Let $R$ be a commutative ring and $I,J$ ideals of $R$ such that $J$ is finitely generated and the rings $R/I$ and $R/J$ are Noetherian. Are the $R$-modules $R/J$, $J/IJ$, $R/IJ$ Noetherian? Is the ring $R/IJ$ is Noetherian?
[Math] Noetherian quotient rings
commutative-algebranoetherian
Related Solutions
Even for algebras over finite fields, “tensor products of Noetherian rings are Noetherian” may fail dramatically. Assume for example that $K=F((x_i)_{i \in B})$ is a function field. When $B$ is finite, then $K \otimes_F K$ is a localization of $F[(x_i)_{i \in B}, (x'_i)_{i \in B}]$, thus noetherian. Now assume that $B$ is infinite. Then $\Omega^1_{K/F}$ has dimension $|B|$. Since it is isomorphic to $I/I^2$, where $I$ is the kernel of the multiplication map $K \otimes_F K \to K, x \otimes y \mapsto x \cdot y$, it follows that $I$ is not finitely generated, hence $K \otimes_F K$ is not noetherian.
The general case treated in the following paper:
P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35
Here is a selection of some results of that paper: Let $K,L$ be extensions of a field $F$.
- If $K$ is a finitely generated field extension of $F$, then $K \otimes_F L$ is noetherian.
- If $K,L \subseteq F^{\mathrm{alg}}$ are separable algebraic extensions of $F$, and $L$ is normal, then $K \otimes_F L$ is noetherian iff $K \otimes_F L$ is a finite product of fields iff $[K \cap L : F] < \infty$.
- If there is an extension $M$ of $F$ which sits inside $K$ and $L$, which has a strictly ascending chain of intermediate fields, then $K \otimes_F L$ is not noetherian.
- If $K \otimes_F L$ is noetherian, then $\min(\mathrm{tr.deg}_F(K),\mathrm{tr.deg}_F(L)) < \infty$.
- $K \otimes_F K$ is noetherian iff the ascending chain condition holds for intermediate fields of $K/F$ iff $K$ is a finitely generated field extension of $F$.
You should rethink the implication "noetherian $\Rightarrow$ finitely generated". Noetherianness only implies that every ideal is finitely generated as an ideal, but not necessarily as a ring. Every field is noetherian, for example, but an uncountable field cannot be finitely generated. In fact, no infinite field is finitely generated as a ring ; see If a ring is Noetherian, then every subring is finitely generated? for details.
Best Answer
Sebastian did the $R/J$ case in the comments. Now $J/IJ$ is $J\otimes R/I$ which, since $J$ is finitely generated and $R/I$ is Noetherian (as an R-module), is a Noetherian R-module. The exact sequence $$ 0\rightarrow J/IJ \rightarrow R/IJ\rightarrow R/J\rightarrow 0$$ and the previous two results show $R/IJ$ is Noetherian.