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In Computing the quotient $\mathbb{Q}_p[x]/(x^2 + 1)$ and Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields it has been claimed that since $x^2+1$ is irreducible mod $4$ it is irreducible in $\mathbb{Q}_2[x]$ and since $x^2+1$ is irreducible mod $p$ for $p \equiv 3 \bmod 4$ it is irreducible in $\mathbb{Q}_p[x]$. I understand (by inspection for $4$ and quadratic reciprocity for $p \equiv 3 \bmod 4$) that $x^2+1$ is indeed irreducible modulo the relevant primes. But why does this imply that it is irreducible in $\mathbb{Q}_p[x]$? I have tried using Hensel's lemma, but I could only get it to show that if there is a root mod $p$, then it will lift to a root in $\mathbb{Z}_p$. This problem seems to require a converse.

user26857
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user443117
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    If there is a root in $\mathbb Q_p$, it must be in $\mathbb Z_p$, and then give a root in $\mathbb Z_p/p\mathbb Z\cong \mathbb Z/p$. – paul garrett May 03 '17 at 22:49
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    @paulgarrett As for your first claim (if there is a root in $\mathbb{Q}_p$, it must be in $\mathbb{Z}_p$), is that because specific to -1 (maybe because it doesn't have a denominator?), or is there some underlying theory I should know? For the second part, I assume you mean $\mathbb{Z}_p/p\mathbb{Z}_p$? Thanks for your help--I'm new to $p$-adic numbers and Hensel's Lemma, and though I suspect this isn't too hard, I still don't quite understand. – user443117 May 03 '17 at 22:57
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    Take the $p$-adic absolute value of both sides of $x^2=-1$ to ascertain $|x|_p\le 1$ so that $x$ must be in $\Bbb Z_p$. And yes, $\Bbb Z_p/p\Bbb Z_p$, which is $\cong \Bbb Z/p\Bbb Z$. – anon May 03 '17 at 23:03
  • Yes, sorry, the quotient should be by $p\mathbb Z_p$... – paul garrett May 03 '17 at 23:05

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