Why is $x^2+1$ irreducible in $\mathbb{Q}_2[x]$. I was trying to prove that $\mathbb{Q}_2 \otimes \mathbb{Q}(i)$ is a field where $\mathbb{Q}_2$ represents the 2-adic numbers. I proceed as follows: $\mathbb{Q}_2 \otimes \mathbb{Q}(i) \cong \mathbb{Q}_2 \otimes \mathbb{Q}[x]/(x^2 + 1) \cong \mathbb{Q}_2[x]/(x^2 + 1)$, but at this point I need to prove that $x^2 + 1$ is irreducible in $\mathbb{Q}_2$. Please if you can help me, I'll really appreciate it.

Another way to see that $i\notin\Bbb Q_2$ is to note that the valuation of $1+i$ must be $\frac12$, since $2v_2(1+i)=v_2\bigl((1+i)^2\bigr)=v_2(2i)=1+v_2(i)$. But $i$ is a root of unity, so we must have $v(i)=0$. Now we’ve shown that $v_2(1+i)\notin\Bbb Z$, and thus $1+i\notin\Bbb Q_2$.
