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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. –  Brad Oct 6 '12 at 2:06
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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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Oh, I see. I guess I was assuming that maximal ideals are prime in all rings, but this only holds for commutative rings. –  user15464 Oct 6 '12 at 2:20
    
Yes indeed. ${ }$ –  Martin Argerami Oct 6 '12 at 5:34
    
@user15464 All maximal ideals are prime in noncommutative rings as well, but you do need the right defintion of prime ideal that correctly generalizes the commutative one. –  rschwieb Oct 6 '12 at 14:00
    
@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") –  rschwieb Oct 6 '12 at 14:26
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