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I'm currently reading a paper of Rene Schoof, and I got stuck in a line. And I'm trying to check the above sentence. Although that seems to be elementary, I hope someone can give me a counterexample or a proof.

Thank you.

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    $\begingroup$ Polynomial Bezout identity? $\endgroup$ – Seva Aug 1 at 4:24
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    $\begingroup$ Your $\mathbf Z_p$ is presumably notation for the integers mod $p$, not the $p$-adic integers. In a PID, if two elements are relatively prime do you know if they generate the ring? $\endgroup$ – KConrad Aug 1 at 4:43
  • $\begingroup$ I'm Sorry for the notation, by $\mathbb{Z}_p$ I mean the ring of p-adic integers. By modulo p, I mean modulo $p\mathbb{Z}_p[X]$. $\endgroup$ – gualterio Aug 1 at 4:45
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If the two polynomials are $f,g$, then the $\mathbb{Z}_{p}$-module $M := \mathbb{Z}_{p}[X]/(f,g)$ is finitely generated (since at least one of $f,g$ is monic) and satisfies $M \otimes_{\mathbb{Z}_{p}} \mathbb{Z}_{p}/(p) = 0$, hence $M = 0$ by Nakayama.

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  • $\begingroup$ Thank you very much ! $\endgroup$ – gualterio Aug 1 at 5:41

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