If $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

Question

Let $R$ be a commutative unitary ring and $I,J$ be two ideals of $R$. I need to prove that if $R/I$ and $R/J$ are Noetherian (resp. Artinian), then so is $R/(I \cap J)$

It seems that a direct approach won't lead anywhere, so I took the homomorphism $R \to R /I \times R/J$, $r \mapsto (r+I, r+J)$; it has a kernel $I \cap J$. So:

$$R/(I \cap J) \cong f(R)$$

Now the problem is that $f(R)$ is not an ideal of $R/I \times R/J$, though it is a subring (but I don't think that this is useful because I don't know any relevant theorems regarding subrings). I am stuck here.

Any hints or suggestions?

Answer 1

Hint. Use some exact sequence, and note that $R/I$ is a noetherian ring iff it is a noetherian $R$-module.

Edit. Your approach is also good: the map $f:R\to R/I\times R/J$ defined by $f(r)=(r+I,r+J)$ is an $R$-module homomorphism, so $R/(I\cap J)$ is isomorphic to an $R$-submodule of $R/I\times R/J$. But a direct product (sum) of finitely many noetherian (artinian) $R$-modules is also noetherian (artinian), and an $R$-submodule of a noetherian (artinian) $R$-module is noetherian (artinian).