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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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    Can you prove that ${p \choose r} = 0$ for $1 \le r \le p-1$?2011-09-28
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    @QiaochuYuan Ah because ${p \choose r} = 0$ is always divisible by $p$ because $p$ is a prime factor of it, and the additive order of any non-zero element of $R$ is $p$?2011-09-28
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    @LHS: Or because $p=0$ in $R$, yes.2011-09-28
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    @ArturoMagidin: Hmm why is $p=0$ sorry?2011-09-28
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    @LHS: "... R is an integral domain with $\mathbf{char} R=p\gt 0$." Because $R$ is of characteristic $p$, which means that $p=0$ in $R$ (where "$p$" denotes the image of the integer $p$ under the canonical map $\mathbb{Z}\to R$).2011-09-28
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    @ArturoMagidin: So $p=1+1+\dots +1$ $p$ times? Ah this is a special case for the value of char $R$ obviously $a\neq 1+1+\dots +1$ $a$ times as this makes no sense2011-09-28
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    @LHS: Since there is a canonical map from $\mathbb{Z}$ into any ring with unity $R$ (the map that sends $1$ to the unity of $R$), it is usual to identify the **integer** $a$ with the image of $a$ in $R$ (which will equal the sum of the unity of $R$ with itself, with $a$ summands). This is true for any *integer* $a$, but in general you cannot express an arbitrary element of $R$ as a sum of $1$s.2011-09-28
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    @ArturoMagidin: Thanks, that makes this much more clear!2011-09-28
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    @LHS This map is so important that it has a name, viz. the [Frobenius map / endomorphism.](http://en.wikipedia.org/wiki/Frobenius_endomorphism)2011-09-28
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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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    Surely your $n$ should be a $p$ (lest the OP get stuck trying to prove that $\binom{n}{k}$ is divisibly by $n$ regardless of whether $n$ is prime or not...)2011-09-28
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    Indeed, this definitely works for when n is prime, and as it is it works fine!2011-09-28
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    Chiming in with Arturo and others. This is not true, unless $n$ is a prime. The smallest counterexample is ${4\choose 2}=6$ that is not divisible by $4$. Such counterexamples exist for any composite $n$, but the proof won't fit into this comment :-)2011-09-28
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    @JyrkiLahtonen: Yes, I see, I'd come to that conclusion as well! It definitely only works generally when $n$ is prime2011-09-28
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    Thanks everyone; I knew this but overlooked it.2011-09-28