Suppose K/Q is a quadratic extension, (so $[K : \mathbb{Q}] = 2$). I want to show that $K/\mathbb{Q}$ is Galois and find it's Galois group.
I have already shown that $K=\mathbb{Q}(\sqrt{d})$ for some square free integer d. Does this help? Can I use the fact that $K/\mathbb{Q}$ is galois if $\mathbb{Q}(Aut(K/\mathbb{Q}))= \mathbb{Q}$?
I think that it's galois group $Gal(K/\mathbb{Q})$ is $\cong Z/2Z$ but I'm not sure how to show this.
Can you find an automorphism of $\Bbb{Q}(\sqrt{d})$ that fixes $\Bbb{Q}$?
I like the identity, $\mathrm{id}$, and also $\sqrt{d} \mapsto -\sqrt{d}$. (This latter map doesn't fix $\Bbb{Q}$ elementwise if $d$ is a square, which is (part of) where squarefreeness comes in. (More below in the edit.)) Let's call this second map $P$.
Are there any other automorphisms fixing the base field?
No.
(It might be helpful to think of this as a map on the generic element $a + b \sqrt{d}$ for $a,b \in \Bbb{Q}$, or alternatively as a linear map on the vector space $\Bbb{Q}^2$ with basis $\{1, \sqrt{d}\}$. In the latter case, the generic element is $(a,b)$ and the map is some $2 \times 2$ matrix. Since we want to fix $\Bbb{Q}$ elementwise, we can't move the first coordinate, so the matrix is $\begin{pmatrix} 1 & u \\ 0 & v \end{pmatrix}$. We easily see $u=0,v=1$ is an automorphism, the identity, and $P$ is $u=0,v=-1$, a map of order $2$.
Can we eliminate other choices of $u$ and $v$? Let $\varphi_{u,v}$ be the map we are considering. Note that $\varphi_{u,v}$ is a ring automorphism, so $$\begin{align*} a^2 - b^2d &= \varphi_{u,v}(a^2-b^2d) \\ &= \varphi_{u,v}((a+b\sqrt{d})(a-b\sqrt{d})) \\ &= \varphi_{u,v}(a+b\sqrt{d})\varphi_{u,v}(a-b\sqrt{d}) \\ &= (a+ub + vb\sqrt{d})(a+ub-vb\sqrt{d}) \\ &= (a+ub)^2 - (vb\sqrt{d})^2 \\ &= a^2 + 2uab + u^2b^2-v^2b^2d \end{align*}$$ and we discover $u = 0$ and $v^2 = 1$, so in the previous paragraph we've actually written down all the choices of $u$ and $v$ yielding an automorphism.)
Do the maps $\{\mathrm{id},P\}$ fix anything outside of $\Bbb{Q}$?
No. Any $(a,b)$ with $b \neq 0$ is moved by $P$. So the fixed field is precisely $\Bbb{Q}$.
Is the extension Galois?
Yes. Since the fixed field is precisely the base field and the extension is algebraic (and/or the extension is normal and separable), the extension is Galois.
What's $\mathrm{Gal}(K/\Bbb{Q})$?
We've shown it's the group $\langle \{\mathrm{id}, P\} \rangle \cong \Bbb{Z}/2\Bbb{Z}$.
Edit: If $d$ is a square, $P$ moves elements of $\Bbb{Q}$. If $d$ is not squarefree, $d = e^2 f$ for some $e,f \in \Bbb{Q}$ with $f$ squarefree, and $\sqrt{d} = e \sqrt{f}$. So the extension is really $\Bbb{Q}(\sqrt{f})$, so we might as well only consider extensions by square roots of squarefree rationals.
