The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Example 10.26.1. In this example we describe $X = \mathop{\mathrm{Spec}}(\mathbf{Z}[x]/(x^2 - 4))$. Let $\mathfrak {p}$ be an arbitrary prime in $X$. Let $\phi : \mathbf{Z} \to \mathbf{Z}[x]/(x^2 - 4)$ be the natural ring map. Then, $ \phi ^{-1}(\mathfrak p)$ is a prime in $\mathbf{Z}$. If $ \phi ^{-1}(\mathfrak p) = (2)$, then since $\mathfrak p$ contains $2$, it corresponds to a prime ideal in $\mathbf{Z}[x]/(x^2 - 4, 2) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x^2)$ via the map $ \mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, 2)$. Any prime in $(\mathbf{Z}/2\mathbf{Z})[x]/(x^2)$ corresponds to a prime in $(\mathbf{Z}/2\mathbf{Z})[x]$ containing $(x^2)$. Such primes will then contain $x$. Since $(\mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x)$ is a field, $(x)$ is a maximal ideal. Since any prime contains $(x)$ and $(x)$ is maximal, the ring contains only one prime $(x)$. Thus, in this case, $\mathfrak p = (2, x)$. Now, if $ \phi ^{-1}(\mathfrak p) = (q)$ for $q > 2$, then since $\mathfrak p$ contains $q$, it corresponds to a prime ideal in $\mathbf{Z}[x]/(x^2 - 4, q) \cong (\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)$ via the map $ \mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, q)$. Any prime in $(\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)$ corresponds to a prime in $(\mathbf{Z}/q\mathbf{Z})[x]$ containing $(x^2 - 4) = (x -2)(x + 2)$. Hence, these primes must contain either $x -2$ or $x + 2$. Since $(\mathbf{Z}/q\mathbf{Z})[x]$ is a PID, all nonzero primes are maximal, and so there are precisely 2 primes in $(\mathbf{Z}/q\mathbf{Z})[x]$ containing $(x-2)(x + 2)$, namely $(x-2)$ and $(x + 2)$. In conclusion, there exist two primes $(q, x-2)$ and $(q, x + 2)$ since $2 \neq -2 \in \mathbf{Z}/(q)$. Finally, we treat the case where $\phi ^{-1}(\mathfrak p) = (0)$. Notice that $\mathfrak p$ corresponds to a prime ideal in $\mathbf{Z}[x]$ that contains $(x^2 - 4) = (x -2)(x + 2)$. Hence, $\mathfrak p$ contains either $(x-2)$ or $(x + 2)$. Hence, $\mathfrak p$ corresponds to a prime in $\mathbf{Z}[x]/(x - 2)$ or one in $\mathbf{Z}[x]/(x + 2)$ that intersects $\mathbf{Z}$ only at $0$, by assumption. Since $\mathbf{Z}[x]/(x - 2) \cong \mathbf{Z}$ and $\mathbf{Z}[x]/(x + 2) \cong \mathbf{Z}$, this means that $\mathfrak p$ must correspond to $0$ in one of these rings. Thus, $\mathfrak p = (x - 2)$ or $\mathfrak p = (x + 2)$ in the original ring.


Comments (2)

Comment #45 by Rankeya on

In the second to last line of this example, Z[x]/(x-2) should be Z[x]/(x+2).

There are also:

  • 2 comment(s) on Section 10.26: Examples of spectra of rings

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