Every finite-dimensional algebra which is not simple contains a maximal ideal whose annihilator is nonzero The following problem is from Chapter 3 of Drozd and Kirichenko's "Finite-Dimensional Algberas" that I am self-studying.
Let $A$ be a finite-dimensional unital algebra that is not simple. Show it contains a maximal two-sided ideal with nonzero annihilator.
The result is clear for the semisimple (and not simple) case. Further, if $Rad(A)\neq 0$ was maximal, it suffices to recall that $Rad(A)$ is nilpotent (for if $Rad(A)^n=0$, any member of $Rad(A)^{n-1}$ would annihilate $Rad(A)$).
I thought I could reduce the remaining problem ($Rad(A)\neq 0$ is not maximal) to the semisimple case by taking the quotient by the Jacobson radical, but I don't know if lifting the two-sided ideals of $A/Rad(A)$ to $A$ will yield two-sided ideals of $A$. I also don't know how I would handle $A/Rad(A)$ being simple while $A$ is not.
 A: $A$ is finite dimensional, so it is left Artinian. Hence $A$ has a minimal nonzero left ideal, say $\mathfrak m$. Let ${\mathfrak a}=\{x\in A\;|\;x\mathfrak m=0\}$ be the left annihilator of $\mathfrak m$. ${\mathfrak a}$ is a 2-sided ideal of $A$ and $\mathfrak m$ is a faithful simple module of $A/{\mathfrak a}$. This implies that $A/{\mathfrak a}$ is a left Artinian primitive ring. Left Artinian primitive rings are simple, so ${\mathfrak a}$ is a maximal $2$-sided ideal of $A$. The fact that ${\mathfrak a}$ annihilates $\mathfrak m$ implies that ${\mathfrak a}$ has a nonzero annihilator.
A: It seems everything you did is good and is also how I would approach this problem. I will help you with the two final steps that are easier than they look.

but I don't know if lifting the two-sided ideals of A/Rad(A) to A will yield two-sided ideals of A.

Let $\phi$ denote the quotient map $\phi: A \to A/J$. Now let $M$ be a two-sided ideal of $A/J$. As always $\phi^{-1}(M)$ denotes the set of all $a \in A$ such that $\phi(a) \in M$.
We show that $\phi^{-1}(M)$ is a two-sided ideal, assuming that $a, b \in \phi^{-1}(M)$ and that $r$ is a random element of $A$.

*

*To see that $a + b \in \phi^{-1}(M)$ we just notice that $\phi(a + b) = \phi(a) + \phi(b)$: the sum of two elements of $M$, hence in $M$.


*To see that $ra \in \phi^{-1}(M)$ we notice that $\phi(ra) = \phi(r)\phi(a)$ so the product of a random element in $A/J$ by an element of $M$, hence an element of $M$.


*To see that $br \in \phi^{-1}(M)$ we notice that $\phi(br) = \phi(b)\phi(r)$ hence the product of an element of $M$ with a random element of $A/J$, hence in $M$.
That's it. Pretty easy. It is just the fact that $\phi$ is a homomorphism over and over.

I also don't know how I would handle A/Rad(A) being simple while A is not.

I understand why you write this, this seems to be a much trickier issue than the previous one. But in practice you do need really worry about it because if $A/J$ is simple, then $J$ is maximal (see if you understand why!), so you already handled this case.
