# The tensor product of two Artinian modules is Artinian

user$xxxxx$ posted (and then deleted) the following question which I think deserves to be here:

Prove that the tensor product of two Artinian modules is Artinian.

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The OP didn't mention if the ring is commutative or not, so I suggest to consider both cases. – user26857 Dec 25 '12 at 17:54
The following paper by George Bergman discusses the case where the base ring is commutative: arxiv.org/abs/1108.2520v1 – Rankeya Dec 26 '12 at 5:05
I think this question is more suitable for MathOverflow. – user18119 Dec 30 '12 at 22:53
This is a very interesting question, as it is so easy to formulate, but seemingly not as easy to answer. Did you by any chance find an answer in the meantime / post this on MO? I'd really like to know more about this :) – InvisiblePanda Jan 15 '13 at 23:46
@Randal'Thor The answer is YES, but I don't know yet an elementary proof. – user26857 Jan 16 '13 at 0:08

The non-commutative version is somewhat broken as tensor products change rings. Let $D$ be a division ring infinite dimensional over its center $K$. Then $D_D$ and ${}_DD$ are Artinian, but $D\otimes_D D = D_K$ is not.

One can already see the issue when one states the hypothesis clearly: one module is right artinian, one module is left artinian, the resulting tensor product is not even over the same ring, and has no associated chain conditions.

The comments note that a tensor product of Artinian modules over a commutative ring is both Artinian and Noetherian. A relatively elementary proof due to Facchini, Faith, and Herbera appears as (an auxiliary result) Prop 6.1 in Faith–Herbera (1997).

If $R$ is a commutative quasi-local ring and $M,N$ are Artinian $R$-modules, then the descending chains $M \geq JM \geq \dots \geq J^nM$ eventually stabilize, say at $n$ for both. Then $J^nM \otimes N = (JJ^nM) \otimes N = J^nM \otimes JN = \dots J^nM \otimes J^k N$, but each element of $N$ is annihilated by a power of $J$, so every generator in $J^nM \otimes N$ is 0, and $J^nM \otimes N = 0$. Since $J^n M \otimes N \to M \otimes N \to M/J^nM \otimes N \to 0$ is exact, we get $M \otimes N \cong M/J^nM \otimes N$. Similarly, we apply $J$ to $N$ to get $M \otimes N \cong M/J^nM \otimes N/J^nN$. Now the factors in the tensor product are both finite length, and so the tensor products length is at most the product of those two lengths, and is in particular, finite.

I am not positive how elementary the reduction to the local case is, but I assume it is well-known.

• Faith, Carl; Herbera, Dolors. “Endomorphism rings and tensor products of linearly compact modules.” Comm. Algebra 25 (1997), no. 4, 1215–1255. MR1437670 DOI:10.1080/00927879708825918
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