# help again

• Dec 1st 2007, 10:18 PM
kalagota
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 $\displaystyle \phi$? is it like this?

let F , F' be fields and define $\displaystyle \phi : F \rightarrow F'$... i am not sure..

thanks..
• Dec 2nd 2007, 04:01 AM
Plato
Quote:

Originally Posted by kalagota
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 $\displaystyle \phi$? is it like this?

let F , F' be fields and define $\displaystyle \phi : F \rightarrow F'$... i am not sure..

thanks..

It is hard to understand your difficulty.
If $\displaystyle \phi$ maps every element to zero, then you are done.
Otherwise, show it is injective. You said that is the easy part.
• Dec 2nd 2007, 04:05 AM
kalagota
Quote:

Originally Posted by Plato
It is hard to understand your difficulty.
If $\displaystyle \phi$ maps every element to zero, then you are done.
Otherwise, show it is injective. You said that is the easy part.

thanks, but how will i define phi?
• Dec 2nd 2007, 04:14 AM
Plato
Quote:

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

I don't see where we are ask to define $\displaystyle \phi$?
• Dec 2nd 2007, 04:18 AM
kalagota
Quote:

Originally Posted by Plato
I don't see where we are ask to define $\displaystyle \phi$?

ok, check out my "proof"..

let $\displaystyle a,b \in F$ such that $\displaystyle \phi (a) = \phi (b)$

then $\displaystyle \phi (a) - \phi (b) = \phi (a-b) = 0$

but by properties of a homomorphism, $\displaystyle a - b = 0 \implies a=b$

therefore $\displaystyle \phi$ is 1-1.. QED.

is that it?
• Dec 2nd 2007, 07:04 AM
ThePerfectHacker
Quote:

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

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

Thus, if $\displaystyle \phi$ is a homomrophism from the field then $\displaystyle \ker \phi$ is an ideal. But a field can only has $\displaystyle \{ 0\}$ 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.