# Using Carlitz's exponential formula to prove an identity

This is a question on the homework for my finite fields class. The beginning of the assignment defines the following notation:

For $i\geq1$, define the following elements of $A=\mathbb{F}_{q}[T]$ :

$[i]=T^{q^{i}}-T$,

$L_{i}=[i][i-1]\cdots[1]$,

$D_{i}=[i][i-1]^{q}\cdots[1]^{q^{i-1}}$

and put $D_{0}=L_{0}=1$.

Let $n\geq1$. Put $A(n):=\{a \in A\mid \text{deg} (a)< n \}$

and $e_{n}(x):=\prod_{a\in A(n)}\left(x+a\right)$.

I am trying to prove: $\sum\frac{1}{b}=\frac{(-1)^{n}}{L_{n}}$ where the LHS ranges over all monic $b$ in $A$ of degree equal to $n$.

Earlier on the assignment, we showed $e_{n}(x)+D_{n}=\prod(x+b)$ where $b$ is monic with degree $n$.

I thought I would be able to take the Carlitz analog of the logarithm $\log_{c}(x)$ and then take the formal derivative and evaluate it at zero. But I've realized it is not true that $\log_{c}(xy)=\log_{c}(x)+\log_{c}(y)$. I've been stuck on this problem for some time, and can't think of any other way to approach it. I would appreciate a nudge in the right direction.

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From $e_n(x)+D_n=\prod(x+b)$ it follows that $\prod b=D_n$ and the $x^1$ coefficient of $e_n(x)$ is $D_n\sum 1/b$. But the $x^1$ coefficient is equal to $\prod'a$ where the product is over all nonzero polynomials $a$ of degree $< n$. All you have to do is to prove this product equals $(-1)^nD_n/L_n$. The leading coefficient should be easy enough. For the rest you need to do a census of the degrees of the irreducible factors of polynomials of degree $< n$. Possibly some earlier results in your course might cover this?

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Thanks! That makes sense. I'm fairly certain this is what we were intended to do because a combination of class notes and an old homework give us the rest of the argument. I was convinced that the point was to find some endomorphism similar to log that would turn products into sums, but I was definitely going in the wrong direction. – WWright Oct 26 '10 at 20:07

Using the previous homework problem we know that: $e_{n}(x)+D_{n}=\prod(x+a)$
Consider $\frac{f'}{f}$ where we are formally differentiating with respect to x. This transforms the LHS to $\frac{e\prime_{n}(x)}{e_{n}(x)+D_{n}}$ where
$e\prime_{n}(x)=\frac{d}{dx}\sum_{i=0}^{n-1}x^{q^{i}}\frac{D_{n}}{D_{i}L_{n-i}}=(-1)^{n}\frac{D_{n}}{L_{n}}+\sum_{i=1}^{n-1}(-1)^{n-i}q^{i}\frac{D_{n}}{D_{i}L_{n-i}^{q}}x^{q^{i-1}}$.
At $x=0$, $e\prime_{n}(x)=(-1)^{n}\frac{D_{n}}{L_{n}}$ and so the LHS becomes $\frac{(-1)^{n}}{L_{n}}$.
Similarly, the RHS gets transformed to $\sum\frac{1}{x+b}$ since $f'=\sum_{b}\frac{\prod(x+a)}{x+b}$ and $f=\prod(x+a)$. At $x=0$ this becomes $\sum\frac{1}{b}$ which is what we wanted.