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How many homomorphisms are there from a finite field to a ring?

I have some basic knowledge of this but I am not able to put it into use. I know

$1)$ If $\phi$$(1)=a$, then $|a|$ should divide both, order of the field as well as the order of the ring. But order of the ring isn't specified here.

$2)$ Also, since $\phi(1.1)=\phi(1).\phi(1)$ so $a^2=a$ i.e. a homomorphism maps 1 to an idempotent.

Please help me on how to proceed further.

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Note that the only possible kernels of a ring homomorphism $F\to R$ are $0$ and $F$ itself – Hagen von Eitzen Jun 6 '14 at 9:45
So either 0 map or an identity map is a homomorphism. So, the answer is two? – user141561 Jun 6 '14 at 11:09
Some definitions of ring homomorphism include the condition $\phi(1)=1$. – Derek Holt Jun 6 '14 at 11:17
@user141561 There may be less than two, for example if $F=\mathbb F_2$ and $R$ has odd order. There may be more than two, for example because $F$ alone might already have a few automorphisms, or because $R$ contains several "copies" of $F$. – Hagen von Eitzen Jun 6 '14 at 13:48
up vote 2 down vote accepted

There can be just the trivial homomorphism, as is the case with $F_3\to \Bbb Z$, or there could be infinitely many rng homomorphisms, as is the case with $F_2\to \prod_{i=1}^\infty F_2$.

There isn't a uniform answer for all pairs of fields and rings: it depends on your rings, or else what you want to require of the homomorphism.

Supposing that the homomorphism maps the identity of $F$ to the nonzero identity of $R$, you have in injection of $F$ into $R$, so $R$ must have the same characteristic as $F$. This is necessary but not sufficient because, for example, $F_4$ can't be injected into $F_2$ even though they have the same characteristic. If the field has a prime number of elements, then it is sufficient as well. The field is additively cyclic, and the image of $1$ will determine the images of everything else.

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Also, a finite field is multiplicatively cyclic and so the image of a generator determines the images of everything else. (Of course, $0$ goes to $0$.) – lhf Jun 6 '14 at 14:14

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