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 $\phi$? is it like this?

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

thanks..

2. 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 $\phi$? is it like this?

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

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

3. Originally Posted by Plato
It is hard to understand your difficulty.
If $\phi$ 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?

4. 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 $\phi$?

5. Originally Posted by Plato
I don't see where we are ask to define $\phi$?
ok, check out my "proof"..

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

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

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

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

is that it?

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