Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Note that the only possible kernels of a ring homomorphism $F\to R$ are $0$ and $F$ itself –  Hagen von Eitzen Jun 6 at 9:45
    
So either 0 map or an identity map is a homomorphism. So, the answer is two? –  user141561 Jun 6 at 11:09
2  
Some definitions of ring homomorphism include the condition $\phi(1)=1$. –  Derek Holt Jun 6 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 at 13:48

1 Answer 1

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.

share|improve this answer
    
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 at 14:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.