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

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?

## 2 Answers

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.

• Ah! Yeah I have seen that Artinian prime rings are simple. I think I was wondering if there a direct proof in this case. But thanks! May 28, 2019 at 23:20

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'.