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Let $S$ be a polynomial ring with $n$ variables and $I$ be an ideal of $S$.

If $S/I$ is regular ring then is $I$ generated by linear forms?

is the converse true?

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$\mathbb{R}[x] / \langle x^2 + 1 \rangle$ is regular.... –  Hurkyl Aug 3 '12 at 7:33
    
Nice answer, @Hurkyl: +1 –  Georges Elencwajg Aug 3 '12 at 7:35
    
thanks Hurkyl. If the codition "$S/I$ be graded" add, above question is true? –  Sang Cheol Lee Aug 3 '12 at 7:42

1 Answer 1

up vote 3 down vote accepted

It is perhaps best to think of your question geometrically. If $I$ is generated by linear forms, then you are asking if the intersection of a collection of hyperplanes through the origin is smooth. The answer is yes (since it is a linear subspace, and these are smooth).

Now there are certainly smooth varieties that are not linear subspaces, so $S/I$ can be regular without $I$ being generated by linear forms.

But if you request that $I$ be graded, then you are looking at the cone over a projective variety (the projective variety described by the graded ideal $I$).
This will be singular unless $I$ will is generated by linear forms.

Here is a proof:

Let $I = L + J$, where $L$ is the graded subideal generated just by the linear forms in $I$, and $J$ is the graded subideal generated by all the non-linear forms in $I$. Then $S/I = (S/L)/(I/L) = (S/L)/\overline{J})$, where I write $\overline{J}$ to denote the image of $J$ in $S/L$. Now if we change variables appropriately, then $S/L$ is just a polynomial ring in some smaller number of variables.

If $I \neq L$, i.e. $I$ is not generated by linear forms, then $\overline{J} \neq 0,$ and so the question reduces to the following:

If $J$ is a graded ideal in $S$ all of whose generators are degree $> 1$, can $S/J$ be regular? The answer is no; the associated conical variety has a singularity at the origin. (E.g. by the Jacobian criterion).

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