Chevalley’s Theorem, K˝ onig–Rados Theorem

We discuss two famous theorems concerning congruences with prime modulus. We consider first a system of congruences

(3.6.1) 𝑓𝑖(𝑥1, 𝑥2, . . . , 𝑥𝑡) ≡ 0 (mod 𝑝) , 𝑖 = 1, 2, . . . , 𝑘 where𝑝is a prime,𝑘 ≥ 1, and

𝑓𝑖(𝑥1, 𝑥2, . . . , 𝑥𝑡), 𝑖 = 1, 2, . . . , 𝑘

are polynomials in𝑡variables with integer coefficients and constant terms0, i.e.

(3.6.2) 𝑓𝑖(0, 0, . . . , 0) = 0, 𝑖 = 1, 2, . . . , 𝑘.

(3.6.2) implies that𝑥1≡ 𝑥2 ≡ ⋯ ≡ 𝑥𝑡≡0(mod 𝑝)satisfies (3.6.1). We call this a trivial solution.

Chevalley’s Theorem asserts that with suitable requirements for the degrees of𝑓𝑖, (3.6.1) has a non-trivial solution, too. (The degree of a term𝑥𝑛11. . . 𝑥𝑛𝑡𝑡is𝑛1+ ⋯ + 𝑛𝑡 and the degree of a polynomial is the maximal degree among its terms having non-zero coefficient.)

Theorem 3.6.1(Chevalley’s Theorem). If the polynomials𝑓𝑖 in(3.6.1)satisfy(3.6.2) and the sum of their degrees is less than the number of variables, i.e.

(3.6.3)

𝑘

∑

𝑖=1

deg 𝑓𝑖< 𝑡,

then(3.6.1)has a non-trivial solution. ♣

Examples. The system of congruences

𝑥1+ 2𝑥2+ 3𝑥3+ 4𝑥4+ 5𝑥5≡ 0 (mod 23) 𝑥31+ 2𝑥1𝑥2+ 3𝑥2𝑥3+ 4𝑥3𝑥24+ 5𝑥52≡ 0 (mod 23)

has a non-trivial solution with not all𝑥𝑖 multiples of23. (Here𝑘 = 2and5 = 𝑡 >

1 + 3 = deg 𝑓1+ deg 𝑓2.)

We can apply Theorem 3.6.1 also for𝑘 = 1, just one polynomial. For example, the divisibility

𝑝 ∣ 𝑥13+ 3𝑥32+ 5𝑥33+ 7𝑥34+ 9𝑥1𝑥2+ 11𝑥3𝑥4

can be satisfied for any prime𝑝so that not all𝑥𝑖 are divisible by𝑝. (Now𝑡 = 4and deg 𝑓 = 3.)

Proof. Assuming that there is only a trivial solution, we shall force a contradiction.

We define two new polynomials in𝑡variables:

𝐹(𝑥1, 𝑥2, . . . , 𝑥𝑡) =

𝑘

∏

𝑖=1

(1 − 𝑓𝑖𝑝−1(𝑥1, 𝑥2, . . . , 𝑥𝑡)) and

𝐺(𝑥1, 𝑥2, . . . , 𝑥𝑡) =

𝑡

∏

𝑗=1

(1 − 𝑥𝑗𝑝−1).

By Fermat’s Little Theorem,

𝑐𝑗≢ 0 (mod 𝑝) ⟹ 𝑐𝑗𝑝−1≡ 1 (mod 𝑝) .

This implies that substituting arbitrary integers𝑐1, . . . ,𝑐𝑡into𝐺, we get (3.6.4) 𝐺(𝑐1, 𝑐2, . . . , 𝑐𝑡) ≡ {1 (mod 𝑝) , if𝑐1≡ ⋯ ≡ 𝑐𝑡 ≡ 0 (mod 𝑝)

0 (mod 𝑝) , otherwise.

We show that the same holds also for𝐹, i.e.

