I would like how could one prove the following
Let $R$ be a ring, $\mathfrak{a}_1,\dots,\mathfrak{a}_r$ ideals of $R$ such that each $R/\mathfrak{a}_i$ is a Noetherian ring. Then
- $\bigoplus_{i=1}^rR/\mathfrak{a}_i$ is a Noetherian $R$-module
- If $\bigcap_{i=1}^r\mathfrak{a}_i=0$, then $R$ is a Noetherian ring
This is proved in A Term of Commutative Algebra, page 325 of the text, 16.36. These are some free lecture notes by Allen Altman and Steven Kleiman which I am studying from, however, I find their answer rather unsatisfactory, because I can't justify most of the things they assume to be true. Is there another way to prove the aforementioned facts?
Thanks in advance for your answers.
Suppose $R$ is a commutative ring and $I$ an ideal thereof. Let $S=R/I$, for simplicity.
Then every $S$-module $M$ becomes an $R$-module, by defining $rx=(r+I)x$. The verification is easy. The set of $R$-submodules of $M$ is the same as the set of $S$-submodules of $M$. Thus, if $M$ is Noetherian as an $S$-module, it is also as an $R$-module.
In your case, each $R/\mathfrak{a}_i$ is Noetherian as a ring, so also as a module over $R/\mathfrak{a}_i$, hence also as a module over $R$.
Finally, use that $R/\bigcap_{i=1}^n\mathfrak{a}_i$ embeds as an $R$-module in the direct sum $\bigoplus_{i=1}^n R/\mathfrak{a}_i$.