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!

  • $\begingroup$ Please edit your question to contain a precise statement of what you're trying to prove. Right now it's unclear. $\endgroup$ – KReiser Sep 19 '18 at 0:49
  • $\begingroup$ @KReiser thanks $\endgroup$ – Sky Sep 19 '18 at 0:58
  • $\begingroup$ No, in this way you can conclude that $\dim S(Y)\geq\dim Y$. What do not you understand in the other thread? $\endgroup$ – Armando j18eos Sep 19 '18 at 11:26
  • $\begingroup$ @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. $\endgroup$ – Sky Sep 19 '18 at 14:21

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