# Product of monic polynomials in a Function Field

I'm trying to prove that given $P$ is a monic irreducible over $\Bbb F_q[t]$ and $d=\deg(P)$, then $$\prod_{\substack{f\text{ monic}\\ 0\leq\deg(f)<d}}f=\pm1\pmod{P}.$$My first thought goes to using the analog of Wilson's Theorem over function fields, but it doesn't require the polynomials to be monic. At this point, I thought that every polynomial can be written as $\alpha f$ where $\alpha\in\Bbb F_q^\ast$ and $f$ is monic. Combining all of these, I got $$\prod_{\substack{f\text{ monic}\\ 0\leq\deg(f)<d}}\left(f^{q-1}\prod_{\alpha\in\Bbb F_q^\ast}\alpha\right)=(-1)^d\left(\prod_{\substack{f\text{ monic}\\0\leq\deg(f)<d}}f\right)^{q-1}=-1\pmod{P}.$$So I can divide the $(-1)^{d-1}$ over, but the $q-1$ power throws me off. Am I going in the right direction? Any helpful hints? I'd appreciate it if I can resolve this on my own, so no full answers (yet) please, unless my solution is correct so far and one small hint finishes it.

• are you sure it's true? for $d=2$ I'm getting $x^q-x$ as the product; if say $q$ is prime then certainly $x^q-x\pm1$ is irreducible, so it can't be a multiple of $P$ (unless $q=2$) Commented Sep 11, 2013 at 20:15
• This is problem 7 of Chapter 1 in Rosen's book Number Theory in Function Fields. Let me make sure I didn't leave anything off that the book states. Commented Sep 11, 2013 at 20:35
• My statements accurately reflect those of the book. Commented Sep 11, 2013 at 20:39
• @user8268: You are absolutely correct; the statement isn't true. See my now-posted (and corrected) solution. Commented Sep 27, 2013 at 13:31

The statement is indeed false. If we look at the case $q=3$ and $P=t^2+1$, for example, then the monic polynomials of degree $1$ are $$t,\quad t+1,\quad t+2.$$ Multiplying these together, we have $$t(t+1)(t+2)=t^3-t=t(t^2+1)+t\equiv t\pmod{P}.$$Therefore, we have a clear counterexample to the exercise.