Originally published at 狗和留美者不得入内. You can comment here or there.

Below we shall restrict to commutative rings.

**Proposition 1** The following conditions of a ring are equivalent.

- is finitely generated.
- Every chain of ideals is such that there exists such that . We call this the ascending chain condition (ACC).

*Proof*: Suppose (2) is not satisfied, which would means that there exists a chain of ideals that is an infinite subchain of the such that for all . If were finitely generated by elements , then since for each there is an such that , we have for , , which satisfies the ascending chain condition, contradicting our hypothesis that it does not. This proves that (1) implies (2).

If is not finitely generated, then there exists an infinite generating set such that no finite subset of it generates . This means we can indefinitely select an element not already included in our ideal and let be the smallest ideal containing , starting from any principal ideal, which violates the (ACC), thereby proving that (2) implies (1).

**Definition 1** A ring that satisfies one of the two equivalent conditions in Proposition 1 is a *Noetherian ring*.

**Definition 2** A *prime ideal* of a ring is an ideal satisfying the following two properties.

- For any , if , either or .
- is a proper ideal of .

**Definition 3** A *primary ideal* of a ring is an ideal such that if , either or for some .

**Proposition 2** A prime ideal is also a primary ideal.

*Proof*: Trivial.

**Proposition 3** The prime ideals of the ring are the prime numbers, and the primary ideals of are the ideals generated by powers of primes as well as by itself.

*Proof*: Bezout’s identity tells us that is a principal ideal ring. Every ideal of of the form , where is prime satisfies the definition of prime ideal directly by the definition of prime number. In the case of , where for non-unit elements , Property 1 would be violated. In the case of , Property 2 would be violated.

Ideal trivially satisfies the criteria for a primary ideal. As for for prime and , any element in it is of the form for some . If a factorization of it in the form , if , then the multiplicity of in is necessarily less than , which means that , which means that . Thus is primary. In the case of In the case of , where , where and are co-prime (such a factorization is guaranteed to exist for composite by the Fundamental Theorem of Arithmetic), then is not in , and no power of , which is not divisible by some prime that divides , is a multiple of . Thus is not primary.

**Definition 4** A *reducible ideal* is an ideal such that there are two strictly larger ideals such that . An *irreducible ideal* is an ideal that is not reducible.

**Definition 5** The *quotient ideal* or *colon ideal* of two ideals of ring is

We leave to the reader verify that the definition of ideal is satisfied in Definition 4.

**Definition 6** The sum of two ideals of a ring is

**Proposition 4** The sum of two ideals is also any ideal.

*Proof*: We verify the following

- Every ideal contains the zero element, the additive identity.
- If , then , where is the additive inverse of the multiplicative identity. Thus, by the distributive property, . This shows closure under multiplicative inverse.
- Suppose and . Then , the sum of an element in and an element in . This shows closure under multiplication by any element of .
- Suppose . Then, by commutativity, , which proves closure under addition.

**Proposition 5** An irreducible ideal of a Noetherian ring is necessarily a primary ideal.

*Proof*: We the contrapositive, namely that if is such that there exists for and disjoint with , then is a reducible ideal. We assume the hypothesis of the contrapositive. Because is Noetherian, for the chain of ideals

there exists such that for all . We define two ideals

which both contain properly. Thus, if we can show that , then we have shown that is reducible. Since , . To prove the other direction of inclusion, suppose that some element of expressed in the form for some , were equal to some element of expressed in the form . That tells that

If suffices to show that , which would imply that this arbitrary element in is necessarily in . To do so, we assume otherwise and then multiply both sides of by to derive

Having assumed , we thus have that when . That means that . This gives that is both in and not in the same set, a contradiction.

**Proposition 6** The intersection of an arbitrary collection of ideals of a ring , expressed as , is an ideal.

*Proof*: It suffices to show the following.

- is a subgroup of
- For every and every , .

