# Primes And Polynomials

*A result on the prime divisors of polynomial values*

Cropped from source |

Issai Schur was a mathematician who obtained his doctorate over a hundred years ago. He was a student of the great group theorist Ferdinand Frobenius. Schur worked in various areas and proved many deep results, including some theorems in basic number theory.

Today we discuss a nice lemma due to Schur. Actually, it’s a theorem.

There are many things named after Schur. Wikipedia has compiled a list of them:

Frobenius-Schur indicator Herz-Schur multiplier Jordan-Schur theorem Lehmer-Schur algorithm Schur algebra Schur class Schur complement method Schur complement Schur decomposition Schur functor Schur index Schur multiplier Schur number Schur orthogonality relations Schur polynomial Schur product theorem Schur test Schur-convex function Schur-Horn theorem Schur-Weyl duality Schur-Zassenhaus theorem Schur’s inequality Schur’s lemma (from Riemannian geometry) Schur’s lemma Schur’s property Schur’s theorem

This lists two Schur lemmas. But when you click on Wikipedia’s “Schur’s lemma” page, there are three Schur lemmas. No, wait—there are four, including one called “Schur’s test.” But the result we are interested in is classed as a *theorem*. Wikipedia’s single page for “Schur’s theorem” lists not just *five* but *six* Schur’s theorems. This is one higher than the count eventually reached in Monty Python’s famous “Spanish Inquisition” skit. We want the sixth one, which is quite useful—like a lemma.

## The Result

Here is Schur’s lemma—no, theorem—published in 1912.

Theorem 1Let be a non-constant polynomial with integer coefficients. Then has infinitely many prime divisors.

Here has infinitely many prime divisors means that the number of primes that arise as divisors of the values as is infinite. Note, this works even if the polynomial is reducible or contains some trivial factors. Thus

is fine. As is .

Schur’s theorem is quite fun, even just to prove. Here is a nice version from a paper by Ram Murty, whose work on extensions of Euclid’s famous proof of the infinitude of primes we covered last summer. We follow Murty’s proof.

*Proof:* Suppose is an integral polynomial of least degree for which the statement fails,

where . If , then is an integral polynomial of lower degree for which it must fail, so we have . By hypothesis, has only finitely many prime divisors, which we can represent as . For each natural number , define and

Now because can take the value or only finitely often, and all sufficiently large give for the same reason. Because includes as a factor, is divisible by , giving

for some integer that by choice of is divisible by all of . But then as in Euclid’s proof, is co-prime to all those primes, so it must furnish a prime divisor of that is not among them. This is a contradiction.

This proof was simple but clever. Here is a concrete version for the polynomial , which may help in understanding the proof: Say is divisible by only say. Let The trick is to look at

where

For some large enough there exists a prime that divides . But this prime divides and so for some . But then modulo is equal to which is a contradiction.

## An Application

As the proof hints, Schur’s theorem is a proper extension of Euclid’s, which is just the case . Here is a less-obvious application:

Theorem 2Suppose that is a polynomial with integer coefficients. Assume that is a perfect square for all integers large enough. Then for some polynomial .

There are many proofs of this theorem. One uses the famous Hilbert Irreducibility theorem, which we also discussed last summer. Another proof uses Schur’s lemma and the fact that if and are products of irreducible integer polynomials that are collectively distinct then the ideal generated by and contains a positive integer—namely, their resultant . Again following Murty’s paper:

*Proof:* We can factor , where is a product of irreducible integer polynomials, and the goal is to show that . If has positive degree, then by Schur’s theorem, there are infinitely many prime divisors of . The square-freeness of implies that it shares no factors with its derivative , so . We need only take dividing for some large enough that is a perfect square but such that does not divide . Then the order of dividing must be even, which implies that the order of dividing must be even. So divides .

By a similar token, since divides as well and is a perfect square, we get that divides . Now is congruent to modulo , and since divides from before, it follows that divides . Since belongs to the ideal generated by and in , divides too, but this contradicts the choice of . So must be a constant. Since the notion of “irreducible” with applies to constants, any square dividing is already part of , so must be a product of distinct primes. This forestalls dividing in the above, so we must have .

## Open Problems

We did mention Schur’s long list of things named after him. Did people name things after others more often years ago? Or is naming them still common-place?

[fixed last proof]

I’m confused… why is it that p^2 dividing h(n) implies p divides h'(n)? This obviously resembles the fact that if f^2 divides h then f divides h’, but with actual numbers plugged in. It’s not obvious why it would still work in that context; because while you can differentiate h, the polynomial, you can’t different h(n), the number…

Thanks—you are right, the shortcut was incorrect.

The patern and shapes of prime Numbers might to follow the polynmial equations and theirs with complex roots with mod(T) that is moviment of the time.the flux of the time is Given by the velocities of the sequênces measured by the imaginary numbers

I don’t see Schur rings in the list. When I began as a DPhil student in Oxford in 1968, my first job was to read Wielandt’s book on Finite Permutation Groups: one of its five chapters was on “The method of Schur”, i.e. Schur rings: these are subrings of the group ring spanned by sums of subsets of the group, these subsets forming a partition. (Wielandt was a student of Schur.)