Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial. Consider $\bar{f} \in \mathbb{F}_p[x].$ Let $\rho_p$ be the number of distinct roots of $\bar{f}$ in $\overline{\mathbb{F}}_p$, the algebraic closure of $\mathbb{F}_p$. Also, let $\rho$ be the number of distinct roots of $f$ in $\mathbb{C}$.

Question 1:

Do you have a reference for showing that $\rho_p \leq \rho$ for all large primes $p$?

If not, then of course, I'd also be happy with an (understandable) argument that shows this. (I'm weak in algebraic number theory.)

Question 2:

Do we only need to assume that $p$ is such that $\bar{f}$ is nonzero in $\mathbb{F}_p$?

My background to the question: I actually have used the first result in counting the maximal subgroups (by their index) of groups of the form $\mathbb{Z}^k \rtimes \mathbb{Z}$ and similar metabelian groups. Now that I'm going back and rereading my work (to convert it into a paper) I've realized that I used this without justification.

  • 1
    $\begingroup$ (1) If $r_1,r_2,...,r_{\rho+1}$ are different roots, then the coefficients are a non-zero solution of a homogeneous system that has the Vandermonde determinant of the $r_i$ as matrix. Therefore, all coefficients would be zero. Take into account that $\rho$ is equal to the degree of $f$. This argument doesn't have much to do with anything but that the coefficients are a domain, and well the fundamental theorem of algebra to know that $\rho$ is the degree. $\endgroup$ – user551819 Apr 25 '18 at 20:01
  • $\begingroup$ The thing that is proper to characteristic $p$ is that $\rho_p$ can be strictly smaller than $\rho$. For example, for $f(x)=x^p-a$ the only root in $\overline{\mathbb{F}}_p$ is $a^{1/p}$, since $f(x)=(x-a^{1/p})^p$. $\endgroup$ – user551819 Apr 25 '18 at 20:06
  • $\begingroup$ @totoro I do not understand your first comment. Would you mind expanding it into an answer and add more details? $\endgroup$ – Andrew Kelley Apr 28 '18 at 0:07

Yes, this is more or less immediate from considering the factorization of $f$ in $\mathbb{Z}[x]$. We can factor $f$ as $$f=af_1^{e_1}\dots f_n^{e_n}$$ where $a\in\mathbb{Z}$ and $f_1,\dots,f_n\in\mathbb{Z}[x]$ are distinct irreducible polynomials of positive degree. By Gauss's lemma, the $f_i$ are also irreducible in $\mathbb{Q}[x]$. An irreducible polynomial always has distinct roots in characteristic $0$, and distinct irreducible polynomials over a field have no roots in common with each other in any extension field. So each $f_i$ contributes $\deg f_i$ distinct roots to $f$ over $\mathbb{C}$, and so $f$ has $\sum_i \deg f_i$ distinct roots over $\mathbb{C}$.

Now, let $p$ be such that $\bar{f}\neq 0$. Reducing our factorization of $f$ mod $p$ we get a factorization $$\bar{f}=\bar{a}\bar{f}_1^{e_1}\dots \bar{f}_n^{e_n}.$$ Since $\bar{f}\neq 0$, none of these factors are $0$. But now it is clear that $\bar{f}$ cannot have more than $\sum_i \deg \bar{f}_i\leq\sum_i\deg f_i$ roots in $\overline{\mathbb{F}}_p$, since each root of $\bar{f}$ is a root of some $\bar{f}_i$. Thus $\rho_p\leq \rho$.

In fact, for all sufficiently large $p$, $\rho_p=\rho$. This can be proven, for instance, by computing $\gcd(f,f')$ in $\mathbb{Q}[x]$ using the Euclidean algorithm. This computation will only involve finitely many rational numbers and thus will be valid in $\mathbb{F}_p[x]$ for all sufficiently large $p$. Since $\frac{f}{\gcd(f,f')}$ has the same roots as $f$ but with no repeated roots (as long as $\deg f$ is less than the characteristic), this implies $\rho_p=\rho$ as long as $p$ is sufficiently large.

Alternatively, this fact is immediate from a bit of model theory. If there are infinitely many different $p$ such that $\rho_p\neq\rho$, take an ultraproduct $F$ of the fields $\overline{\mathbb{F}}_p$ for all such $p$. This ultraproduct $F$ will be a field of characteristic $0$. The statement "$f$ splits and does not have exactly $\rho$ distinct roots" can be expressed in the first-order language of rings and is true in each factor $\overline{\mathbb{F}}_p$, and thus it is also true in the ultraproduct $F$. But since $F$ has characteristic $0$, $f$ must have exactly $\rho$ distinct roots in $F$, so this is a contradiction.

  • $\begingroup$ References added: Corollary 34 on page 547 of Dummit and Foote tells us that an irreducible polynomial over a field of characteristic 0 is separable (i.e. has no repeated roots). In their proof of Cor. 34, they state at the top of page 548 that distinct irreducible polynomials never have zeros in common (by Proposition 9 [on page 520]). $\endgroup$ – Andrew Kelley Apr 28 '18 at 16:06

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