How many field homomorphisms? Let $F$ and $F′$ be two finite fields with nine and four elements respectively.
How many field homomorphisms are there from $F$ to $F′$?
 A: Recall that any finite field has a finite characteristic--- a smallest positive integer $n$ with the property that for all $x$ in the field, $x + x + ... + x \;(n \text{ times}) = 0$. One can show from the field axioms that such an $n$ must be prime, and a divisor of the number of elements in the field.
So here we see that $F$ must have characteristic $3$ and $F^{\prime}$ must have characteristic $2$. So let $f$ be any map from $F$ to $F^{\prime}$ satisfying
$f(x+y) = f(x) + f(y) \quad \forall x , y \in F\tag{1}$
and
$f(xy) = f(x) f(y) \quad \forall x , y \in F \tag{2}$
It is easy to show that property $(1)$ implies that $f(0) = 0$. So
$0 = f(0)
= f(1+1+1)$
[... since $1+1+1 = 0$ in $F$, since $F$ has characteristic $3$]
$= f(1) + f(1) + f(1)$
[... by property $(1)$]
$= 0 + f(1)$
[... $f(1) + f(1) = 0$ in $F^{\prime}$ since $F^{\prime}$ has characteristic $2$]
$= f(1)$
[... this is a fundamental property of 0)
and hence for any $x \in F$ we have
$f(x) = f(x*1)$
[... this is what $1$ does in any field]
$= f(x) f(1)$
[... by property $(2)$]
$= f(x) 0$
[... as we just showed f(1) = 0]
$= 0$
[... an elementary consequence of the field axioms is that $y*0 = 0$ for any $y$ in any field].
We conclude that any map from $F$ to $F^{\prime}$ satisfying $(1)$ and $(2)$ must satisfy $f(x) = 0$ for all $x$. It is easy to check that the map given by this formula does satisfy $(1)$ and $(2)$.
Now: the term "field homomorphism" is usually defined so that not only must $(1)$ and $(2)$ be satisfied above but that $1$ must be mapped to $1$. So we have shown that by this definition there are no field homomorphisms whatsoever from $F$ to $F^{\prime}$ and the answer to your question is zero. 
A: Hint $1$: A homomorphism of fields is injective.  Can you see why?
Hint $2$:  Hint $1$ answers your question.  Can you see why?
