Let $R$ be a noetherian ring. Let $A$ be an $R$-algebra finitely generated as $R$-module, which is an artinian ring. Then $R/\operatorname{ann}(A)$ is artinian?
My first attempt is,
Q.1. Since $A$ is finitely generated over $R$, $A$ is also artinian $R$-module?
Since $R$ is noetherian, we can write $A = (x_1, \cdots ,x_n)_{R}$. Consider a homomorphism
$f:R \to A\oplus A \cdots \oplus A$
(n-times) as $ r \mapsto (rx_1,\cdots, rx_n)$. This homomorphism has kernel $\operatorname{ann}(A)$. So we have an injective $R$-module homomorphism $\bar{f} : R/\operatorname{ann}(A) \to A\oplus A\cdots \oplus A$. If the above question is true, then $A\oplus A\cdots \oplus A$ is also artinian $R$=module. So $R/\operatorname{ann}(A)$ is also an artinian $R$-module.
Q.2. Furthermore, $R/ \operatorname{ann}(A)$ is an artinian ring? ; i.e., is it artinian $R/ \operatorname{ann}(A)$-module?
Best Answer
I assume that the rings are unitary.
Let $f:R→A$ be a ring morphism. Then $ra:=f(r)a$. If $rA=0$ then $f(r)a=0$ for all $a∈A$. In particular, for $a=1$ we get $f(r)=0$, so $r∈\ker f$. The converse is obvious, and thus $\mathrm{Ann}_R(A)=\ker f$.
Since $R/\ker f⊂A$ and the extension is integral (in fact, it is finite) we get that $\dim R/\ker f=0$ and we are done.