Lemma 15.9.7. Let $A$ be a ring, let $I \subset A$ be an ideal. Let $f \in A[x]$ be a monic polynomial. Let $\overline{f} = \overline{g} \overline{h}$ be a factorization of $f$ in $A/I[x]$ and assume that $\overline{g}$, $\overline{h}$ generate the unit ideal in $A/I[x]$. Then there exists an étale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ and a factorization $f = g' h'$ in $A'[x]$ lifting the given factorization over $A/I$.

Proof. Say $f = x^ d + a_1 x^{d - 1} + \ldots + a_ d$ has degree $d$. Write $\overline{g} = \sum \overline{b}_ j x^ j$ and $\overline{h} = \sum \overline{c}_ j x^ j$. Then we see that $1 = \sum \overline{b}_ j \overline{c}_{d - j}$. It follows that $\mathop{\mathrm{Spec}}(A/I)$ is covered by the standard opens $D(\overline{b}_ j \overline{c}_{d - j})$. However, each point $\mathfrak p$ of $\mathop{\mathrm{Spec}}(A/I)$ is contained in at most one of these as by looking at the induced factorization of $f$ over the field $\kappa (\mathfrak p)$ we see that $\deg (\overline{g} \bmod \mathfrak p) + \deg (\overline{h} \bmod \mathfrak p) = d$. Hence our open covering is a disjoint open covering. Applying Lemma 15.9.3 (and replacing $A$ by $A'$) we see that we may assume there is a corresponding disjoint open covering of $\mathop{\mathrm{Spec}}(A)$. This disjoint open covering corresponds to a product decomposition of $A$, see Algebra, Lemma 10.24.3. It follows that

$A = A_0 \times \ldots \times A_ d, \quad I = I_0 \times \ldots \times I_ d,$

where the image of $\overline{g}$, resp. $\overline{h}$ in $A_ j/I_ j$ has degree $j$, resp. $d - j$ with invertible leading coefficient. Clearly, it suffices to prove the result for each factor $A_ j$ separatedly. Hence the lemma follows from Lemma 15.9.6. $\square$

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