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 $i$th elementary symmetric function in $\alpha _1, \ldots , \alpha _ n$. By the usual theory of elementary symmetric polynomials (details omitted) the ring $S$ is finite free over $R$ with basis the elements $\alpha _1^{e_1} \alpha _2^{e_2} \ldots \alpha _ n^{e_ n}$ with $0 \leq e_ i \leq n - i$. A fortiori, the fibre rings of $R \to S$ are finite and hence have dimension $0$. On the other hand, $S$ is generated by $n$ elements over $R$ subject to $n$ equations. Hence $S$ is a relative global complete intersection over $R$. Since the rank of $S$ over $R$ is positive, we also see that $S$ is faithfully flat over $R$.

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