Example 10.136.8. Consider the ring map

\begin{eqnarray*} R = \mathbf{Z}[a_1, \ldots , a_ n] & \longrightarrow & S = \mathbf{Z}[\alpha _1, \ldots , \alpha _ n] \\ a_1 & \longmapsto & \alpha _1 + \ldots + \alpha _ n \\ \ldots & \ldots & \ldots \\ a_ n & \longmapsto & \alpha _1 \ldots \alpha _ n \end{eqnarray*}

In other words this is the unique ring map of polynomial rings as indicated such that

$x^ n + a_1 x^{n - 1} + \ldots + a_ n = \prod \nolimits _{i = 1}^ n (x + \alpha _ i)$

holds in $\mathbf{Z}[\alpha _ i, x]$. Another way to say this is that $a_ i$ maps to the $i$th elementary symmetric function in $\alpha _1, \ldots , \alpha _ n$. Note that $S$ is generated by $n$ elements over $R$ subject to $n$ equations. Hence to show that $S$ is a relative global complete intersection over $R$ we have to show that the fibre rings $S \otimes _ R \kappa (\mathfrak p)$ have dimension $0$. This follows as in Example 10.136.7 because the ring map $\mathbf{Z}[a_1, \ldots , a_ n] \to \mathbf{Z}[\alpha _1, \ldots , \alpha _ n]$ is actually finite since each $\alpha _ i \in S$ satisfies the monic equation $x^ n - a_1 x^{n - 1} + \ldots + (-1)^ n a_ n$ over $R$.

## Comments (0)

There are also:

• 2 comment(s) on Section 10.136: Syntomic morphisms

## 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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00SR. Beware of the difference between the letter 'O' and the digit '0'.