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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.2012-10-06

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As you said, $R$ is simple, so your maximal ideal $\mathfrak m$ is 0 and it is not prime.

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    @user15464 and I should have added that 0 *is* prime with that definition. Sometimes the "commutative prime" definition is renamed to "completely prime ideal" in noncommutative texts. Completely prime implies prime, but not conversely, as in your example. The real thing at issue here is that nilpotent elements no longer have that nice relationship with the "commutative prime" definition that they used to have. Instead, they are replaced with "nilpotent right ideals". The analogous statement is that this ring has no nilpotent right ideals (not "has no nilpotent elements")2012-10-06