Ring homomorphism defined on a field

the kernel of $\phi$ (referred to in the comments) is the set $\{r\in R : \phi(r)=0\}$. exercise 1: this is an ideal of $R$ (an additive subgroup $I$ of $R$ with the property that $rI\in I$ for every $r\in R$, need to be a little more specific if $R$ is noncommutative etc.). exercise 2: a field $F$ only has two ideals, $0$ and $F$. so if $R$ is a field then the kernel of $\phi$ is either $0$ (in which case $\phi$ is injective) or all of $R$ (in which case $\phi$ is the zero map)

Here's a proof sketch from first principles. This should work even if your professor "skips around".

THEOREM $\ $ TFAE for a field $\rm\:R\:$ and a ring hom $\rm\ f\:: R\to R'$

$\rm (1)\ \ \ f\:$ is not one-one

$\rm (2) \ \ \ f(r) = 0\ $ for some $\rm\ r\ne 0,\ \ r\in R$

$\rm (3) \ \ \ f(1) = 0$

$\rm (4) \ \ \ f(R) = 0$

Proof $\rm\ (1\Rightarrow 2)\ \ \ a\ne b,\ f(a) = f(b)\ \Rightarrow\ f(a-b) = f(a)- f(b) = 0$

$\rm\ (2\Rightarrow 3)\ \ \ r\ne 0\ \Rightarrow 1/r\in R\ \Rightarrow\ f(1) = f(r\cdot 1/r) = f(r)\ f(1/r) = 0$

$\rm\ (3\Rightarrow 4)\ \ \ f(r) = f(1\cdot r) = f(1)\ f(r) = 0$

$\rm\ (4\Rightarrow 1)\ \ \ R$ a field $\rm\Rightarrow 1\ne 0\:,\:$ so $\rm\ f(1) = f(0) = 0\ \Rightarrow\ f\:$ is not one-one.