For (1), we first note that since the additive identity is found every ideal, it is also found in any intersection of ideals. For the existence of additive inverse, any is in each of the s. With uniqueness of inverse, is found in all the s and is thus in . With similar logic, one proves the closure under addition.

For (2), we again have that if for all , for all , which implies that .

**Proposition 7** Every ideal of a ring is either irreducible, or it is the intersection of all ideals in which properly contain it, which is equal to the intersection of all ideals which properly contain and do not properly contain any ideal that properly contains .

*Proof*: Every ideal is obviously the intersection of all ideals in which contain it. We index this collection of ideals via a set , which maps bijectively to the collection of all ideals in containing , and we shall call the element in corresponding to itself as . As for , we take to be and observe that

In this case, we must have or as a singleton set. If on the other hand were a proper superset of and there were multiple elements in , then would have to be in as well, which would violate the definition of . In the case of as a singleton set, we have every ideal properly containing would also contain , which would imply that any two ideals strictly larger than when intersected, would result in some superset of , which shows that if , then must be irreducible.

**Corollary 1** An ideal in is irreducible if and only if the ideal represented by the intersection all ideals in the ring strictly larger than it is strictly larger than .

*Proof*: In the proof of Proposition 7, we showed that the right hand side of the propositional equivalence we are trying to prove implies that the left hand side of it, which is that the ideal is irreducible. Otherwise, as one can tell from the definition of in the proof of Proposition 7, any distinct represent ideals strictly larger than such that , which is a statement of reducibility of .

**Proposition 8** Every ideal of a Noetherian ring is the intersection of a finite collection of irreducible ideals.

*Proof*: If is irreducible, this proposition is trivially true, so we assume otherwise, letting . Let be the set of ideals which cannot be expressed as the intersection of a finite collection of irreducible ideals. By the ACC, every chain in constructed such that if the successor element of any ideal in the chain for which there exists an ideal in strictly larger than , eventually terminates at some . Since , must not be irreducible, which means we can express it in the form , wherein both properly contain . By the aforementioned qualification on chains of ideals in , , which means that can both be expressed as the intersection of some finite collection of irreducible ideals, which would in turn imply the same for , which implies that must be empty. This completes our proof.

**Definition 7** The *radical* of an ideal of a ring , denoted by , is given by

**Proposition 9** The radical of primary ideal is necessarily a prime ideal.

*Proof*: Let be a primary ideal. Suppose by contradiction that there exists such that neither nor are in or equivalently, , for all . Let be such that . and for all . Thus violates the definition of primary ideal.

**Definition 7** A decomposition of an ideal as an intersection of primary ideals is a *primary decomposition*. A *Lasker ring* is a ring for which every ideal has a finite primary decomposition. An *irredundant primary decomposition* into primary ideals is one such that

- Removing any of the intersected primary ideals changes the intersection.
- The radicals of the primary ideals intersected are all distinct (prime) ideals.

**Theorem 10 (Lasker-Noether theorem)** A Noetherian ring is necessarily a Lasker ring.

*Proof*: Apply Proposition 5 to Proposition 8, using the definition of Lasker ring in the process.

Interestingly, Emanuel Lasker, who proved the Lasker-Noether theorem for the special case of polynomial rings and converge power series rings in 1905, was if not the top at least one of the top chess players of his generation. Fleeing Nazism in the 1930s, he spent a couple years in his 60s in the USSR, after which he spent his remaining four years or so in America. In addition to being top at chess and a mathematician who proved a foundational theorem in abstract algebra, he also wrote philosophical works and co-wrote a drama. He was in some sense a polymath and another major data point in favor of the g-factor as well as non-diminishing returns of g with respect to intellectual achievement. As far as I am aware now, the historical chain of top chess players is along the likes of Wilhelm Steinitz -> Emanuel Lasker -> Alexander Alekhine -> Mikhail Botvinnik -> Boris Spassky -> Bobby Fischer -> Garry Kasparov -> Magnus Carlsen.