# Hartshorne Exercise 2.6.

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. Commented Feb 9, 2014 at 23:02
• @BrianFitzpatrick I added the proposition and the proof as they appear in Hartshorne!
– V-B
Commented Feb 9, 2014 at 23:07
• What about the exercise? Commented Feb 9, 2014 at 23:07
• @BrianFitzpatrick It's its counterpart for projective varieties, i added that too.
– V-B
Commented Feb 9, 2014 at 23:09

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.