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.

  • $\begingroup$ Surely you mean $\operatorname{Spec} A$, not $\operatorname{Spec} A[t]$? $\endgroup$
    – KReiser
    Commented Dec 30, 2020 at 21:43
  • $\begingroup$ Oh yes, you are right! This question needs to be edited. I will do that now $\endgroup$ Commented Dec 30, 2020 at 21:45
  • 7
    $\begingroup$ "The grading given by $\deg t=1$...." surely means that $\deg t^2=2$ and so on, so that the grading is $A[t] = A \oplus tA \oplus t^2A \oplus \cdots$. $\endgroup$ Commented Dec 30, 2020 at 21:45
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    $\begingroup$ 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. $\endgroup$
    – Sera Gunn
    Commented Dec 30, 2020 at 22:15

1 Answer 1


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.

  • $\begingroup$ 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 $A \cong A[t]/(t)$. $\endgroup$
    – Sera Gunn
    Commented Dec 30, 2020 at 23:55
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    $\begingroup$ @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 $(t)$. $\endgroup$
    – KReiser
    Commented Dec 31, 2020 at 0:04

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