Consider a polynomial $$f(X) = X^{(p-1)p^{n-1}} + X^{(p-2)p^{n-1}} + \cdots + X^{p^{n-1}} + 1$$

Now I need to show $f(X+1)$ safeties Eisenstein's criterion. My argument is that
$$f(X+1) = (X+1)^{(p-1)p^{n-1}} + (X+1)^{(p-2)p^{n-1}} + \cdots + (X+1)^{p^{n-1}} + 1$$

I consider term by term, obviously, the coefficient of the highest order term is $1$, so $p$ does not divide $1$, the coefficient of the constant term is $p$, so $p^2$ does not divide $p$, for the middle terms, I observe that $p$ can mostly divide them, but for the first terms of each term of $f(X+1)$, coefficients for them are all $1$ ($p$ does not divide $1$) and I cannot easily find other equal order terms with them to observe the total coefficient, how should I solve it?

  • $\begingroup$ "but for the first terms of each term of $f(X+1)$, coefficients for them are all $1$ ($p$ does not divide $1$)", are you sure about that? Do you have some example of this? $\endgroup$ – Sil Apr 20 at 22:41
  • $\begingroup$ look at the second term, bu using binomial theorem, the first term of it is x^(p^n - 2p^(n-1)). so the coefficient of it is 1. p does not divide it. $\endgroup$ – Jonny Apr 20 at 22:46
  • $\begingroup$ I am not sure what you mean by second term and first term here, can you write down the full polynomial that satisfies this? For example for $p=3,n=2$ we have $f(x+1)=x^6+6x^5+15x^4+21x^3+18x^2+9x+3$, I don't see the thing you describe. $\endgroup$ – Sil Apr 20 at 22:47
  • $\begingroup$ I mean, if I take p=3, the first term of (X+1)$^{(P-2)P^{n-1}}$ is x$^{3^n - 2*3^{n-1}}$ $\endgroup$ – Jonny Apr 20 at 22:54
  • $\begingroup$ I see, well that is just a part of the polynomial, you need to collect all coefficients (possibly by Binomial theorem), then it won't be just $1$, and Eisenstein will be applicable. $\endgroup$ – Sil Apr 20 at 22:56

Here is something that could get you started (too long for a comment), write out the polynomial as:

\begin{align*} f(X+1) &= (X+1)^{(p-1)p^{n-1}} + (X+1)^{(p-2)p^{n-1}} + \cdots + (X+1)^{p^{n-1}} + 1\\ &= \sum_{k=0}^{p-1}(X+1)^{kp^{n-1}}\\ &= \sum_{k=0}^{p-1}\sum_{i=0}^{kp^{n-1}}\binom{kp^{n-1}}{i}X^i\\ &= \sum_{i=0}^{(p-1)p^{n-1}}\sum_{k=\lceil i/p^{n-1}\rceil}^{p-1}\binom{kp^{n-1}}{i}X^i\\ \end{align*} where we just swapped the indices by writing the conditions $0\leq k \leq p-1$, $0 \leq i \leq kp^{n-1}$, and rewrote into equivalent $0\leq i \leq (p-1)p^{n-1}$,$i/p^{n-1}\leq k \leq p-1$. So coefficient of $X^i$ for $i>0$ can be seen to be $\sum_{k=\lceil i/p^{n-1}\rceil}^{p-1}\binom{kp^{n-1}}{i}$. Now for Eisenstein you need to show divisibility of this sum by $p$ for $i<(p-1)p^{n-1}$.

However I have noticed that in some cases it does not seem to be straightforward to prove, for example for $p=3$, $n=2$ and coefficient at $x^3$ we get get sum $1+20=21$, which is divisible by $3$ but individual terms in sum are not. So something else is needed.

  • $\begingroup$ thanks for your comments! $\endgroup$ – Jonny Apr 21 at 4:01

My way of looking at this very special problem:

Showing that all lower coefficients are divisible by $p$ is the easy half of the problem.

Your original polynomial is $$\frac{X^{p^n}-1}{X^{p^{n-1}}-1}\,. $$ and after the substitution it becomes $$ \frac{(X+1)^{p^n}-1}{(X+1)^{p^{n-1}}-1}\,. $$ Now look at this modulo $p$, i.e. as a polynomial over $\Bbb F_p$, the prime field. The numerator becomes $X^{p^n}+1-1=X^{p^n}$, and similarly the denominator becomes $X^{p^{n-1}}$, so that the quotient has only one term not divisible by $p$.

To see that the constant term is $p$, look at numerator and denominator in the second display: both have constant term zero, so that you have a polynomial of the form $$ \frac{X^{p^n}+\cdots+p^nX}{X^{p^{n-1}}+\cdots+p^{n-1}X}\,, $$ and this polynomial clearly has constant term $p$.

(I’m guessing that you came across this as the cyclotomic polynomial $\Phi_{p^n}(X)$. )


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