The Stacks project

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 10.142.12. 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 10.142.12 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 10.142.8). Hence there exists an idempotent $e \in B_ g/I B_ g$ such that $A/I = (B_ g/I B_ g)_ e$ (Algebra, Lemma 10.142.9). 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. $\square$


Comments (2)

Comment #1334 by JuanPablo on

I do not understand why .

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

Say the resultant of the polynomials as in example 10.139.13 (tag 00UA) works to make étale and maps to an invertible in (that is, modulo ).

But looking at the universal property of it seems to suggest that in any algebra with a second factorization , generating the unit ideals, monic and of the same degree then , 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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