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I am looking at Hartshorne Example III.9.8.4., p260. He says that $a$ is not a zero divisor in $k[a,x,y,z]/I$, where $$ I = (a^2(x+1) -z^2, ax(x+1)-yz, xz-ay,y^2-x^2(x+1)). $$ Is there a good way to see this?

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Perhaps one way would be to show that $a$ is not contained in a minimal prime, since the set of zero divisors is a union of minimal primes – Bruno Joyal Apr 25 '12 at 23:57
Not very elegant or insightful, but you could compute $J := (a) \cap I$ and then show that $a^{-1}J = I$. Both the intersection and the equality check can be computed using Groebner bases. – m_l Apr 26 '12 at 8:23
@Bruno -- The set of zero divisors is the union of the associated primes, and the set of associated primes is the set of minimal primes if the ring is reduced, I think. Is it clear that this quotient is reduced? – paragon Apr 26 '12 at 20:58
@m_l -- What do you mean by $a^{-1}J$? Are you computing this in the localization at $a$? – paragon Apr 26 '12 at 21:01
@Fredrik: Why can we assume that? In $k[x]/x^2$, $x$ is a zero divisor, but $xf \in (x^2)$ implies $x$ divides $f$. – Hurkyl Jun 10 '12 at 3:07
up vote 1 down vote accepted

Let me replace $a$ with $w.$ First, consider the domain $S = k[x,y]/(y^2-x^2(x+1)).$ Then $S[z,w]$ is a graded ring ($\mathbb N$-grading) generated in degree $1$ over $S$ by $\{z,w\}.$ Now $w^2(x+1)-z^2,wx(x+1)-yz,xz-wy$ are homogeneous elements of $S[z,w]$ of respective degrees $2,1,1.$ Thus the ideal generated by these elements is homogeneous, and the quotient $R = S[z,w]/(w^2(x+1)-z^2,wx(x+1)-yz,xz-wy)$ is graded. In particular, both $z,w$ still have degree $1$ in the quotient ($x,y$ are in degree $0$) and so cannot possibly be zerodivisors, since their products with any nontrivial elements must have degree at least $1.$ But $R = k[x,y,z,w]/(w^2(x+1)-z^2,wx(x+1)-yz,xz-wy,y^2-x^2(x+1))$ is exactly our original ring, so we are done.

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