# Maximal ideals in matrix rings

Let $F$ be a field, $R$ the ring of matrices over $F$. I am running into an apparent contradiction with regard to the maximal ideals of $R$. On one hand, we know that $R$ is simple, so its Jacobson radical is trivial.

On the other hand, $R$ possesses nilpotent elements (e.g. strictly upper triangular matrices). If $A\in R$ is nilpotent and $\mathfrak m \subset R$ is a maximal ideal, then $A^n = 0\in \mathfrak m$ for some integer $n$, so since $\mathfrak m$ is prime, either $A^{n-1}$ or $A$ is in $\mathfrak m$. Inductively, we infer that $A$ is in $\mathfrak m$. Therefore $A$ is in the intersection of the maximal ideals of $R$, namely the Jacobson radical of $R$.

How can I resolve this apparent contradiction? Is the Jacobson radical trivial or is it not?

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Note that the Jacobson radical is not always equal to the intersection of all maximal two-sided ideals, though it is equal to the intersection of all maximal left ideals, and it is equal to the intersection of all maximal right ideals. –  Brad Oct 6 '12 at 2:06

As you said, $R$ is simple, so your maximal ideal $\mathfrak m$ is 0 and it is not prime.
Yes indeed. ${ }$ –  Martin Argerami Oct 6 '12 at 5:34