Chains of Modules, Quotient Modules are Noetherian/Artinian

Quotient and Kernel

Let M be a noetherian module, or left module if you prefer, and let Q be a quotient module of M with kernel K. Both K and Q are noetherian. An infinite chain in Q lifts to an infinite chain in M, and an infinite chain in K also lives in M. Thus acc or dcc propagates down to quotient and kernel.

It is possible to put this in reverse, but first we need a lemma. Assume, for a moment, that each submodule of M corresponds uniquely to its image in Q and its intersection with K. Map an ascending chain in M onto an ascending chain in Q, and intersect with K to find an ascending chain in K. At each step the image in Q or the intersection with K must increase. Thus K and Q noetherian implies M is noetherian, and similarly for artinian. As a bonus, the length of the chain in M, ascending or descending, is no longer than the sum of the lengths of the chains in K and in Q. The length of M, if it has a well defined length across all possible chains, is bounded by the length of K plus the length of Q.

And now for the all important lemma. Assume an image Q0 and an intersection K0, and build the submodule from the ground up.

The submodule includes K0, and at least one coset of K0 for each coset of K representing Q0. Our submodule cannot contain anything beyond K0 in K, and if it contains x+K0 and y+K0 in a particular coset of K, it contains x-y in K, which is a contradiction. Therefore the submodule is determined by Q0 and K0, and the prior results apply.

Short Exact

In the following short exact sequence, M is noetherian iff K and Q are noetherian. This is merely a restatement of the kernel quotient theorem, using different notation.

0 → K → M → Q → 0

Noetherian Quotient Rings

If R is a noetherian ring and Q is the quotient ring, Q is a noetherian R module. (If R is not commutative, assume it is left noetherian, and work with left modules.) Now the action of x on Q, for some x in R, depends only on the image of x in Q. In other words, Q is a noetherian Q module, and a noetherian ring. Divide a noetherian ring by any ideal and find another noetherian ring. Similarly, the image of an artinian ring is artinian.

Combining Noetherian Quotients

Let A and B be submodules of M, such that M/A and M/B are noetherian. If H = A∩B, is M/H noetherian?

To simplify notation, mod out by H. With H factored out, A and B are disjoint, except for 0, and M/A and M/B are still noetherian.

If two distinct elements u and v are in A, and are in the same coset of B, then their difference gives an element that is in both A and B, which is a contradiction. Every element in A represents a unique coset of B, and by symmetry, every element of B represents a unique coset of A.

Suppose S is an infinite ascending chain of submodules in M. Divide through by A and the sequence eventually becomes constant. Divide through by B and the sequence eventually becomes constant. Beyond some point, both image sequences are constant. In other words, Sj and Sj+1 have the same image mod A, and the same image mod B.

Suppose x is in Sj+1, but not in Sj. This because Sj+1 properly contains Sj. Now x does not create any new elements in the image M/A. Thus Sj already contains y, which is in the same coset as x. Subtract to find z, an element in the kernel A. Clearly z is a new element, just as x is a new element. When Sj+1 is intersected with A, we find a larger submodule, containing z. This goes on forever, hence A is not noetherian.

Let Tj = Sj∩A. We just showed that T is an infinite ascending sequence of submodules inside A. However, each Tj is its own submodule in M/B. This because each x in Tj represents a unique coset of B. Therefore M/B is not noetherian, and that is a contradiction. Our original sequence S cannot exist, and M is noetherian after all.

Direct Product

Let M be the direct product of two modules. Let the first component act as kernel, while the second component is the quotient. Thus M is noetherian/artinian iff the same holds for both components. Use induction to generalize this to a finite direct product of modules.

Finitely Generated

Let R be noetherian and let M be finitely generated over R. Write M as the homomorphic image of F, a free R module. Since F is a finite direct product it is noetherian, and M, the quotient of F, is also noetherian. The same proof works for artinian.

In summary, finitely generated over a noetherian/artinian ring remains noetherian/artinian.

Beyond this, finitely generated over a noetherian ring is finitely presented. Write M as the quotient of a finitely generated free R module F, which is noetherian. The kernel K is a submodule of F, and is also noetherian, hence finitely generated, which makes M finitely presented.

Localization

Let S be a multiplicatively closed set in the center of R. An infinite chain of left ideals in R/S pulls back to an infinite chain of left ideals in R. Thus the properties of noetherian or artinian pass to the fraction ring.

Matrix Ring

The n×n matrices over R are noetherian or artinian, according to R. After all, the matrix ring is a free R module of rank n2.