Let $l$ be a prime number and let $x$ be an integer which is an $l$-th power modulo every prime. The claim is that $x$ is the $l$-th power of some integer.
I've read the following proof, but I don't understand it:
"Consider $K=\mathbb{Q}(x^{1/l})$ and its Galois closure $L / \mathbb Q$. Then for each prime $p$ which is unramified in $L$, there is a prime ideal $P$ in $K$ above $p$ such that $f(P/p)=1$. So the corresponding Frobenius fixes some $l$-th root of $x$. By Cebotarev theorem, $Gal(L/\mathbb{Q})$ has no element of degree $l$ and so it fixes some $l$-th root of $x$, which lies in $\mathbb{Q}$."
My questions:
Why does the Frobenius element $Frob_{Q/p} \in Gal(L/\mathbb Q)$ fix some $l$-th root of $x$? Here $Q$ is a prime of $L$ above $P$.
Why does $Gal(L/\mathbb{Q})$ have no element of degree $l$ ? A priori, we don't know whether $T^l - x$ is irreducible or not over $\Bbb Z$.
Why does it imply that $Gal(L/\mathbb{Q})$ fix some $l$-th root of $x$?
I know the statement of Cebotarev theorem but don't know how to apply it here. I would like some details about this, please.
