# If $A$ is a ring, then why is $\operatorname{Proj} A[t] \cong \operatorname{Spec} A$?

The following is an example I saw in a book on Algebraic Geometry.

Example: Let $$A$$ be a ring and consider $$A[t]$$, with the grading given by $$\deg t = 1$$ and $$\deg a = 0$$ for $$a \in A$$. Then the structure map gives an isomorphism $$\operatorname{Proj} A[t] \cong \operatorname{Spec} A$$

By structure map, the authors mean a morphism of schemes $$\varphi : \operatorname{Proj} A[t] \to \operatorname{Spec} A$$. I am not sure exactly what this morphism would be however, I assume that we start off with the standard embedding $$A \hookrightarrow A[t]$$ and then apply the functor $$\operatorname{Spec}$$ to get a map $$\operatorname{Spec} A[t] \to \operatorname{Spec} A$$ and then pre-compose that with the inclusion $$i : \operatorname{Proj} A[t] \to \operatorname{Spec} A[t]$$ to get the desired morphism $$\varphi$$.

Now firstly, the $$\operatorname{Proj}$$ construction only works for graded rings. So $$A[t]$$ should be of the form $$A[t] = \bigoplus_{d \geq 0} A_d$$. The authors say above "the grading given by $$\deg t = 1$$ and $$\deg a = 0$$ for $$a \in A$$", and I assume that this means that $$A_0$$ is the image of the standard inclusion of $$A$$ in $$A[t]$$, and that $$A_1 = (t)$$, the ideal generated by $$t$$ so that "essentially" $$A[t] = A_0 \oplus A_1 = A \oplus (t)$$. Is this the correct interpretation?

Now onto the claim that the $$\varphi$$ yields an isomorphism. I don't even see why $$\varphi$$ is a bijection, let alone a homeomorphism. If I label the embedding $$A \hookrightarrow A[t]$$ as $$j$$, then $$\varphi$$ is given by $$\varphi(\mathfrak{p}) = j^{-1}(\mathfrak{p})$$ for any $$\mathfrak{p} \in \operatorname{Proj} A[t]$$ and to say that $$\varphi$$ is bijective would mean that every prime ideal $$\mathfrak{q}$$ of $$A$$ is of the form $$\mathfrak{q} = j^{-1}(\mathfrak{p})$$ for some unique $$\mathfrak{p} \in \operatorname{Proj} A[t]$$ and for starters I do not see why this would be the case.

Furthermore I would need to show that the morphism of sheaves $$\varphi^\sharp : \mathcal{O}_{\operatorname{Spec} A} \to \varphi_* \mathcal{O}_{\operatorname{Proj} A[t]}$$ is also isomorphic and the way one usually does is by looking at basic open sets on $$\operatorname{Spec} A$$. In more words - if I wanted to see what $$\varphi^\sharp$$ would be in this case, then if I chose an $$f \in A$$ and considered the open subset $$D(f) \subseteq \operatorname{Spec} A$$ then $$\Gamma(D(f), \mathcal{O}_{\operatorname{Spec} A}) \cong A_f$$. But $$\Gamma(D(f), \varphi_*\mathcal{O}_{\operatorname{Proj} A[t]}) = \Gamma(\varphi^{-1}(D(f)), \mathcal{O}_{\operatorname{Proj} A[t]})$$ and this may not have such a nice form as far as I know, since $$\varphi^{-1}(D(f))$$ might not be a basic open subset of $$\operatorname{Proj} A[t]$$. So I cannot express $$\varphi^\sharp_{D(f)} : \Gamma(D(f), \mathcal{O}_{\operatorname{Spec} A}) \to \Gamma(\varphi^{-1}(D(f)), \mathcal{O}_{\operatorname{Proj} A[t]})$$ in any nice way to allow me to check that $$\varphi^\sharp$$ would indeed be an isomorphism.

• Surely you mean $\operatorname{Spec} A$, not $\operatorname{Spec} A[t]$? Dec 30, 2020 at 21:43
• Oh yes, you are right! This question needs to be edited. I will do that now Dec 30, 2020 at 21:45
• "The grading given by $\deg t=1$...." surely means that $\deg t^2=2$ and so on, so that the grading is $A[t] = A \oplus tA \oplus t^2A \oplus \cdots$. Dec 30, 2020 at 21:45
• The idea is that $\operatorname{Spec} A[t]$ is a "line over $\operatorname{Spec} A$" and $\operatorname{Proj}$ is the set of lines in $\operatorname{Spec} A[t]$ (of which there is just one). So $\operatorname{Proj} A[t]$ is a "point over $\operatorname{Spec} A$" and a "point over $\operatorname{Spec} A$" is just $\operatorname{Spec} A$. I'll try to write a more rigorous explanation when I have more time, if nobody else gets to it. Dec 30, 2020 at 22:15

For $$R$$ a graded ring, $$\operatorname{Proj} R$$ has as its underlying set the homogeneous prime ideals of $$R$$ which do not contain the irrelevant ideal. In our case, the irrelevant ideal is $$(t)$$, and so no homogeneous prime ideal in $$\operatorname{Proj} A[t]$$ contains $$t$$. So $$D_+(t)\subset\operatorname{Proj} A[t]$$ is all of $$\operatorname{Proj} A[t]$$, and therefore by the identification $$D_+(t)\cong \operatorname{Spec} A[t]_{(t)}$$ (where this denotes the homogeneous localization, or $$(A[t]_t)_0$$) and the observation that $$A[t]_{(t)}=A$$, we see that $$\operatorname{Proj} A[t]\cong\operatorname{Spec} A$$.
If you want to get more hands-on, you can verify that any homogeneous prime $$P\subset A[t]$$ not containing the irrelevant ideal is exactly specified by it's intersection with $$A$$. For every homogeneous element $$at^n\in P$$, we get that either $$a\in P$$ or $$t\in P$$ by primality of $$P$$. By the assumption that $$(t)\not\subset P$$, the former must be the case, so $$P$$ is the ideal generated by $$P\cap A$$. Thus there's a bijection as sets between $$\operatorname{Proj} A[t]$$ and $$\operatorname{Spec} A$$, and one can verify that this induces a bijection between basic open sets ($$D(a)\subset \operatorname{Spec} A$$ and $$D_+(f)\subset \operatorname{Proj} A[t]$$ where $$f$$ is homogeneous of positive degree), so it is even a homeomorphism.
• There are a few things wrong with the first half of your answer and I'm not sure what the best way to fix it is. First, $D(t)$ should be the localization at $t$ not at $(t)$. So that should be $A[t,t^{-1}]$. Second, taking Spec of that isn't right because that considers in-homogeneous primes as well. And lastly, I don't believe one can obtain $A$ (generally) as a localization of $A[t]$; the correct identity would be $A \cong A[t]/(t)$. Dec 30, 2020 at 23:55
• @TrevorGunn I mean the homogeneous localization - that is, the degree-zero portion of $A[t]_t$. It's unfortunate that the notation is so close to the localization at the ideal $(t)$. Dec 31, 2020 at 0:04