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If $R$ is an integral domain with char $p$ where $p>0$ and $f:R\to R$ where $f(x)=x^p$

How would one go about showing addition is preserved? e.g. $f(a+b)=f(a)+f(b)$? Multiplication is obvious. So far for addition I have:

$f(a+b)=(a+b)^p=\sum\limits_{r=0}^p {p\choose r}a^rb^{p-r}=a^p+\sum\limits_{r=1}^{p-1} {p\choose r}a^rb^{p-r}+b^p=f(a)+\sum\limits_{r=1}^{p-1} {p\choose r}a^rb^{p-r}+f(b)$

So i'm currently having difficulty proving $\sum\limits_{r=1}^{p-1} {p\choose r}a^rb^{p-r}=0$

Also can anyone think of an example such that $f$ isn't surjective?

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    @BillDubuque Thank you so much for showing me that!2011-09-28

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As for an example where $f$ isn't surjective, let $R = F[X]$ (the polynomials in one indeterminate $X$ over a field $F$ of characteristic $p$). Then for example $X$ is not in the range.

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Prove that for prime $p$, ${p \choose r}$ is divisible by $p$ for all $1 \leq r < p$. Can you see why you'd be done after this?

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    Thanks everyone; I knew this but overlooked it.2011-09-28