This is a proposition in Auslander's book (Representation Theory of Artin algebras). I want proof that:
If $f$ is an essential epimorphism then Ker$f \subset rad A$, where $f: A\rightarrow B$, and $A$, $B$ are finitely generated modules over a left Artin ring.
I try to prove this proposition unsuccessfully. In the book says that is an easy consequence of the following fact: $rad A= rad\Lambda*A$, when $A$ is a finitely generated module over a left Artin ring $\Lambda$. I really appreciate any help for proving this, I use this result every time in my new studies.
Take $X$ a maximal submodule of $A$. Suppose there is $a\in$ Ker$f$ and $a\notin X$. So we have $X+$ Ker$f=A$ and consider the following:
$g:X\rightarrow A$ an imbedding and $f: A\rightarrow B$.
Since $f$ is an epimorphism, given $b\in B$ there is $a\in A$ that $b=f(a)=f(x+d)$, where $x\in X$ and $d\in$Ker$f$. Then $b=f(a)=f(x+d)=f(x)=f(g(x))$, and $fg$ is epimorphism, so on $g$ is epimorphism since $f$ is essential. But this is a contradiction with definition of $g$.
Then Ker$f\subset$ rad$A$.