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Suppose we have an integral domain, S, of characteristic p, where p is prime. If we have the ring homomorphism θ: S → S by θ(a) = a^p, prove that θ is onto.

Since the commutative ring S is arbitrary, my plan is to show that θ is one-to-one, and since it maps S to S, it must also be onto. However, I'm not sure how to show this. Any help would be greatly appreciated.

Edit: Disregarding whether θ is onto, how could you show that Kerθ contains only 0, so that you could make the statement that θ is one-to-one?

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  • $\begingroup$ This need not even be true if $S$ is a field. $\endgroup$ – Tobias Kildetoft Nov 8 '16 at 18:28
  • $\begingroup$ Your idea would work if $S$ were finite. $\endgroup$ – Tobias Kildetoft Nov 8 '16 at 18:31
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$S$ contains no nilpotent element for $S$ is integral domain and nilpotent is zero divisor. So $a^p=0$ implies $a=0$ and $Ker({\theta})=0$. Thus $\theta$ is one to one.

If $S$ is finite, $\theta$ is onto.

If $S$ is infinite, $\theta$ may not be onto. Here is an example. Let $K = F_p(t)$ be the finite field of $p$ elements together with a single transcendental element. Thus $K$ is the field of rational functions with coefficients in $F_p$. Then $t\notin\theta(K)$. For if so, let $\theta(q(t)/r(t))=q^p(t)/r^p(t)=t$. But $$ \deg\frac{q^p(t)}{r^p(t)}=p(\deg q-\deg r)\neq 1=\deg(t) $$ This means that $\theta$ is not onto.

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  • $\begingroup$ Is there a way to prove that Ker(θ) contains only 0 and no other elements? Since Ker(θ) = {0} is what allows us to say that θ is one to one. $\endgroup$ – Randy Nov 8 '16 at 18:39
  • $\begingroup$ @Randy The homomorphism must be injective, because we have an integral domain. That means that $a^p = 0$ implies $a = 0$, as said in the answer above. $\endgroup$ – Arthur Nov 8 '16 at 19:10

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