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How would I prove this?

I know that I must show $f(a)=f(b) \Rightarrow a = b$

I also know I must use the definition of homomorphism, ie:

$f(a+b)=f(a)+f(b)$

$f(ab)=f(a)f(b)$

$f(1)=1$

I am assuming that a contradiction would be a good approach to solve this, but not quite sure on specifics.

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  • $\begingroup$ Duplicate. See here $\endgroup$ – Troy Woo Oct 4 '14 at 13:13
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    $\begingroup$ I don't believe that this is a duplicate (or more accurately, I don't believe that the OP, given the way this question was asked, could possibly regard it as a duplicate). $\endgroup$ – John Hughes Oct 4 '14 at 13:26
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Suppose $f(a) = f(b)$, then $f(a-b) = 0 = f(0)$. If $u = (a-b) \ne 0$, then $f(u)f(u^{-1}) = f(1) = 1$, but that means that $0 f(u^{-1}) = 1$, which is impossible. Hence $a - b = 0$ and $a = b$.

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  • $\begingroup$ possibly a silly question, but why does 0=f(0) and why does f(u)=0? $\endgroup$ – Mark Oct 4 '14 at 13:16
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    $\begingroup$ I've defined $u = a-b$, and just established that $f(a-b) = 0$. So $f(u) = 0$. As for why $f(0) = 0$, it's because $f(0+1) = f(0) + f(1)$, which gives $f(1) = f(0) + f(1)$, so $1 = f(0) + 1$, so $0 = f(0)$. $\endgroup$ – John Hughes Oct 4 '14 at 13:23
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A field homomorphism must in particular be a ring homomorphism, so its kernel is an ideal. The only ideals of a field are the zero ideal and the field itself.

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  • $\begingroup$ I have not studied ideals, can you explain this concept? $\endgroup$ – Mark Oct 4 '14 at 13:12

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