Example 15.9.7 (07M2)—The Stacks project

Example 15.9.7. This example shows that one cannot drop the condition on the leading coefficient in Lemma 15.9.6. Namely, let $A = \mathbf{Z}$, $I = 4\mathbf{Z}$, $f = 1$ and $\bar{g} = \bar{h} = 2x^2 + 2x + 1$ in $A/I[x]$. If $A \to A'$ is étale and $A'/I A' = A/I$, then the $2$-adic completion of $A'$ is isomorphic to $\mathbf{Z}_2$. Now any polynomial $g'$ in $\mathbf{Z}_2[x]$ congruent to $2x^2 + 2x + 1$ mod $4$ will not have a polynomial multiplicative inverse in $\mathbf{Z}_2[x]$, whence the conclusion of the lemma does not hold.

Comments (2)

Comment #10053 by Branislav Sobot on March 18, 2025 at 03:55

I believe this lemma is wrong as stated. I think you need an additional condition that $\bar{g}$ and $\bar{h}$ are at most the degree of $f$ . The counterexample I have in mind is: Take ring $A:=\mathbb{Z}$ , $I:=4\mathbb{Z}$ , $f=1$ and $\bar{g}=\bar{h}=2x^2+2x+1$ .

Comment #10557 by Stacks Project on July 22, 2025 at 10:52

Thanks very much for catching this! I don't know what I was thinking. I've replaced the lemma by your counterexample in this commit. We could fix this in the way you said, but it would have to be something like: for every prime $\mathfrak p \in V(I)$ the images of $\overline{g}$ and $\overline{h}$ in $(A/I)_\mathfrak p[x]$ have sum of degrees at most the degree of $f$ ; I think this would be essentially useless in practice so I refrained from doing this.

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