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Let $R$ be a polynomial ring (in finitely many, say 8, indeterminates) over an algebraically closed field $k$. Suppose $I=(f,g) \neq R$ is a proper ideal of $R$ and $I = J \cap L$ for two proper ideals $J,L \neq I$. How do I show $J+L \neq R$?

I'd like an algebraic proof.

If $f,g$ are homogenous polynomials, I believe this follows from Hartshorne (1977) II.8 exercise 4.c, which I do not understand.

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I thought of the Chinese Remainder theorem when I looked at your question. I don't know how to prove your question in general, but if $I$ is prime, then assuming that $J + L = R$ and applying Chinese Remainder gives you a contradiction. In fact, if $Spec(R/I)$ is connected, then by the Chinese Remainder, you get a contradiction. –  Rankeya Jul 23 '12 at 20:34
@Rankeya: right. This will be used to show $(f,g)$ is prime, using some additional hypotheses. It is fine to assume $(f,g)$ is an intersection of primes, but I specifically need the case where it is not itself prime. –  Jack Schmidt Jul 23 '12 at 20:43
Sorry if my comment gave you the impression that I was giving a solution. I was merely making an observation. –  Rankeya Jul 23 '12 at 20:45
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2 Answers

up vote 1 down vote accepted

Let $J=(1-x-y)$, $L=(x,y)$. Then $J\cap L =JL$ is proper and two generated but $J+L=R$.

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Thanks! This exposes in an error in my translation. I should have required that when $I=(f,g)$, that $f,g$ have no common factor. –  Jack Schmidt Jul 25 '12 at 14:08
Strange comment by OP! I'm affraid you had to ask first to find a counterexample, and then, if you like, to impose the condition that $\text{gcd}(f,g)=1$. –  user26857 Jul 25 '12 at 15:14
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Another counter-example due to Sathaye: Let $R=k[x,y]$, $f=1-x^2$, and $g=y$, then $I=(f,g)$ has height 2, so is a complete intersection. It is obviously non-singular since $(f,g)+J = (1-x^2,y,-2x) = (1,y,x) = R$. However it is obviously not connected (being the disjoint union of two points).$\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\Proj}{\operatorname{Proj}}$

In particular, $J=(1-x,y)$ and $L=(1+x,y)$ have intersection $I=J\cap L = (1-x^2,y)$ and $J+L=R$, but $J \neq I \neq L$. Hence $\Spec(R/I)$ is disconnected, non-singular, and a complete (ideal theoretic and set theoretic) intersection.

We can homogenize this example to $f=z^2-x^2$, $g=y$, then for $I=(z^2-x^2,y)$, we have $\Proj(R/I)$ is a nonsingular, disconnected, complete intersection of dimension 0. This shows that the dimension requirements in the Hartshorne exercise are necessary.

If we increase the dimension by considering the example in the projective space associated to $R=k[x,y,z,w]$ and set $f=w^2-x^2$, $g=y$, then the affine patch $w=1$ is two parallel lines ($z$ free), and so is still a disconnected, non-singular, complete intersection. However, the two parallel lines meet at infinity, $[x=0:y=0:z:w=0]$ (where $z$ is free, but WLOG is $z=1$), and so the result is a connected, non-singular, complete intersection of dimension 1.

In particular, an affine patch of a non-singular connected projective algebraic set need not itself be connected.

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