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.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: