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..

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- Dec 1st 2007, 10:18 PMkalagotahelp 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.. - Dec 2nd 2007, 04:01 AMPlato
- Dec 2nd 2007, 04:05 AMkalagota
- Dec 2nd 2007, 04:14 AMPlato
- Dec 2nd 2007, 04:18 AMkalagota
- Dec 2nd 2007, 07:04 AMThePerfectHacker
**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.