I've been trying to prove this fact and would be interested in someone could help me out. I have included an attempt at a proof that I don't think is completely correct but I think it could possibly be made to work.
Let $R$ be a left Artinian (not assumed to be commutative) ring with $P$ a prime ideal of $R$ and let $J$ be the Jacobson radical of $R$. Then $J$ is nilpotent since $R$ is left Artinian and so there exists some $k$ with $J^{k} = 0 \subseteq P$, and so $J \subseteq P$ since $J$ is a two-sided ideal of $R$ and $P$ is prime. Now, let $\chi$ be the set of finite intersections of maximal left ideals of $R$. Then since $R$ is left Artinian and $\chi$ is non-empty, it contains a minimal element $M = M_1 \cap \dots \cap M_k$ say, for $M_i$ some maximal ideals of $R$. Then if $M'$ is any other maximal ideal, $M' \cap M \in \chi$ and $M' \cap M \subseteq M$ and so in fact $M' = M$. But then $M$ is the intersection of all maximal ideals of $R$ since it contains every maximal ideal and is an intersection of maximal ideals. But this is exactly the statement that $M = J$ and so then
$$
M_1 \cdot \ldots \cdot M_k \subseteq M_1 \cap \dots\cap M_k = M = J \subseteq P.
$$
Now I would like to conclude, using primality of $P$, that at least one of the $M_i \subseteq P$ but the definition of prime ideal in a non-commutative ring states that whenever $I,J$ are two-sided ideals with $IJ \subseteq P$ we have either $I \subseteq P$ or $J \subseteq P$. The issue here is that the $M_i$ are only assumed to be left ideals. The wikipedia article for prime ideals states that we can drop the condition that $I,J$ are two-sided and instead only require that they are ideals on the same side, and the resulting definition is equivalent, but I haven't seen this and I'm unsure how to prove this fact. Is there a way to avoid this?
Edit: I think there is another issue with this argument. I don't think I can argue that $M_1 \cdot \ldots \cdot M_k \subseteq M_1 \cap \dots \cap M_k$ if we are only assuming that the $M_i$ are left ideals. So it seems like this style of attempt is doomed?
Best Answer
You’re right, the intersection can be different from the product. Just consider the left ideals of $R=M_2(F)$ given by $Re_{11}$ and $Re_{22}$. So that does not hold at all, even in a simple ring.
But surely you’ve covered the proposition that an Artinian prime ring is simple? From that, it is easy, because for any prime ideal $P$ in a left Artinian ring, $R/P$ is simple, hence $P$ is maximal.