# 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.

• – Watson Jan 14 '18 at 14:13

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:

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

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

• 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? – Peter May 1 '17 at 6:02
• @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 $\mathbb{Q}_2$? – user443117 May 5 '17 at 20:27
• @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 $4$. – Starfall May 5 '17 at 20:41