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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 $a+b\sqrt 2\mapsto a+2b\sqrt 2$?

Arthur
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Mark Schultz-Wu
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1 Answers1

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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.

quid
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