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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.

Hunter Batley
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Mark
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5 Answers5

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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, unless $f$ is trivial. Hence $a - b = 0$ and $a = b$ or $f$ is trivial.

Jaxiak
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John Hughes
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    possibly a silly question, but why does 0=f(0) and why does f(u)=0? – Mark Oct 04 '14 at 13:16
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    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 $0 = f(0)$. – John Hughes Oct 04 '14 at 13:23
  • Nice answer! Am I right that $f$ cannot be trivial because it is a field homomorphism and therefore must map the unit to the unit by definition? – RFZ Mar 11 '25 at 11:00
  • Excellent question, one that can be resolved by looking up the definition and (perhaps) memorizing it. – John Hughes Mar 11 '25 at 13:19
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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.

Parcly Taxel
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ncmathsadist
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To prove it is an injective we must show that if $f(a) = f(b)$ then $a = b$.

Now suppose $f(a) = f(b)$ but $a \ne b$, then we have that $f(a) - f(b) = 0$ and $u = a - b \ne 0$.

So $f(u u^{-1}) = f(u)f(u^{-1}) = 1$, but $f(u) = f(a - b) = f(a) - f(b) = 0$. We have that $f(u)f(u^{-1}) = 1 = 0 f(u^{-1})$ in which for field is impossible. Thus $a = b$.

Ari Royce Hidayat
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The kernel of $f$ must be equal to $\{0\}$. Suppose that $f(a)=0$ for $a\ne0$; then for all $b$ one has $f(ab)=f(a)f(b)=0$. Since, for $a\ne0$ the function $a\rightarrow ab$ is onto in a field we have $f=0$ which is not possible because $f(1)=1$. It follows the kernel of $f$ is $\{0\}$.

Ataulfo
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Let us assume that the homomorphism is not injective that is there exists $f(a)=f(b)$, where $a\neq b$ which implies $f(a-b)=0$ as it is a homomorphism. This means that the kernel of $f$ has a non zero element say $d$. $f(\frac{1}{d})= \frac{f(1)}{f(d)}$, the image of which is not defined because of division by zero. This is a contradiction as every element in the domain of the homomorphism has to have a well defined image.

Daniel Aricatt
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