(3.6.5) 𝐹(𝑐1, 𝑐2, . . . , 𝑐𝑡) ≡ {1 (mod 𝑝) , if𝑐1≡ ⋯ ≡ 𝑐𝑡≡ 0 (mod 𝑝) 0 (mod 𝑝) , otherwise.

We consider first

𝑐1≡ ⋯ ≡ 𝑐𝑡≡ 0 (mod 𝑝) . By (3.6.2),

𝑓(𝑐1, . . . , 𝑐𝑡) ≡ 0 (mod 𝑝)

for every𝑖, so every factor of𝐹(𝑐1, . . . , 𝑐𝑡)and so𝐹(𝑐1, . . . , 𝑐𝑡)itself is congruent to1 modulo𝑝.

Now, we turn to the other case when at least one of the integers𝑐1, . . . ,𝑐𝑡is not a multiple of𝑝. We assumed that (3.6.1) has only a trivial solution, hence𝑐1, . . . ,𝑐𝑡is not a solution, so

𝑓𝑖(𝑐1, . . . , 𝑐𝑡) ≢ 0 (mod 𝑝)

for at least one𝑖. Applying Fermat’s Little Theorem again, this implies 𝑓𝑖𝑝−1(𝑐1, 𝑐2, . . . , 𝑐𝑡) ≡ 1 (mod 𝑝) .

This means that a factor of𝐹(𝑐1, . . . , 𝑐𝑡)and so𝐹(𝑐1, . . . , 𝑐𝑡)itself is divisible by𝑝. Here- with we have proven (3.6.5).

By (3.6.4) and (3.6.5),

(3.6.6) 𝐹(𝑐1, . . . , 𝑐𝑡) ≡ 𝐺(𝑐1, . . . , 𝑐𝑡) (mod 𝑝) for arbitrary integers𝑐1, . . . ,𝑐𝑡.

3.6. Chevalley’s Theorem, Kőnig–Rados Theorem 93

From now on, we shall consider all polynomials as polynomials in𝑡variables over the modulo𝑝field.

In this interpretation, (3.6.6) tells us that𝐹and𝐺assume the same values for every substitution (the same polynomialfunctionsbelong to𝐹and𝐺; however, this does not imply in general the equality of the polynomials themselves, that is, the equality of the coefficients in the case of a finite field).

Let𝐻∗be the reduced form of the polynomial𝐻obtained by replacing every𝑥𝑝𝑖 in𝐻 with𝑥𝑖 as long as possible. The exponents of𝑥𝑖 in the terms of𝐻∗are at most 𝑝 − 1, and𝐻 and𝐻∗assume the same values everywhere. It can be easily proven by induction on the number of variables that if the polynomials𝐻and𝐾assume the same values everywhere, then the (formal) polynomials𝐻∗and𝐾∗are equal (so they have the same coefficients).

We saw that𝐹and𝐺assume the same values everywhere, therefore the polyno- mials𝐹∗and𝐺∗are equal. Hence,deg 𝐺∗= deg 𝐹∗. However, by𝐺 = 𝐺∗and (3.6.3), this leads to a contradiction:

deg 𝐺∗= deg 𝐺 = (𝑝 − 1)𝑡 > (𝑝 − 1) (

𝑘

∑

𝑖=1

deg 𝑓𝑖) = deg 𝐹 ≥ deg 𝐹∗. □

In the second half of this section, we express the number of solutions of a con- gruence𝑓(𝑥) ≡ 0 (mod 𝑝)in an exact formula with the help of the coefficients. This theorem by Kőnig and Rados is rather only of theoretical significance; it can be hardly applied for computing the number of solutions in practice.

