In general, it is usually not the case that for a ring $R$, the Jacobson radical of $R$ has to be equal to the intersection of the maximal ideals of $R$.
However, what I do like to know is, if we are given a finite-dimensional algebra $A$ over the field $F$, is it true that the Jacobson radical of $A$ is precisely the intersection of the maximal ideals of $A$?
Best Answer
You're probably familiar with the basic result that among Artinian rings, the semiprime rings (intersection of all prime ideals $P(R)$ is zero) and semiprimitive rings (intersection of all maximal right ideals $J(R)$ is zero) coincide. These are precisely semisimple rings. So, it makes sense that they should coincide for Artinian rings. By the same basic facts, the prime ideals of Artinian rings coincide with the maximal ideals, so that justifies us using $P(R)$ above as notation for the intersection of maximal ideals.
Here's an easy way to see they coincide for a finite dimensional algebra $R$:
Since $P(R)$ is a nil ideal and $R$ is Artinian, $P(R)$ is nilpotent, and hence is contained inside $J(R)$. Since $R/P(R)$ is semisimple, and $J(R)$ is minimal among ideals which make a semisimple quotient of $R$, $J(R)\subseteq P(R)$. Thus, the two sets are equal.
Denoting the intersection of maximal ideals as $M(R)$, there's an easy way to see why $J(R)\subseteq M(R)$ for all rings. If $I$ is a maximal ideal of $R$, then $I+J(R)$ is also an ideal, but it cannot be $R$. Thus $I+J(R)=I$ implies $J(R)\subseteq I$. Taking the intersection over all $I$, $J(R)\subseteq M(R)$.