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Let $R$ be the polynomial ring in $n$ variables over an algebraically closed field $k$. I'm trying to prove that for all ideals $I$ of $R$ it holds that

$R/I$ is a finite-dimensional $k$-vector space of and only if $R/\sqrt{I}$ is as well.

I need to avoid using algebraic varieties, so my approach was considering the isomorphism $$R/\sqrt{I} \cong \frac{R/I}{\sqrt{I}/I}$$ so that the claim follows if $\sqrt{I}/I$ is a finite-dimensional vector space, however I'm not sure if this is true. My thoughts are that the module extension $I\subseteq \sqrt{I}$ is integral, so it is modulo-finite if it is finitely generated, but I'm having trouble proving the latter and I'm not really sure it helps (I'm not even sure it is true).

Any ideas on this please? Thanks in advance

Daniel
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1 Answers1

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Because of $(\sqrt I)^n \subset I$ and an induction argument it suffices to show that $I/I^2$ is finite dimensional if $R/I$ is. To that account note that $I/I^2$ is finitely generated over $R/I$ and this is finitely generated over $k$. Finite generation is of course transitive.

For an whole other approach, you can use that $R/I$ is finite dimensional if and only if it is artinian. Being artinian does not depend on whether one has divided out the nilradical.

MooS
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  • Thanks for the answer MooS. So, I tried to use what you said, here's what I got: There is $N$ such that $(\sqrt{I})^N\subseteq I$, which is due to the fact that $\sqrt{I}$ is finitely generated. Since $\sqrt{I}/I$ is isomorphic to $$\frac{\sqrt{I}/(\sqrt{I})^N}{I/(\sqrt{I})^N},$$ it suffices to see that $\sqrt{I}/(\sqrt{I})^N$ is finite-dimensional. For this matter take $J =\sqrt{I}$, as you claim it follows that $J/J^2$ is f-dim (assuming $R/J$ is), then $J/J^3 = (J/J^3)/(J^2/J^3)$ is f-dim as well, and inductively it holds for $J/J^m$. How does it look? Thanks! – Daniel Mar 24 '17 at 22:17
  • First part looks fine to me. For the second part, note that induction was maybe the wrong word. The same argument yields that $I^n/I^{n+1}$ is finite-dim. for all $n$ and this of course yields that $I/I^n$ is finite-dim. by a standard argument. – MooS Mar 25 '17 at 06:45