## Prove that a given map is a field automorphism

Prove directly that the map $\varphi : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$ given by $a+b\sqrt{2} \mapsto a-b\sqrt{2}$ is a field isomorphism.

Note that elements of $\mathbb{Q}(\sqrt{2})$ can be written uniquely in the form $a+b\sqrt{2}$ where $a,b \in \mathbb{Q}$, so that this rule indeed defines a function.

Now $\varphi((a+b\sqrt{2}) + (c+d\sqrt{2})) = \varphi((a+c) + (b+d)\sqrt{2})$ $= (a+c) - (b+d)\sqrt{2}$ $= (a-b\sqrt{2}) + (c-d\sqrt{2})$ $= \varphi(a+b\sqrt{2}) + \varphi(c+d\sqrt{2})$, so that $\varphi$ preserves addition.

Similarly, we have $\varphi((a+b\sqrt{2})(c+d\sqrt{2})) = \varphi((ac+2bd) + (ad+bc)\sqrt{2})$ $= (ac+2bd) - (ad+bc)\sqrt{2}$ $= (a-b\sqrt{2})(c-d\sqrt{2})$ $= \varphi(a+b\sqrt{2})\varphi(c+d\sqrt{2})$, so that $\varphi$ preserves multiplication.

Thus $\varphi$ is a ring homomorphism, and hence a field homomorphism.

Note that if $\varphi(a+b\sqrt{2}) = 0$, then we have $a-b\sqrt{2} = 0$, and so if $b = 0$ we have $a/b = \sqrt{2}$, where $a$ and $b$ are rational numbers- a contradiction since $\sqrt{2}$ is not rational. So $b = 0$, and thus $a = 0$, and we have $\mathsf{ker}\ \varphi = 0$. So $\varphi$ is injective.

Finally, $\varphi$ is surjective since we have $\varphi(a+(-b)\sqrt{2}) = a+b\sqrt{2}$ for all $a$ and $b$.

So $\varphi$ is a field isomorphism.

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