Homomorphisms between fields are injective.

Question

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.

Answer 1

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 $a = b$.

Answer 2

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.

Answer 3

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 $a = b$.

Answer 4

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 $\{0\}$.