multiplying this by an appropriate constant yields a further element of irreducible in .

Let be irreducible in . We form the ideal and define . We claim that is prime. Indeed, if , then and factor in into irreducible components. Since is a unique factorisation domain, occurs in at least one of those two factorisations.

Assume there is a nonzero element contained within all the , where is irreducible over . factors in uniquely into finitely many irreducible components, leading to a contradiction to the infinitude of irreducible elements of . Hence,

- ,

where each is prime and . Hence, by the previous case, each can be written as the intersection of maximal elements, and thus, so can .

Now for the general case where is an arbitrary Jacobson ring and is a general prime ideal of . Set . is a prime ideal, since if , where , then or , and hence or . We further set . Then we have

via the isomorphism

- .

Set

- and .

Then is an integral domain and a Jacobson ring (lemma 14.2), and is a prime ideal of with the property that . Hence, by the previous case,

- .

Thus, since ,

- ,

which is an intersection of maximal ideals due to lemma 12.4 and since isomorphisms preserve maximal ideals.

**Theorem 14.5 (Goldman's second criterion)**:

A ring is Jacobson if and only if for every maximal ideal , is maximal in .

**Proof**:

The reverse direction is once again easier.

Let be a prime ideal within , and let . Set

- .

Assume . Then there exist , such that

- .

By shifting parts of to , one may assume that does not have any coefficients contained within . Furthermore, if follows . Further, , since if , , , then annihilates all higher coefficients of , which is why equals the constant term of times and thus . Hence and let be the leading coefficient of . Since the nontrivial coefficients of the polynomial must be zero for it being constantly one, , contradicting the primality of .

Thus, let be maximal containing . Assume contains . Then and thus . contracts to a maximal ideal of , which does not contain , but does contain . Hence the claim.

The other direction is more tricky, but not as bad as in the previous theorem.

Let thus be a Jacobson ring. Assume there exists a maximal ideal such that is not maximal within . Define

- and . is a prime ideal, since if such that , or and hence or . Further

via the isomorphism

- .

According to lemma 12.5, is a maximal ideal within . We set

- and .

Then is a Jacobson ring that is not a field, is a maximal ideal within (isomorphisms preserve maximal ideals) and , since if is any element of which is not mapped to zero by , then at least one of must be nonzero, for, if only , then , which is absurd.

Replacing by and by , we lead the assumption to a contradiction where is an integral domain but not a field and .

is nonzero, because else would be a field. Let have minimal degree among the nonzero polynomials of , and let be the leading coefficient of .

Let be an arbitrary maximal ideal of . can not be the zero ideal, for otherwise would be a field. Hence, let be nonzero. Since , . Since is maximal, . Hence, , where and . Applying the general division algorithm that was described above in order to divide by and obtain

for suitable and such that . From the equality holding for we get

- .

Hence, , and since the degree of was minimal in , . Since all coefficients of are contained within (since they are multiplied by ), . Thus (maximal ideals are prime).

Hence, is contained in all maximal ideals of . But since was assumed to be an integral domain, this is impossible in view of lemma 12.3 applied to the set , yielding a prime ideal which is separated from by a maximal ideal since is a Jacobson ring. Hence, we have obtained a contradiction.