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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3$\begingroup$ Polynomial Bezout identity? $\endgroup$– SevaCommented Aug 1, 2019 at 4:24
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1$\begingroup$ 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? $\endgroup$– KConradCommented Aug 1, 2019 at 4:43
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$\begingroup$ I'm Sorry for the notation, by $\mathbb{Z}_p$ I mean the ring of p-adic integers. By modulo p, I mean modulo $p\mathbb{Z}_p[X]$. $\endgroup$– gualterioCommented Aug 1, 2019 at 4:45
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1 Answer
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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.
answered Aug 1, 2019 at 5:27