# Is $\sqrt{I}/I$ finite dimensional?

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.

• What if $I=(x_1^2)$? – Angina Seng Dec 28 '17 at 9:25
• 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). – 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.

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 $$R$$.