Automorphisms of $\mathbb Q(\sqrt 2)$ I have a really quick question in Galois theory:
If I have a field such as $\mathbb Q(\sqrt2)$, and I want to look at the automorphisms of it, it seems clear that $a+b\sqrt 2\mapsto a+x\sqrt 2$ for some $x$ (as $a\mapsto a$ to ensure that $\sigma(1) =1$, which ensure's its a homomorphism).
My question is why can $x$ only be $\pm b$?  What's the barrier with a field automorphism $a+b\sqrt 2\mapsto a+2b\sqrt 2$?

You need that $\sigma(\sqrt{2})^2 = \sigma(\sqrt{2}^2) = \sigma(2) = 2$.
Thus,  $\sigma(\sqrt{2})$ also needs to be a root of $X^2 - 2$. 
This can be generalized for any algebraic number: its image must have the same minimal polynomial, so the choice are the various roots of the minimal polynomial.  
