The Stacks project

Proof. Suppose $\overline{g} = \overline{b}_0 x^ n + \overline{b}_1 x^{n - 1} + \ldots + \overline{b}_ n$, and $\overline{h} = \overline{c}_0 x^ m + \overline{c}_1 x^{m - 1} + \ldots + \overline{c}_ m$ with $\overline{b}_0, \overline{c}_0 \in \kappa (\mathfrak p)$ nonzero. After localizing $R$ at some element of $R$ not contained in $\mathfrak p$ we may assume $\overline{b}_0$ is the image of an invertible element $b_0 \in R$. Replacing $\overline{g}$ by $\overline{g}/b_0$ and $\overline{h}$ by $b_0\overline{h}$ we reduce to the case where $\overline{g}$, $\overline{h}$ are monic (verification omitted). Say $\overline{g} = x^ n + \overline{b}_1 x^{n - 1} + \ldots + \overline{b}_ n$, and $\overline{h} = x^ m + \overline{c}_1 x^{m - 1} + \ldots + \overline{c}_ m$. Write $f = x^{n + m} + a_1 x^{n + m - 1} + \ldots + a_{n + m}$. Consider the fibre product

where the map $\mathbf{Z}[a_ k] \to \mathbf{Z}[b_ i, c_ j]$ is as in Examples 10.136.7 and 10.143.12. By construction there is an $R$-algebra map

which maps $b_ i$ to $\overline{b}_ i$ and $c_ j$ to $\overline{c}_ j$. Denote $\mathfrak p' \subset R'$ the kernel of this map. Since by assumption the polynomials $\overline{g}, \overline{h}$ are relatively prime we see that the element $\Delta = \text{Res}_ x(g, h) \in \mathbf{Z}[b_ i, c_ j]$ (see Example 10.143.12) does not map to zero in $\kappa (\mathfrak p)$ under the displayed map. We conclude that $R \to R'$ is étale at $\mathfrak p'$. In fact a solution to the problem posed in the lemma is the ring map $R \to R'[1/\Delta ]$ and the prime $\mathfrak p' R'[1/\Delta ]$. Because $\text{Res}_ x(f, g)$ is invertible in this ring the Sylvester matrix is invertible over $R'[1/\Delta ]$ and hence $1 = a g + b h$ for some $a, b \in R'[1/\Delta ][x]$ see Example 10.143.12. $\square$