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I was looking at results that give necessary (and possibly sufficient) conditions on an ideal $I$ in the ring $R=k[x_1,x_2,...,x_n]$ such that $R/I$ is normal. I don't know if there is a standard result about this, but any references either way would be appreciated.

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I don't think there is a non-tautological necessary and sufficient on $I$ for the quotient $A\stackrel {def}{=}k[x_1,x_2,...,x_n]/I$ to be normal. I'll describe a sufficient condition, which actually characterizes a stronger condition: that $Spec(A)$ be smooth over $k$.

Since your polynomial ring $R=k[x_1,x_2,...,x_n]$ is noetherian, you may choose a finite number of generators $F_1,F_2,...,F_r$ of $I$ .
For every closed point $x \in Spec (A)$ , equivalently every maximal ideal ${\mathfrak m}_x \subset A$, you can consider the jacobian matrix $J(x)=(\frac {\partial F_i}{\partial x_j}(x))_{i,j}$, whose entries are in $\kappa(x)=A/{\mathfrak m}_x$.
The jacobian criterion says that if $rank J(x)=n-dim(A_ {{\frak m}_x})\quad \quad (JAC(x))$

then $Spec(A)$ will be smooth at $x$, hence $A_ {{\frak m}_x}$ will be normal.
And if $JAC(x)$ holds for all closed points $x\in Spec(A)$, then $Spec(A)$ will be smooth over $k$ and hence $A$ will be normal.

To sum up, the jacobian criterion gives a sufficient condition for normality ( because smooth $\Rightarrow$ regular $\Rightarrow$ normal), but some non smooth schemes are nevertheless normal and so failure of the jacobian test does not allow you to conclude anything about normality.

A simple example is $A=k[x,y,z]/(xy-z^2)\;$ ( $char.(k)\neq2$), the coordinate ring of a quadratic cone in $\mathbb A^3_k$. The ring $A$ is normal although the cone is not smooth at the origin $O$: the jacobian $(y,x,-2z)$ has rank $0$ at $O$and the right hand side of $JAC(O)$ is $3-2=1$, so $JAC(O)$ fails.

To end on an optimistic note, if $char(k)=0$ and $dim(A)=1$, then the the jacobian criterion gives a necessary and sufficient condition for normality. [Because in characteristic zero, smooth=regular and in dimension one, regular =normal]

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    Thanks for the very elaborate answer. This helps a lot. On a related note, is there a particular name for the ideal $I$ such that $R/I$ is normal (just asking based on your use of "non-tautological")2011-10-06