Noetherian Ring where all primes are maximal I am trying to show that a commutative noetherian ring where all prime ideals are maximal is artinian. 
I know that every ideal contains a finite product of primes and also as all primes are maximal we have that $Kdim=0$ 
I think that I should be able to use that every ideal contains a  finite product of primes to do something but I am unsure?
Thanks for any help
 A: I thought I would post the solution that I now have (which I think is essentially just what is said in comments/ answer above) for completeness.
We will use the following three results.

If we have the following short exact sequence:
$0\xrightarrow{\ \ \ } M_1 \xrightarrow {\ \alpha \ } M_2 \xrightarrow{\ \beta\ } M_3 \xrightarrow{\ \ \ } 0$ of $R$ modules then if $M_2$ is artinian then so are $M_1$ and $M_3$

To see this we just note that if we had a chain that did not terminate in $M_1$ then its image under $\alpha$ would not terminate in $M_2$ ( as $\alpha$ is injective) similarly if we had a non-terminating decending chain in $M_3$ then we could just lift it back to $M_2$ via $\beta^{-1}$. Both of these lead to a contradiction and hence result.

If $R$ is Artinian and $M$ is a f.g. $R$ module then $M$ is Artinian

To see this we note that if $R$ is artinian then so is $R^n$ and so for any f.g. module $M=x_1R+\ldots +x_nR$ we have the following short exact sequence:
$$0\xrightarrow{\ \ \ } Ker \xrightarrow{\ \ \ } R^n\xrightarrow{\ \ \ } M \xrightarrow {\ \ \ }0$$
and so from above $M$ is artinian.

Let $R$ be a ring with ACC on two sided ideals. Let $I\neq R$ be an ideal of $R$. Show that >there exists prime ideals $P_1,\ldots , P_n$ such that $P_1\ldots P_n\subset I$

Let $\mathcal{F}=\{L\lhd R| \mbox{L does not contain a product of primes}\}$. Then each chain has a max element by the ACC condition and so from Zorn's Lemma we have that this family has a max element. Denote it as $M$.
Now as $M$ is not prime (else it would not be in $\mathcal{F}$) we have that there exists two ideals $A,B\lhd R$ such that $AB\not\lhd M$ but $AB\lhd M$
This then gives that $M\subset (A+M)$ and $M\subset (B+M)$ so that both $(A+M)$ and $(B+M)$ are not in $\mathcal{F}$, which means that they contain a product of primes
However from the above condition we have that $(A+M)(B+M)=AB+AM+MB+M^2\subseteq M$ and as from above we have that $(A+M)(B+M)$ contains a product of primes so does $M$ hence the family $\mathcal{F}$ is empty and we are done.

Showing that $R$ is Artinian.

From above we know that there exists $P_1\ldots P_n=0$ and so we have the following sequence:
$$0=P_1\ldots P_n \subseteq P_1\ldots P_{n-1} \subseteq \ldots \subseteq P_1\subseteq R$$
Now as each $P_i$ is maximal we have that $R/P_i$ is a field and hence it is Artinian.
We also have that $P_1\ldots P_j/P_1\ldots P_jP_{j+1}$ is finitley generated (as $R$ is Noetherian) and has the natural module structure over $R/P_{j+1}$ and hence from above it is Artinian.
We then move up the chain noting that at each stage $P_1\ldots P_j/P_1\ldots P_jP_{j+1}$ is Artinian and $P_1\ldots P_{j+1}$ is artinian (from the previous step) and so $P_1\ldots P_J$ is artinian.
Then we have that $P_1$ is artinian and $R/P_1$ is artinian and hence $R$ is Artinian as required.
A: Let $R$ be the ring.
Proof: (which isn't too long, but is too high powered)


*

*Looking at a minimal primary decomposition of $\{0\}$, we find there are finitely many minimal primes of $R$, and these are maximal. Hence there are only finitely many maximal ideals, and $R$ is semilocal.

*since maximals=primes, the Jacobson radical and nilradical coincide.

*By Levitzky's theorem, the Jacobson radical is nilpotent, so the ring is semiprimary

*By the Hopkins-Levitzki theorem, $R$ is Artinian.



You mentioned this: 

I think that I should be able to use that every ideal contains a finite product of primes to do something but I am unsure?

That is essentially a piece of the puzzle for some formulation of Hopkins-Levitzki, where you need that $\{0\}$ is a finite product of prime ideals. This is the approach you can find in Atiyah-Macdonald where they give another proof in theorem 8.5. It's a little harder to write out, and already well-written there, so I think I will not duplicate their work.
