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All rings are assumed to be associative and have a 1. Let $k$ be a commutative artininan ring and $R$ a finitely generated $k$-algebra. Is it true that if $R$ is connected and hereditary, then $k$ is a field?

EDIT: As pointed out by YCor and Benjamin Steinberg, we must also require $k$ to be faithfully embedded in the center of $R$.

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    $\begingroup$ Is $R$ also finitely generated as a $k$-module or can it be infinitely generated? $\endgroup$
    – Benjamin Steinberg
    Commented Feb 15 at 15:29
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    $\begingroup$ ($R$ hereditary means that submodules of projective $R$-modules are projective.) $\endgroup$
    – YCor
    Commented Feb 15 at 15:35
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    $\begingroup$ For me the question is strange: if $k$ is say, local and not a field, e.g., $k=L[x]/x^2$ with $L$ a field, and $R=L$ is the residual field, then $R$ is connected and hereditary (it is a field!) but $k$ is not a field. $\endgroup$
    – YCor
    Commented Feb 15 at 15:37
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    $\begingroup$ Do you want to assume that k is faithfully embedded in the center of R? Otherwise, as @YCor points out you can just take any connected hereditary algebra R over k/m with m a maximal ideal. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 15 at 15:40
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    $\begingroup$ If R is finitely generated as a right k-module and J is the radical of k, then if R is hereditary, then I claim JR=0 and so it must really be an R/J-algebra. $\endgroup$
    – Benjamin Steinberg
    Commented Feb 15 at 15:48

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$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need to be finitely generated.

So assume $k$ is connected and not a field. Then $\operatorname{rad}k\neq0$ and so, since the Jacobson radical of an artinian ring is nilpotent, we can pick a nonzero nilpotent element $a\in k$. Let $n>1$ be the smallest integer such that $a^n=0$.

Now consider the short exact sequence $$0\to\operatorname{ann}_R(a)\to R\to aR\to0.$$

If $R$ is right hereditary then $aR$, as a right ideal of $R$, is projective, and so this short exact sequence splits (as a sequence of right $R$-modules).

So $R\cong\operatorname{ann}_R(a)\oplus aR$ as a right $R$-module.

But $\operatorname{ann}_R(a)$ is annihilated by $a$, and $aR$ is annihilated by $a^{n-1}$, so $R$ is annihilated by $a^{n-1}$, which is not true, since $a^{n-1}\neq0$.

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