Example 10.27.1. 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.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #45 by Rankeya on

Comment #51 by Johan on

There are also: