Is $\sqrt{I}/I$ finite dimensional over $\mathbb{C}$, where $I$ is a (non-zero) ideal in $R=\mathbb{C}[x_1,\dots,x_n]$?

I know the ring is Noetherian, so they are finitely generated.

The question is from an attempt to prove $R/I$ is finite dimensional if and only if $I$ is contained in finitely many maximal ideals.

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    $\begingroup$ What if $I=(x_1^2)$? $\endgroup$ – Angina Seng Dec 28 '17 at 9:25
  • $\begingroup$ Yeah, you are right, so how do you prove the right to left direction? I can only prove it when $I$ is radical (and using Nullstellensatz). $\endgroup$ – FunctionOfX Dec 28 '17 at 9:37

To prove that if $I$ is contained in only finitely many maximal ideals, then $R/I$ has finite dimension over $\Bbb C$ note that $V(I)$ is a finite set, so that $I(V(I))=\sqrt I$ is the intersection of finitely many maximal ideals $M_1,\ldots,M_k$ (this all comes from the Nullstellensatz). Then as $\sqrt I$ is finitely generated, $I\supseteq \sqrt I^m=\bigcap_j M_j^m$ for some $m$. Then $R/I$ is a quotient of $R/\sqrt I^m\cong \prod_j R/M_j^m$. But all of the $R/M_j^m$ are finite-dimensional.

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Since you managed to prove it for radical ideals, I will give you a proof of the following:

Lemma 1. Let $R=k[x_1, \dotsc, x_n]$ and $I \subset R$ an ideal. If $R/\sqrt I$ is finite-dimensional as $k$-vector space, so is $R/I$.

Before proving this, let me state

Lemma 2. Let $R=k[x_1, \dotsc, x_n]$ and $I,J \subset R$ two ideals. If $R/I$ and $R/J$ are finite-dimensional as $k$-vector space, so is $R/IJ$.

By choosing $I=J$ and iterating, we of course get that $R/I^n$ is also finite-dimensional for all $n \geq 1$.

Proof of Lemma 2.

We have an exact sequence $$0 \to I/IJ \to R/IJ \to R/I \to 0,$$ hence it suffices to show that $I/IJ$ is finite-dimensional. $I$ is finitely generated, thus we get a surjection $$R^{\oplus n} \twoheadrightarrow I.$$ Tensoring with $R/J$ yields a surjection $$(R/J)^{\oplus n} \twoheadrightarrow I/IJ.$$ The LHS is clearly a finite-dimensional $k$-vector space, hence $I/IJ$ is also.

Proof of Lemma 1.

Since $R$ is notherian, we have $(\sqrt I)^n \subset I$ for some $n$ and thus have a surjection $$R/(\sqrt I)^n \twoheadrightarrow R/I.$$ By Lemma 1 we have that $R/(\sqrt I)^n$ is finite-dimensional and hence the same holds for $R/I$.

Note that the proof holds for any noetherian $k$-algebra $R$.

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