Let $p$ be a prime and let $\Bbb{Q}(ζ_p)/\Bbb{Q}$ be cyclotomic extension. Let $\Bbb{Q}_p(ζ_p)/\Bbb{Q}_p$ be cyclotomic extension of p adic field.
I wonder why $Gal(\Bbb{Q}(ζ_p)/\Bbb{Q})$ and $Gal(\Bbb{Q}_p(ζ_p)/\Bbb{Q}_p)$ are the same.
I think if I could prove the restriction map $Gal(\Bbb{Q}_p(ζ_p)/\Bbb{Q}_p)$→ $Gal(\Bbb{Q}(ζ_p)/\Bbb{Q})$, $σ→σ｜\Bbb{Q}(ζ_p)$ is injective, then, it is done because it is easy to see order of $Gal(\Bbb{Q}(ζ_p)/\Bbb{Q})$ is smaller than $p-1$ and locally cyclotomic extension is just Lubin tate and it's galois group's order is p-1.
Thank you for your help.
Lubin-Tate theory is overkill.
Method 1: Let $F$ be a field of characteristic $0$ and $\Phi_n(x)$ in $\mathbf Q[x]$ (in fact $\mathbf Z[x]$) be the $n$th cyclotomic polynomial. If $\Phi_n(x)$ is irreducible in $F[x]$ and $\zeta$ is a root of $\Phi_n(x)$ then $F(\zeta)/F$ is Galois with ${\rm Gal}(F(\zeta)/F) \cong (\mathbf Z/n\mathbf Z)^\times$ by the exact same proof as used when $F = \mathbf Q$, because the only property we need about $\Phi_n(x)$ for that proof is its irreducibility.
So all you need to do to answer your question is show $\Phi_p(x)$ is irreducible over $\mathbf Q_p$, which is done in the same way as over $\mathbf Q$: $\Phi_p(x+1)$ is Eisenstein at the prime $p$ in $\mathbf Z_p$ (instead of the prime $p$ in $\mathbf Z$).
Method 2: Consider the extensions $\mathbf Q(\zeta_p)/\mathbf Q$ and $\mathbf Q_p/\mathbf Q$. The first one is a finite Galois extension. From Galois theory, if $L/K$ is Galois and $M/K$ is arbitrary, then $LM/M$ is Galois with restriction giving an isomorphism ${\rm Gal}(LM/M) \to {\rm Gal}(L/L \cap M)$. So if $L \cap M = K$ then restriction gives an isomorphism ${\rm Gal}(LM/M) \to {\rm Gal}(L/K)$. Take $K = \mathbf Q$, $L = \mathbf Q(\zeta_p)$, and $M = \mathbf Q_p$. You want to show $L \cap M = \mathbf Q(\zeta_p) \cap \mathbf Q_p$ is equal to $\mathbf Q$. To do that, we use some ramification in algebraic number theory. Since $\mathbf Q(\zeta_p)/\mathbf Q$ is totally ramified at $p$, every intermediate field is Galois over $\mathbf Q$ and totally ramified at $p$. At the same time, a Galois extension of $\mathbf Q$ that has an embedding into $\mathbf Q_p$ is unramified at $p$. That makes $\mathbf Q(\zeta_p) \cap \mathbf Q_p$ both totally ramified and unramified at $p$, so that intersection must be $\mathbf Q$.