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 $R$.

There are also:

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

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).