As I posted just thought that to show is also a root of you would say something along the lines of factors into for some and somehow this ends up with as a root... I know this is obviously not correct but am I along the right lines??
Hi,
I've been set a question and think I've managed about half of it but stuck on some other bits..
Show that is irreducible over . Let be given as a root of . Show that is also a root of . Hence, find a splitting field for over . Identity the Galois group . (You must justify any claim that a particular map is an isomorphism. Be sure to state carefully
any facts that you assume in doing this.)
first part is easy just reducing by to get \ which has no zeros in and thus irreducible over and hence over also.
I think I can identify the Galois group using Prop 22.4 which gives , a perfect square thus the Galois Group is
I can't think how to show that is also a root of . For the splitting field all I can think to do is say that as is a root of then splits in and so is the splitting field for over .
Any help for this last bit or if you can point out any errors I've already made would be greatly appreciated!
As I posted just thought that to show is also a root of you would say something along the lines of factors into for some and somehow this ends up with as a root... I know this is obviously not correct but am I along the right lines??