How to show $\sqrt{p}$ is not in $\mathbb{Q}_p$?

Question

I am pretty sure that $\sqrt{p}$ is not in $\mathbb{Q}_p$, the $p$-adic numbers. But I just can't quite show it. Could someone please explain how I could show this? Thanks!

Answer 1

HINT: Suppose there existed $x\in\Bbb Q_p$ such that $x^2=p$. What would be then $|x|_p$?

Answer 2

Suppose otherwise. Then

$$ 2 v(\sqrt{p}) = v((\sqrt{p})^2) = v(p) = 1 $$

which contradicts the fact the $p$-adic valuation on $\mathbf{Q}_p^\times$ is always an integer.

Answer 3

You can show (or you already know) that the p-adic valuation $v_p$ extends from $\mathbb{Q}^{\times}$ to $\mathbb{Q}_p^{\times}$ and is continuous, hence the image of $v_p$ is still $\mathbb{Z}$. If $\sqrt{p}$ was in $\mathbb{Q}_p$, then $v_p(\sqrt{p})$ would have to be half of $v_p(p)$ and at the same time in $\mathbb{Z}$.