# What does it mean geometrically for a graded ring to be generated in degree $1$?

This is a rather basic question that I have never really thought about until now, when I was forced to think about a sheafy version. This condition is needed for example to define the relative Proj construction, and I don't understand what exactly about it is making something 'projective.' Here's an example result who's proof I understand algebraically, but not geometrically. I would like to explain geometrically what this proof is about.

Consider the well-known result that if $$f: Z \to X$$ is a projective, birational map between quasiprojective varieties, there is a coherent sheaf of ideals $$\mathscr{I}$$ on $$X$$ so that $$Z \simeq \tilde{X}$$, the blow up with respect to $$I$$. This is proven in Hartshorne and the proof sketch is as follows.

Since $$f$$ is projective we have a map $$i: Z \hookrightarrow \mathbb{P}^n_X$$ and an invertible sheaf $$\mathscr{L} = i^\ast \mathcal{O}_X(1)$$ on $$Z$$. We define $$\mathscr{S} = \bigoplus_{d \geq 0} \mathscr{L}^d$$. Since $$\mathscr{L}$$ is coherent, each power is and the sum is quasicoherent. The condition to make $$Z$$ the Proj of something is that $$\mathscr{S}$$ needs to be generated in degree $$1$$. This isn't the case, so we work affinely and find a related algebra corresponding to a degree $$D$$ Veronese embedding so that after composing $$i$$ above with the Veronese embedding, the condition holds.

Some more technicalities follow in this argument concerning whether or not we have a coherent sheaf of ideals, and arranging this to be the case, but the primary issue has already arisen.

What is this Veronese embedding doing, geometrically? I understand how it takes degree $$D$$ things to hyperplanes - this isn't quite what I'm looking for and is a little bit on-the-nose. Stated another way, what is composing with this embedding fixing about this object in projective space that is detected about being generated in degree $$1$$, that isn't there in general? Can we see visually/geometrically what fails in the Proj construction when this condition isn't met?

I apologize if this question is still not clear, and I worry that more rambling is only making it worse, so I'll stop here and try to clarify if anyone doesn't understand what I'm asking.

When you have a graded ring $$A$$ finitely generated in degree 1 as a $$A_0$$-algebra, then what you'll have is a surjection of graded rings $$A_0[X_1,\ldots,X_n] \twoheadrightarrow A$$. This is what gives you the closed embedding $$\mathrm{Proj} A \hookrightarrow \mathbb{P}^n_{A_0}$$.
You can then also realize $$A$$ as a quotient of $$A_0[X_1,\ldots,X_n]$$ (so really cut out by equations!). Under this embedding, $$A_1$$ remembers the restriction of linear forms to $$A$$.
• Great answer. Possible minor typo, unless I've misunderstood something: I think you have $\mathbb{P}^{n}_{A}$ written where you should have $\mathbb{P}^{n}_{A_{0}}$. – Alex Wertheim Feb 23 at 2:35