Define the graded rings $R_\bullet =k[u,v,w]/(uw-v^2)$ and $S_\bullet=k[x,y]$, where $u,v,w,x,y$ all have degree $1$. Then it is obvious that $\text{Proj}~R_\bullet$ and $\text{Proj}~S_\bullet$ are both $\mathbb{P}^1_k$, while there is an isomorphism $\phi:\text{Proj}~R_\bullet \rightarrow \text{Proj}~S_\bullet$ which is just the inverse of Veronese embedding morphism. In exercise 16.4G, it claims that this isomorphism $\phi$ is not induced by a graded homomorphism of graded rings $S_\bullet \rightarrow R_\bullet$ with degree $d$, where $d$ is arbitrary.

Edit: since $\text{Proj}~S_\bullet=\mathbb{P}_k^1$, we only need to show that the map $\phi$ cannot be a closed embedding! But $\phi$ must pull-back $\mathcal{O}_{\text{Proj}~S_\bullet}(1)$ to $\mathcal{O}_{\text{Proj}~R_\bullet}(d)$ for some $d$. However we have $h^0(\text{Proj}~R_\bullet,\mathcal{O}_{\text{Proj}~R_\bullet}(d)) \geq h^0(\text{Proj}~R_\bullet,\mathcal{O}_{\text{Proj}~R_\bullet}(1))=3$ and $h^0(\text{Proj}~S_\bullet,\mathcal{O}_{\text{Proj}~S_\bullet}(1))=2$. For $\phi$ to be an embedding, we need $h^0(\text{Proj}~S_\bullet,\mathcal{O}_{\text{Proj}~S_\bullet}(1)) \geq h^0(\text{Proj}~R_\bullet,\mathcal{O}_{\text{Proj}~R_\bullet}(1))$, and this cannot happen. So $\phi$ cannot be an embedding!

  • $\begingroup$ A graded ring homomorphism $S_{\bullet}\to R_{\bullet}$ would carry $S(1)$ to $R(1)$, which is part what you're trying to do with your statement here about pullbacks. In particular, the generators $x$ and $y$ in degree one would have to be carried to some linear combination of $u,v,w$, say $x\mapsto\ell_1(u,v,w)$ and $y\mapsto \ell_2(u,v,w)$. Some linear combination of $\ell_1 $ and $\ell_2$ must then yield $u,v,w$. But then some linear algebra will tell you this is not going to work... Maybe I made some silly error somewhere, that seems too easy... $\endgroup$ – John Martin Mar 22 '16 at 18:38
  • $\begingroup$ Thank you and sorry for late reply. What stuck me is to show the pull-back of global sections is actually given by the map $S_1 \rightarrow R_1$, is this trivial? $\endgroup$ – Wenzhe Mar 24 '16 at 23:07
  • $\begingroup$ I wouldn't say that it is trivial. In fact I would typically argue that nothing is trivial. In any case, the argument I presented above is an attempt at proving the fact that the pull back of global sections cannot come from a graded ring morphism, precisely because it would have to take $S_1$ to $R_1$, which seems to present a problem. $\endgroup$ – John Martin Mar 25 '16 at 15:27
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    $\begingroup$ Thank you, check the question! math.stackexchange.com/questions/1712304/… So it is indeed the case! $\endgroup$ – Wenzhe Mar 25 '16 at 20:12
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    $\begingroup$ I have edited, suppose $S_\bullet \rightarrow R_\bullet$ is of degree $d$, then $\phi$ pulls $\mathcal{O}(1)$ to $\mathcal{O}(d)$. Since $\phi$ is a morphism to $\mathbb{P}^1$, if it is an isomorphism, it must be a closed embedding a priori, thus if we show it is not a closed embedding, it cannot be an isomorphism. However, if $\phi$ is a closed embedding, it must map the global sections of $\mathcal{O}(1)$ isomorphicly to that of $\mathcal{O}(d)$, which cannot happen from the dimension arguments. $\endgroup$ – Wenzhe Mar 25 '16 at 21:38

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