Does $ \Bbb{Q}_2$ has $\sqrt{-1}$?

Question

Does $ \Bbb{Q}_2$ has $\sqrt{-1}$?

I tried to use Hensel lemma as usual. Let $f(x)＝x^2＋1$. But if some $a∈\Bbb{Z}$, $f(a)＝0$, then $f'(a)$ can always divide by $2$. So I cannot use Hensel lemma.

Could you help me?

Answer 1

An alternative approach: consider $1+i$, and its minimal $\Bbb Q_2$-polynomial. You see that it’s $X^2-2X+2$, clearly $\Bbb Q_2$-irreducible by Eisenstein. So $1+i\notin\Bbb Q_2$, and thus $i\notin\Bbb Q_2$.

Answer 2

No: if the monic polynomial $x^2+1$ had a root in $\Bbb Q_2$ then that root would be in the ring of integers $\Bbb Z_2$; but a root in $\Bbb Z_2$ implies that there is a root modulo $2^k$ for all $k\ge1$, contradicting the fact that $x^2+1$ has no root modulo $4$.