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Tag 00JQ

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

Example 10.56.4. Let $R$ be a ring. If $S = R[X]$ with $\deg(X) = 1$, then the natural map $\text{Proj}(S) \to \mathop{\mathrm{Spec}}(R)$ is a bijection and in fact a homeomorphism. Namely, suppose $\mathfrak p \in \text{Proj}(S)$. Since $S_{+} \not \subset \mathfrak p$ we see that $X \not \in \mathfrak p$. Thus if $aX^n \in \mathfrak p$ with $a \in R$ and $n > 0$, then $a \in \mathfrak p$. It follows that $\mathfrak p = \mathfrak p_0S$ with $\mathfrak p_0 = \mathfrak p \cap R$.

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 12945–12954 (see updates for more information).

    \begin{example}
    \label{example-proj-polynomial-ring-1-variable}
    Let $R$ be a ring. If $S = R[X]$ with $\deg(X) = 1$, then the natural map
    $\text{Proj}(S) \to \Spec(R)$ is a bijection and in fact a homeomorphism.
    Namely, suppose $\mathfrak p \in \text{Proj}(S)$. Since
    $S_{+} \not \subset \mathfrak p$ we see that $X \not \in \mathfrak p$.
    Thus if $aX^n \in \mathfrak p$ with $a \in R$ and $n > 0$, then
    $a \in \mathfrak p$. It follows that $\mathfrak p = \mathfrak p_0S$
    with $\mathfrak p_0 = \mathfrak p \cap R$.
    \end{example}

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