Theorem 3.6.2(Kőnig–Rados Theorem). Let𝑝be a prime and𝑓 = 𝑎0+ 𝑎1𝑥 + ⋯ + 𝑎𝑝−2𝑥𝑝−2 be a polynomial with integer coefficients having𝑎0 ≢ 0 (mod 𝑝). Then the number of solutions of the congruence𝑓(𝑥) ≡ 0 (mod 𝑝)is𝑝 − 1 − 𝑟where𝑟 = 𝑟(𝐴)is the rank of the cyclic(𝑝 − 1) × (𝑝 − 1)matrix𝐴over the modulo𝑝field,

𝐴 =

⎛

⎜

⎜

⎝

𝑎0 𝑎1 . . . 𝑎𝑝−2

𝑎𝑝−2 𝑎0 . . . 𝑎𝑝−3

⋮ ⋮ ⋱ ⋮

𝑎1 𝑎2 . . . 𝑎0

⎞

⎟

⎟

⎠

. ♣

Remarks: (1) The theorem immediately implies that𝑓(𝑥) ≡ 0 (mod 𝑝)is solvable if and only if the rank of𝐴is less than𝑝 − 1, i.e.det 𝐴 ≡ 0 (mod 𝑝).

(2) The requirements imposed on𝑓are not serious restrictions; we can obtain the number of solutions for an arbitrary polynomial𝑓by a simple reduction to the Kőnig–Rados Theorem, see Exercise 3.6.11.

Proof. We shall need the following elementary results from linear algebra. They all refer to𝑛 × 𝑛matrices over a field𝐹where𝑟(𝐵)denotes the rank of matrix𝐵; in our case,𝑛 = 𝑝 − 1and𝐹is the modulo𝑝field.

(i) Let𝑡1,𝑡2, . . . ,𝑡𝑛be distinct elements in𝐹. Then the Vandermonde matrix

𝑉 = 𝑉(𝑡1, 𝑡2, . . . , 𝑡𝑛) =

⎛

⎜

⎜

⎜

⎜

⎝

1 1 1 . . . 1

𝑡1 𝑡2 𝑡3 . . . 𝑡𝑛 𝑡12 𝑡22 𝑡23 . . . 𝑡𝑛2

⋮ ⋮ ⋮ ⋱ ⋮

𝑡𝑛−11 𝑡2𝑛−1 𝑡3𝑛−1 . . . 𝑡𝑛𝑛−1

⎞

⎟

⎟

⎟

⎟

⎠ has rank𝑟(𝑉) = 𝑛.

(ii) If𝑟(𝐵) = 𝑛, so𝐵has an inverse, then𝑟(𝐶𝐵) = 𝑟(𝐶)for an arbitrary𝐶.

Assertion (ii) follows from the inequality

𝑟(𝑀𝑁) ≤ min(𝑟(𝑀), 𝑟(𝑁))

valid for arbitrary matrices𝑀and𝑁. On the one hand,𝑟(𝐶𝐵) ≤ 𝑟(𝐶), and on the other hand,𝑟(𝐶) = 𝑟((𝐶𝐵)𝐵−1) ≤ 𝑟(𝐶𝐵).

Turning to the proof of Theorem 3.6.2, let𝑠denote the number of solutions of the congruence𝑓(𝑥) ≡ 0 (mod 𝑝). Consider the matrix𝐷 = 𝐴𝑉where𝑉=𝑉(1, 2, . . . , 𝑝−1).

By (i) and (ii),

(3.6.7) 𝑟(𝐷) = 𝑟(𝐴) = 𝑟.

Performing the multiplication𝐴𝑉, the𝑗th element of the first row in𝐷is 𝑑1𝑗= 𝑎0+ 𝑎1𝑗 + 𝑎2𝑗2+ ⋯ + 𝑎𝑝−2𝑗𝑝−2= 𝑓(𝑗).

For a simple form of the𝑗th element in the second row, we use also𝑗𝑝−1≡ 1 (mod 𝑝):

𝑑2𝑗 = 𝑎𝑝−2+ 𝑎0𝑗 + 𝑎1𝑗2+ ⋯ + 𝑎𝑝−3𝑗𝑝−2 ≡

≡ 𝑎𝑝−2𝑗𝑝−1+ 𝑎0𝑗 + 𝑎1𝑗2+ ⋯ + 𝑎𝑝−3𝑗𝑝−2= 𝑗𝑓(𝑗) (mod 𝑝) . Similarly, for the𝑗th element in the𝑖th row, we obtain

𝑑𝑖𝑗≡ 𝑗𝑖−1𝑓(𝑗) (mod 𝑝) .

