The Stacks project

Proof. There exists an automorphism $\Psi$ of $A[x_1, \ldots , x_ n]$ over $A$ such that $\Psi (P) = ax_ n^ d +$ lower order terms in $x_ n$ with $a \not= 0$, see Algebra, Lemma 10.115.2. We may replace $P$ by $\Psi (P)$ and we may replace $A$ by $A_ a$. Thus we may assume that $P$ is monic in $x_ n$ of degree $d > 0$. For $i = 1, \ldots , n - 1$ let $d_ i$ be the degree of $P$ in $x_ i$. Note that this implies that $P^\varphi$ is monic of degree $d$ in $x_ n$ and has degree $\leq d_ i$ in $x_ i$ for every homomorphism $\varphi : A \to \kappa$ where $\kappa$ is a field. Thus if $P^\varphi$ is reducible, then we can write

with $Q_1, Q_2$ monic of degree $e_1, e_2 \geq 0$ in $x_ n$ with $e_1 + e_2 = d$ and having degree $\leq d_ i$ in $x_ i$ for $i = 1, \ldots , n - 1$. In other words we can write

where the sum is over the set $\mathcal{L}$ of multi-indices $L$ of the form $L = (l_1, \ldots , l_{n - 1})$ with $0 \leq l_ i \leq d_ i$. For any $e_1, e_2 \geq 0$ with $e_1 + e_2 = d$ we consider the $A$-algebra

where the $(\text{relations})$ is the ideal generated by the coefficients of the polynomial

with $Q_1$ and $Q_2$ defined as in (37.27.4.1). OK, and the assumption that $P$ is irreducible over $\overline{K}$ implies that there does not exist any $A$-algebra homomorphism $B_{e_1, e_2} \to \overline{K}$. By the Hilbert Nullstellensatz, see Algebra, Theorem 10.34.1 this means that $B_{e_1, e_2} \otimes _ A K = 0$. As $B_{e_1, e_2}$ is a finitely generated $A$-algebra this signifies that we can find an $f_{e_1, e_2} \in A$ such that $(B_{e_1, e_2})_{f_{e_1, e_2}} = 0$. By construction this means that if $\varphi : A_{f_{e_1, e_2}} \to \kappa$ is a homomorphism to a field, then $P^\varphi$ does not have a factorization $P^\varphi = Q_1 Q_2$ with $Q_1$ of degree $e_1$ in $x_ n$ and $Q_2$ of degree $e_2$ in $x_ n$. Thus taking $f = \prod _{e1, e_2 \geq 0, e_1 + e_2 = d} f_{e_1, e_2}$ we win. $\square$