Show that a given extension of QQ contains i

Show that i \in \mathbb{Q}(\sqrt{2} + i).

Recall that in our proof that F(\alpha,\beta) = F(\theta) for some \theta, we found that \theta may be taken to be \alpha + c\beta so long as \alpha + c\beta \neq \alpha^\prime + c\beta^\prime for any conjugates \alpha^\prime of \alpha and \beta^\prime of \beta. The minimal polynomials of \sqrt{2} and i over \mathbb{Q} are x^2 - 2 and x^2+1, and their conjugates are \pm\sqrt{2} and \pm i, respectively. Evidently, \sqrt{2} - i, -\sqrt{2}+i, and -\sqrt{2}-i are all distinct from \sqrt{2}+i. Thus \mathbb{Q}(\sqrt{2},i) = \mathbb{Q}(\sqrt{2}+i). In particular, i \in \mathbb{Q}(\sqrt{2} + i).

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: