Example 10.136.7. Let $n , m \geq 1$ be integers. Consider the ring map

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

holds. Note that $S$ is generated by $n + m$ elements over $R$ (namely, $b_ i, c_ j$) and that there are $n + m$ equations (namely $a_ k = a_ k(b_ i, c_ j)$). In order to show that $S$ is a relative global complete intersection over $R$ it suffices to prove that all fibres have dimension $0$.

To prove this, let $R \to k$ be a ring map into a field $k$. Say $a_ i$ maps to $\alpha _ i \in k$. Consider the fibre ring $S_ k = k \otimes _ R S$. Let $k \to K$ be a field extension. A $k$-algebra map of $S_ k \to K$ is the same thing as finding $\beta _1, \ldots , \beta _ n, \gamma _1, \ldots , \gamma _ m \in K$ such that

Hence we see there are at most finitely many choices of such $n + m$-tuples in $K$. This proves that all fibres have finitely many closed points (use Hilbert's Nullstellensatz to see they all correspond to solutions in $\overline{k}$ for example) and hence that $R \to S$ is a relative global complete intersection.

Another way to argue this is to show $\mathbf{Z}[a_1, \ldots , a_{n + m}] \to \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m]$ is actually also a *finite* ring map. Namely, by Lemma 10.38.5 each of $b_ i, c_ j$ is integral over $R$, and hence $R \to S$ is finite by Lemma 10.36.4.

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