Hartshorne Exercise 2.6.

Question

This might sound dumb to many people, but if I were to blindly try and carbon-copy the proof of Proposition 1.7 in Hartshorne to Exercise 2.6 where would I go wrong? Basically, I don't see where exactly that proof is failing?

$\bf{Propositon 1.7}$ If $Y$ is an affine algebraic set, then the dimension of $Y$ is equal to the dimension of its affine coordinate ring $A(Y).$

$\bf{Proof.}$ If $Y$ is an affine algebraic set in $\mathbb{A}^n$, then the closed irreducible subsets of $Y$ correspond to prime ideals of $A=k[x_1,...,x_n]$ containing $I(Y).$ These in turn correspond to prime ideals of $A(Y)$. Hence $\dim Y$ is the length of the longest chain of prime ideals in $A(Y)$, which is it's dimension.

$\bf{Exercise 2.6.}$ If $Y$ is a projective variety with homogeneous coordinate ring $S(Y)$, show that $\dim S(Y)=\dim Y+1.$

Thanks!

It'd be a lot easier to help you with this if you provided us with an outline of the proof yourself. — Brian Fitzpatrick, Feb 09 '14 at 23:02
@BrianFitzpatrick I added the proposition and the proof as they appear in Hartshorne! — V-B, Feb 09 '14 at 23:07
@BrianFitzpatrick It's its counterpart for projective varieties, i added that too. — V-B, Feb 09 '14 at 23:09

score 6 · Answer 1 · answered Feb 09 '14 at 23:30

6

The problem is that the closed irreducible subsets of projective space correspond to homogenous prime ideals not containing the irrelevant ideal.

As a hint for your problem, you know that dimension on a variety can computed affine locally (i.e. if you dehomogenize your definining equations in one coordinate chart, the resulting affine variety has the same dimension as your projective variety). Now, consider how the coordinate ring of the dehomogenization corresponds to the coordinate ring of the full projective variety.

answered Feb 09 '14 at 23:30

Alex Youcis

54,059

Thanks, I overlooked the part "homogeneous prime ideals"! I was just getting on to say that I resolved the issue! – V-B Feb 09 '14 at 23:55
@V-B No problem--good luck! – Alex Youcis Feb 09 '14 at 23:59

score 1 · Answer 2 · answered Nov 09 '17 at 07:51

Actually, if you follow the proof of affine version, you can only conclude $\dim S(Y)\geq \dim Y$, since for a chain of irreducible closed sets $$ Z_0 \supsetneq Z_1 \supsetneq \cdots \supsetneq Z_n$$ You can have corresponding prime ideal chain: $$ I(Z_0) \subsetneq I(Z_1) \subsetneq \cdots \subsetneq I(Z_n).$$ However, the converse doesn’t hold, since the definition of dimension of ring doesn’t change, which is just same with the affine case. So for every prime (not necessarily homogeneous) ideal $\mathfrak p$, we may not have the $Z(\mathfrak p)$ is an algebraic set in $\mathbf P^n$ generally.

Hartshorne Exercise 2.6.

2 Answers2