Any field containing the nth roots of unity for odd n also contains the 2nth roots of unity

Let F be a field over which x^n-1 splits where n is odd. Show that x^{2n}-1 also splits over F.

Note that x^{2n}-1 = (x^n)^2-1 = (x^n+1)(x^n-1).

Since n is odd, if \zeta is a root of x^n-1, then (-\zeta)^n+1 = -\zeta^n+1 = 0. That is, the roots of x^n+1 are precisely the negatives of the roots of x^n-1 (note that these are all distinct, since the derivative of x^n-1 has no nonzero roots). So any field containing the roots of x^n-1 also contains the roots of x^{2n}-1.

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: