Show that each homomorphism of a field is either one to one or maps everything to 0..

the one to one part is easy, i need the second one..

also, how will i define ? is it like this?

let F , F' be fields and define ... i am not sure..

thanks..

Results 1 to 6 of 6

- December 1st 2007, 11:18 PM #1
## help again

Show that each homomorphism of a field is either one to one or maps everything to 0..

the one to one part is easy, i need the second one..

also, how will i define ? is it like this?

let F , F' be fields and define ... i am not sure..

thanks..

- December 2nd 2007, 05:01 AM #2

- December 2nd 2007, 05:05 AM #3

- December 2nd 2007, 05:14 AM #4

- December 2nd 2007, 05:18 AM #5

- December 2nd 2007, 08:04 AM #6

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

**Theorem:**The kernel of a ring homomorphism is an ideal of ring.

Thus, if is a homomrophism from the field then is an ideal. But a field can only has or itself as an ideal. Thus, in the first case the map is one-to-one because the kernel is trivial. And in the other case the map collapses everything into 0 because it is the full field.