Ring homomorphism defined on a field 
Let $\phi\colon R \to R'$ be a ring homomorphism. Prove that if $R$ is a field then either $\phi$ is an isomorphism or $\phi(r) = 0$ for all $r \in R$.

I am stuck on this problem and don't know where to begin. I feel like I'm very weak in writing proofs.
 A: The kernel of $\phi:R\to R'$ (referred to in the comments) is the set $\{r\in R : \phi(r)=0\}$.

*

*Exercise 1: The kernel is a two-sided ideal of $R$, i.e. an additive subgroup $I$ of $R$ with the property that $rI, Ir\subseteq I$ for every $r\in R$.

*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).

A: 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.
A: I hope I'm not giving away too much, but I did this problem only recently! Note if $R$ is a field, then it's simple i.e. it's ideals are only the ${0}$ ideal or $R$, Kernel is an ideal also... I hope you get my drift... if you need any further help let me know! (though I doubt you do!) 
A: Suppose $R$ is a field.  If $\phi(r) = 0$ for all $r \in R$, we're done.  So suppose this is not the case.  If we can now show that $\phi$ is an isomorphism, we'd be done.
Let $r \in R$ with $\phi(r) \neq 0$.  Let $a$ be nonzero.  Then it has an inverse $a^{-1}$ since $R$ is a field.  Further, $$\phi(r) = \phi(aa^{-1}r) = \phi(a)\phi(a^{-1})\phi(r).$$  Can you conclude from this that $\phi$ is injective?  That is, can you show that $\phi(a) \neq 0$ for our arbitrary nonzero $a$?
