Non-Commutative Rings: Prime ideals in a left Artinian Ring are maximal

Question

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.

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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?

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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?

Answer 1

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.

Answer 2

How about this: If $P$ is prime, let $M$ be a maximal ideal containing $P$. By the Artinian condition, for some $n$, $M^n=M^{n+1}$, so because Artinian implies Noetherian, Nakayama's lemma gives $M^n=0$. So for some $k>0$, $M^k\subseteq P$, hence $M\subseteq P$. By maximality, $P=M$.

A good source for this kind of ring theory is Lam's book `A First Course in Noncommutative Rings'.