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I'm working on this problem:

Let $K=Z/pZ$ where p is a prime. For $n\in Z_+ $, use induction on n to show that for $f=a_0+a_1x+a_2x^2+...+a_nx^n \in K[x]$, it is the case that

$f^p=a_0+a_1x^p+a_2x^{2p}+...+a_nt^{np}$

So far I think I have a rough idea of how the base case works, using the fact that $a^p=a$ for any element of K:

Let $n=1$. Then $f=a_0+a_1x$ and so $f^p=(a_0+a_1x)...(a_0+a_1x)$. Using the binomial expansion I know that for a prime p, the coefficients of everything except powers of $a_0$ and $a_1x$ are multiples of that prime, and so these elements are cancelled out. This leaves $f^p=a_0^p+a_1^px^p=a_0+a_1x^p$.

What I don't understand is the induction step. Suppose it is true for $n=k$. Let $n=k+1$. Then $f=a_0+a_1x+...+a_kx^k+a_{k+1}x^{k+1}$.

Any hints about where to go from there? Thank you!

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Atlas
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2 Answers2

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Hint: In $\Bbb Z_p$, we have $(x+y)^p=x^p+y^p$.

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If $f$ has degree $n+1$, you can write $f=g+a_nx^n$ where $g$ has degree $n$, apply the binomial formula to $(g+a_nx^n)^p$ and remark that $(g+a_nx^n)^p=g^p+(a_nx^n)^p$ since $\pmatrix{p\cr i}$ is divisible by $p$ if $0<i<p$.

Tsemo Aristide
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