This proof was part of my lecture notes in Commutative algebra and I am having trouble following it. So, I am asking the question here.
Statement: If A is noetherian and every prime ideal of A is maximal then A is artinian.
Proof: Since A is noetherian, the 0 ideals has a primary decomposition.( Can you please explain why?) Let $\sqrt(q_i)=M_i => {M_i}^{n_i} \subseteq q_i$ for some $n_i$. ${M_1}^{n_1}...{M_r}^{n_r} \subseteq {M_1}^{n_1} \cap... \cap {M_n}^{n_r}$ ( Can you please tell how to deduce this?) $\subseteq q_1 \cap ... \cap q_r =(0) => A $is artinian as A is given noetherian.( How does A being noetherian proves that A is artinian?)
I shall be really thankful for the help provided!
All ideals in Noetherian rings are the finite intersection of primary ideals and $(0)$ is an ideal. This is a very important theorem here is the link to the wikipedia page https://en.wikipedia.org/wiki/Primary_decomposition
Another result is that in a Noetherian ring $R$ for every ideal $I\subseteq r$ there is an $n\in \mathbb{Z}_+$ such that $\sqrt{I}^n\subseteq I$ (just take a finite set of generators of the radical and observe that each of them have a power in $I$)
Product of ideals is contained in the intersection of ideals
The last step requires some context in the theorems you covered. You have that a ring is Artinian if and only if it is Noetherian. Or that a local ring with nilpotent maximal ideal is artinian. each of the rings $R/M_i^{n_i}$ are local artinian and then you use the chinese remainder to show that $R$ is isomorphic to their product (you need to take a special primary decomposition though)