Why is $x^2+1$ irreducible in $\mathbb{Q}_2[x]$.

Question

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.

Answer 1

It's enough to show that $-1$ is not a square in $\mathbb{Q}_2$, and to prove this, you can use the fact that $\mathbb{Z}_2^{\times}=1+2\mathbb{Z}_2$ to show that any unit $x\in\mathbb{Z}_2^{\times}$ which is a square must lie in $1+8\mathbb{Z}_2$.

Answer 2

A monic quadratic polynomial over any field $K$ is irreducible if and only if it has no root. Now $x^2-a$ has no root in $\mathbb{Q}_2$, if and only if $c\not\equiv 1\bmod 8$, where $a=4^{\mu(a)}c$, with $4\nmid c$. Hence $x^2+1$ has no root in $\mathbb{Q}_2$.

Answer 3

This is just Hensel's lemma. Since $(\Bbb Z/8)^\times\cong \Bbb Z/2\oplus\Bbb Z/2$ is a group of exponent $2$, squaring gives the identity on every element, in particular $-1$ is not a square mod $8$ and so is not in $\Bbb Q_2$ either.

Answer 4

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 $1+i\notin\Bbb Q_2$.