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I am reading the book linear algebraic groups by Springer. I have a question on Page 53, on line 3, it is said that $d-h \geq p$ implies that ${d-h \choose p} \not\equiv 0 \pmod p$ by Lemma 3.4.2. But it seems that this is not true. For example, let $d=p^2+p, h=p-1$. Then $p$ divides $d$ and does not divide $h$. $p \leq d-h$ but ${d-h \choose p} \equiv 0 \pmod p$. Thank you very much.

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  • $\begingroup$ What are $c_h$ and $j$? $\endgroup$ – RghtHndSd Sep 21 '13 at 14:59
  • $\begingroup$ @rghthndsd, I am sorry that I forgot to attach Pages 51 and 52. $c_h$ is the coefficients of $f(T, U)$ and $j$ is an integer. $\endgroup$ – LJR Sep 22 '13 at 7:01

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