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I am trying to solve the following problem:

For each rational prime $p$, describe the decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields, where $\mathbb{Q}_p$ is the field of $p$-adic numbers.

I know that the tensor product of $2$ extensions of a field one of which is finite is Artinian and is therefore a product of Artinian local rings, but I do not know how to compute these Artinian local rings. Please if you can help me with this, I'll really appreciate it.

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In general, tensoring a number field with $ \mathbf Q_p $ gives you an appropriate direct product of the completions of it at the different primes lying over $ p $. The key isomorphism is

$$ \mathbf Q_p \otimes \mathbf Q(i) \cong \mathbf Q_p \otimes \mathbf Q[x]/(x^2 + 1) \cong \mathbf Q_p[x]/(x^2 + 1) $$

Now, we have three cases. If $ p = 2 $, then $ X^2 + 1 $ remains irreducible in $ \mathbf Q_2[X] $, since it is irreducible modulo $ 4 $. If $ p \equiv 3 \pmod{4} $, $ X^2 + 1 $ is irreducible modulo $ p $, thus it is also irreducible in $ \mathbf Q_p $. Finally, if $ p \equiv 1 \pmod{4} $, then $ X^2 + 1 $ has a root modulo $ p $ and its derivative does not vanish at this root, so Hensel's lemma gives a root in $ \mathbf Q_p $. Therefore:

$$ \mathbf Q_p \otimes_{\mathbf Q} \mathbf Q(i) \cong \mathbf Q_p(\sqrt{-1}) \textrm{ if } p = 2 \textrm{ or } p \equiv 3 \pmod{4} $$

$$ \mathbf Q_p \otimes_{\mathbf Q} \mathbf Q(i) \cong \mathbf Q_p \times \mathbf Q_p \textrm{ if } p \equiv 1 \pmod{4} $$

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  • $\begingroup$ thank you for your quick response. I just have a question about your solution: the last isomorphism, namely, $\mathbb{Q}_p \otimes \mathbb{Q}[x]/(x^2 + 1) \equiv \mathbb{Q}_p[x]/(x^2 + 1)$ follows because $\mathbb{Q}_p$ is a localization of $\mathbb{Z}_p$ and the natural map $R[U^{-1}] \otimes_R M \to M[U^{-1}]$, defined by sending $r/u \otimes m$ to $rm/u$ is an isomorphism, right? $\endgroup$ – Peter May 1 '17 at 6:02
  • $\begingroup$ @Starfall, could you please explain your statement "if $p=2$, then $x^2+1$ remains irreducible in $\mathbb{Q}_2[x]$ since it is irreducible modulo $4$". Why don't we look modulo $2$, where it factors as $x^2+1=(x+1)^2$, and get that $x^2+1$ is reducible in $\mathbb{Q}_2$? $\endgroup$ – user443117 May 5 '17 at 20:27
  • $\begingroup$ @user443117 Reducibility modulo $ 2 $ only implies reducibility in $ \mathbf Q_2[X] $ if the polynomial is separable. Indeed, $ X^2 + 1 $ can't be reducible in $ \mathbf Q_2[X] $, because it is irreducible modulo $ 4 $. $\endgroup$ – Starfall May 5 '17 at 20:41

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