# Exercise 2.6 in Hartshorne's book.

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

I don't know where is wrong about the following argument.

My idea: copy the proof of the affine case. (There is also the same question in Mathematics. Hartshorne Exercise 2.6. but I don't understand.)

As we know, algebraic sets of $$\mathbb P^n$$ is $$1 \, \text{to} \, 1$$ correspondence homogeneous prime ideal of $$S=k[x_0, x_1, \cdots, x_n]$$ not containing irrelevant maximal ideal $$S_{+}$$. On the other hand, the homogeneous prime ideal is generated by homogeneous elements, by $$S$$ is UFD, we can show the number of homogeneous prime ideals is finite. We can calculate that the length of the longest ascending chain of homogeneous prime ideals is equal to $$\dim{S}$$. So, I think this is a proof.

Where am I wrong? Thank you in advance!

• Please edit your question to contain a precise statement of what you're trying to prove. Right now it's unclear. – KReiser Sep 19 '18 at 0:49
• @KReiser thanks – Sky Sep 19 '18 at 0:58
• No, in this way you can conclude that $\dim S(Y)\geq\dim Y$. What do not you understand in the other thread? – Armando j18eos Sep 19 '18 at 11:26
• @Armandoj18eos thanks,Now I understand.the homogeneous prime ideal is not simple as I thought before.It is difficult to calcalate the asscending chain of prime ideals. – Sky Sep 19 '18 at 14:21