Lemma 15.9.5. 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 $A/I[x]$ such that $\overline{g}$ and $\overline{h}$ are monic and 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 $A'[x]$ with $g'$, $h'$ monic lifting the given factorization over $A/I$.

Proof. We will deduce this from results on the universal factorization proved earlier; however, we encourage the reader to find their own proof not using this trick. Say $\deg (\overline{g}) = n$ and $\deg (\overline{h}) = m$ so that $\deg (f) = n + m$. Write $f = x^{n + m} + \sum \alpha _ i x^{n + m - i}$ for some $\alpha _1, \ldots , \alpha _{n + m} \in A$. Consider the ring map

$R = \mathbf{Z}[a_1, \ldots , a_{n + m}] \longrightarrow S = \mathbf{Z}[b_1, \ldots , b_ n, c_1, \ldots , c_ m]$

of Algebra, Example 10.143.12. Let $R \to A$ be the ring map which sends $a_ i$ to $\alpha _ i$. Set

$B = A \otimes _ R S$

By construction the image $f_ B$ of $f$ in $B[x]$ factors, say $f_ B = g_ B h_ B$ with $g_ B = x^ n + \sum (1 \otimes b_ i) x^{n - i}$ and similarly for $h_ B$. Write $\overline{g} = x^ n + \sum \overline{\beta }_ i x^{n - i}$ and $\overline{h} = x^ m + \sum \overline{\gamma }_ i x^{m - i}$. The $A$-algebra map

$B \longrightarrow A/I, \quad 1 \otimes b_ i \mapsto \overline{\beta }_ i, \quad 1 \otimes c_ i \mapsto \overline{\gamma }_ i$

maps $g_ B$ and $h_ B$ to $\overline{g}$ and $\overline{h}$ in $A/I[x]$. The displayed map is surjective; denote $J \subset B$ its kernel. From the discussion in Algebra, Example 10.143.12 it is clear that $A \to B$ is etale at all points of $V(J) \subset \mathop{\mathrm{Spec}}(B)$. Choose $g \in B$ as in Lemma 15.9.4 and consider the $A$-algebra $B_ g$. Since $g$ maps to a unit in $B/J = A/I$ we obtain also a map $B_ g/I B_ g \to A/I$ of $A/I$-algebras. Since $A/I \to B_ g/I B_ g$ is étale, also $B_ g/IB_ g \to A/I$ is étale (Algebra, Lemma 10.143.8). Hence there exists an idempotent $e \in B_ g/I B_ g$ such that $A/I = (B_ g/I B_ g)_ e$ (Algebra, Lemma 10.143.9). Choose a lift $h \in B_ g$ of $e$. Then $A \to A' = (B_ g)_ h$ with factorization given by the image of the factorization $f_ B = g_ B h_ B$ in $A'$ is a solution to the problem posed by the lemma. $\square$

Comment #1334 by JuanPablo on

I do not understand why $A/I\cong A'/IA'$.

In the proof as currently written it does not seem to work.

Say $g=\Delta$ the resultant of the polynomials $x^n+\Sigma b_{i}x^{n-i}, x^m+\Sigma c_{j}x^{m-j}$ as in example 10.139.13 (tag 00UA) works to make $A\rightarrow B_g$ étale and maps to an invertible in $B\rightarrow A/I$ (that is, modulo $J$).

But looking at the universal property of $B_{\Delta}/IB_{\Delta}$ it seems to suggest that in any $A/I$ algebra with a second factorization $\bar{g}\bar{h}=\bar{g}'\bar{h}'$, $(\bar{g},\bar{h}),(\bar{g}',\bar{h}')$ generating the unit ideals, monic and of the same degree then $(\bar{g},\bar{h})=(\bar{g}',\bar{h}')$, which is in general false.

Comment #1354 by on

Thanks very much for pointing this out. I have fixed it here.

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