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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$\begingroup$ 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 $\endgroup$– Bruno JoyalCommented Apr 25, 2012 at 23:57
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$\begingroup$ 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. $\endgroup$– m_lCommented Apr 26, 2012 at 8:23
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$\begingroup$ @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? $\endgroup$– paragonCommented Apr 26, 2012 at 20:58
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$\begingroup$ @m_l -- What do you mean by $a^{-1}J$? Are you computing this in the localization at $a$? $\endgroup$– paragonCommented Apr 26, 2012 at 21:01
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2$\begingroup$ @Fredrik: Why can we assume that? In $k[x]/x^2$, $x$ is a zero divisor, but $xf \in (x^2)$ implies $x$ divides $f$. $\endgroup$– user14972Commented Jun 10, 2012 at 3:07
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1 Answer
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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.
Edit:
This argument about degrees isn't very convincing. How about the following:
The quotient ring $S = k[x,y,z] / \langle y^2 - x^2(x+1) \rangle$ is a domain, so injects into its quotient field $Q(S)$. This yields an injection $S[a] \hookrightarrow Q(S)[a]$. Now let $J \subset S[a]$ be defined by the three remaining relations. Over $Q(S)[a]$, these all reduce to $a = xz/y$, so $Q(S)[a] / JQ(S)[a] = Q(S)$. Thus there is a canonical map $S[a]/J \to Q(S)$ such that $a \mapsto xz/y$. It is clear that if $a$ is a zero-divisor, then this map must send $a$ to $0$, which is false. Hence $a$ cannot be a zero-divisor.
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$\begingroup$ Could you explain further why, e.g., $w$ can't be a zerodivisor? I don't understand the relevance of the degree being $1$. $\endgroup$– rj7k8Commented Jun 22, 2018 at 23:52
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1$\begingroup$ Dear @rj7k8, I think the original argument is flawed. I've proposed another one. Cheers $\endgroup$– AndrewCommented Jun 27, 2018 at 11:26
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1$\begingroup$ Thanks for returning to such an old question. $\endgroup$– rj7k8Commented Jul 8, 2018 at 8:19