Proving Injectivity of a Ring Homomorphism over a Field

[annoymage]

Homework Statement

Let R be a field and f : R->R be a ring homomorphism

prove that f(r)=0, for all r in R, or f is injective

Homework Equations

n/a

The Attempt at a Solution

or alternative ways i have to prove (Kernel of f)=R or (kernel of f)={0}

i've tried but stuck somewhere, hmm and also seems i can't make any connection with "field" like unit or zero divisor or something like that T_T, help clue pls

 

[JonF]

Suppose for some a,b,c where a and b are distinct in R f(a) =c = f(b). I.e. f is not injective.

With this assumption, can you show that f(1) = 0? (which is equivlent to f(r)=0, for all r in R)

hint: f(ab^-1 - ba^-1) = 0 and ab^-1 - ba^-1 is non zero (you will have to argue this).

 

[JonF]

(Kernel of f)=R or (kernel of f)={0}

This is actually way easier than my other idea.

Suppose any non zero element r satisfies f(r) = 0 what would f(r*r^-1) look like.

 

[annoymage]

sorry a bit late, i think i get the point,

check me please,

Suppose it is not injective, so exist x in R\{0} such that f(x)=0,

then f(x)=f(xe)=f(x)f(e)=0, since R is field, inverse of f(x) exist, then imply, f(e)=0

so, for any y in R f(e)=f(yy^-1)=f(y)f(y^-1)=0, then inverse of f(y^-1) exist, so, f(y)=0 for any y in R,

is my argument correct? and also, is it ok if i not state which R belongs to?

and also this hint, "hint: f(ab^-1 - ba^-1) = 0 and ab^-1 - ba^-1 is non zero (you will have to argue this)."

hmmmm, i can show how f(ab^-1 - ba^-1) = 0, and to show f(e)=0

is this correct?

ab^-1 - ba^-1 non zero, so inverse exist, let D be it's inverse, so f(ab^-1 - ba^-1)f(D)=0,

f((ab^-1 - ba^-1)D)=f(e)=0, hmm it's correct right?