Decomposition of the tensor product $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i]$ into a product of fields

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.

Solution 1:

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:

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

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