Proposition 1.9

I'm trying to understand chapter 1 of Michael Rosen's Number Theory in Function Fields, and on this particular proposition I have questions just about everything (see Proposition 1.9).

  1. The author just straight up puts the claim in an equivalent form, and says its due to corolary of proposition 1.8 (which is just the analogue Fermats Little Theorem for polynomials in a finite field.) Why does this occur?

  2. How do we know both sides have the same set of roots? RHS has the roots explicit, LHS not so much.

  3. Why does the difference have degree less than |P|-1 and why would that imply that the difference must be zero?

  • 1
    $\begingroup$ A is not defined. Is A a polynomial ring over a finite field? $\endgroup$ – Gunnar Sveinsson Jul 8 '19 at 9:05

I will assume $A=\mathbb{F}_q[T]$ is a polynomial ring over a finite field $\mathbb{F}_q$. Since $P(T)$ is irreducible, the quotient ring $A/P$ will be a field, so its units are all the nonzero terms, and so their cardinality is


(I assume $|P|$ means $|A/P|$, which will be $q^d$.) Group theory tells us then that $Q^{|P|-1}=1$ for all of the elements $Q$ in the group $(A/P)^{\times}$, which is the same logic used to argue Fermat's little theorem. One can treat this as a corollary to Lagrange's Theorem: if $G$ is any finite group and $g$ is an element of $G$, then the order $d$ of the cyclic subgroup $\langle g\rangle$ divides $n=|G|$; then $g^n=(g^d)^{n/d}=e^{n/d}=e$. Therefore, $g^n=e$ for all group elements $g$ in $G$.

As for how Rosen does it:

(1) I don't know what proposition 1.8 is, so I don't know what you're asking here,

(2) all polynomials that are not zero in $A/P$ are roots of $X^{|P|-1}-1$ for the same reason I described above: the order of the group $(A/P)^{\times}$ is $|P|-1$, so $Q^{|P|-1}=1$ for all $Q$ in $A/P$, i.e. all nonzero $Q$ in $A/P$ are roots of the polynomial $X^{|P|-1}-1$,

(3) both sides of the equation are monic polynomials of the same degree, so if you subtract them their leading terms will cancel out, leaving only terms of lower degree, hence the degree of the difference is strictly smaller, but it still has $|P|-1$ roots in the field $A/P$ and it is impossible for a nonzero polynomial to have more roots than its degree over a field so the difference is zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.