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The Stacks project

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 ith 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.


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