The Stacks project

Lemma 15.9.6. 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

  1. the leading coefficient of $\overline{g}$ is an invertible element of $A/I$, and

  2. $\overline{g}$, $\overline{h}$ generate the unit ideal in $A/I[x]$.

Then there exists an ├ętale 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. Applying Lemma 15.9.1 we may assume that the leading coefficient of $\overline{g}$ is the reduction of an invertible element $u \in A$. Then we may replace $\overline{g}$ by $\overline{u}^{-1}\overline{g}$ and $\overline{h}$ by $\overline{u}\overline{h}$. Thus we may assume that $\overline{g}$ is monic. Since $f$ is monic we conclude that $\overline{h}$ is monic too. In this case the result follows from Lemma 15.9.5. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07M1. Beware of the difference between the letter 'O' and the digit '0'.