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Let $F=\mathrm{GF}\left(p^k\right)$ be any finite field. Let $G$ be the group of all affine permutations on $F$ (i.e. permutations of form $x\mapsto ax+b$). Then the set of all functions from $F$ to $\bar{F}$ is a linear representation of $G$, where $g(f)(x)=f(gx)$.

What are all sub-representations of this representation? Is it possible to characterize them?

Note: that in this case $\mathrm{gcd}\left(\left|G\right|,F\right)$ not equal to $1$.

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    $\begingroup$ For any $n\geq 0$, the functions representable by a polynomial of degree less than or equal to $n$ clearly form a subrepresentation. The successive quotients are isomorphic to the symmetric powers of the permutation representation of the $x\to ax$ group (the normal subgroup of translations acts trivially) and can be completely described. $\endgroup$
    – Victor Protsak
    Commented Jun 30, 2010 at 9:40
  • $\begingroup$ Victor: I don't follow. Your successive quotients are one dimensional or I am not a hare. $\endgroup$
    – Bugs Bunny
    Commented Jun 30, 2010 at 11:31
  • $\begingroup$ Sorry for being a but unclear, but you, guys and hares, have figured it out all by yourselves. $\endgroup$
    – Victor Protsak
    Commented Jul 2, 2010 at 4:41

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As Victor explained consider the functions $X^m$ where $X^m(\alpha)=\alpha^m$. As $m$ runs between $0$ and $p^k-1$, these functions form a basis of your space of functions. This is a nice wavy basis, i.e., its elements span one-dimensional subrepresentations under the multiplicative group.

Now you have to take the additive group into account. All you need to do is to use binomial formula on $(X+\alpha)^m$ and observe which non-zero $X^t$-s, you can get out. This depends on the $p$-th power in $m$.

In particular, as Victor pointed out, polynomials of degree less than $m$ will span a submodule. But there are more, for instance, polynomials of degree $p$ and zero. In general, you will be getting spans of $X^t$ with $t\leq m$ and $t$ is divisible by the $p$-th power present in $m$ as well as the sums of these gadgets.

Hint: $(X+\alpha)^{p^sn}=(X^{p^s}+\alpha^{p^s})^n$

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  • $\begingroup$ Does it characterize all sub-representations? Let $c_1, \ldots c_k$ be integer numbers. Consider the set $S=\{t=a_0+a_1p+a_2p^2 \ldots a_k p^k | a_i<c_i\}$. Then span of $X^t$ for $t\in S$ will be sub-representation. Does it all sub-representations? If yes, how can you prove it? $\endgroup$
    – Klim Efremenko
    Commented Jun 30, 2010 at 13:33
  • $\begingroup$ Yes, all sub-representations looks like this. Proof: Let $a=a_0+a_1p+a_2p^2+a_k p^k$ then ${a \choose b}$ non zero mod p only if $b_i< a_i$. $\endgroup$
    – Klim Efremenko
    Commented Jun 30, 2010 at 14:13
  • $\begingroup$ I think these are all subrepresentations. To prove this, use complete reducibility as a representation of the multiplicative group. This will tell you that any subrepresentation (of affine group) is spanned by some of $X^t$ and all it remains is to invert some funny determinant, which proves that the span of $(X+\alpha)^m$ is what I said it is. Judging by you comment, you have done it already. $\endgroup$
    – Bugs Bunny
    Commented Jun 30, 2010 at 16:00

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