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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #1334 by JuanPablo on
Comment #1354 by Johan on