This means that (working with equality in the modulo𝑝field instead of congruences)

𝐷 = 𝐴𝑉 =

⎛

⎜

⎜

⎜

⎜

⎝

𝑓(1) 𝑓(2) 𝑓(3) . . . 𝑓(𝑝 − 1)

𝑓(1) 2𝑓(2) 3𝑓(3) . . . (𝑝 − 1)𝑓(𝑝 − 1) 𝑓(1) 22𝑓(2) 32𝑓(3) . . . (𝑝 − 1)2𝑓(𝑝 − 1)

⋮ ⋮ ⋮ ⋮

𝑓(1) 2𝑝−2𝑓(2) 3𝑝−2𝑓(3) . . . (𝑝 − 1)𝑝−2𝑓(𝑝 − 1)

⎞

⎟

⎟

⎟

⎟

⎠

In column𝑗of𝐷every element is0if and only if𝑓(𝑗) ≡ 0 (mod 𝑝). Thus,𝐷has exactly 𝑠columns with all 0s. The other columns are distinct columns of𝑉multiplied by a non- zero scalar, so they are linearly independent according to (i). Hence,𝑟(𝐷) = 𝑝 − 1 − 𝑠.

Combined with (3.6.7), this proves the theorem. □

Exercises 3.6 95

Exercises 3.6

1. Which well-known theorem is obtained as a special case of Chevalley’s Theorem when each polynomial𝑓𝑖has degree one?

2. Verify that the congruence𝑎𝑥2+ 𝑏𝑦2+ 𝑐𝑧2≡ 0 (mod 𝑝)has a non-trivial solution for every prime𝑝and any integers𝑎,𝑏, and𝑐.

3. Prove.

(a) For any𝑛 > 1, there exist three integers such that taking the sum𝑠of their squares,𝑛 ∣ 𝑠but𝑛2 ∤ 𝑠.

(b) Moreover, even(𝑛, 𝑠/𝑛) = 1can be attained.

4. Show that every prime𝑝has a positive multiple less than𝑝4/4that can be written as the sum of at most five fourth powers.

* 5. (a) Let𝑞1, . . . ,𝑞𝑘 be distinct primes and𝑐1, . . . , 𝑐𝑡 distinct positive integers not divisible by any other primes than the𝑞𝑖. Prove that if𝑡 ≥ 2𝑘 + 1, then we can select some distinct numbers from the𝑐𝑗(maybe just one, maybe all of them) so that their product is a cube.

(b) Generalize (a) for𝑝th powers where𝑝is an arbitrary prime.

Remark: The analogous result can be proven (using different methods) for 𝑚th powers where𝑚is a prime power, but it is not proved for any other values of𝑚.

* 6. Show that from any2𝑛 − 1integers, we can select𝑛whose sum is divisible by𝑛.

7. Prove the following generalization of Chevalley’s Theorem. We omit the assump- tion that the constant terms of the polynomials are0and leave the other conditions unchanged. Then the following hold for the system of congruences in question:

(a) If it is solvable, then there are at least two solutions.

* (b) The number of solutions is divisible by𝑝.

8. Let 𝑝 > 2be a prime and(𝑎𝑏, 𝑝) = 1. As an illustration of the Kőnig–Rados Theorem, determine the number of solutions of the congruence𝑓(𝑥) ≡ 0 (mod 𝑝) for the following polynomials𝑓:

(a) 𝑎𝑥 − 𝑏

(b) 1 + 𝑥 + ⋯ + 𝑥𝑝−2 (c) 𝑥𝑝−2− 𝑎.

9. Deduce the solvability of the following congruences from the Kőnig–Rados Theo- rem:

(a) 𝑥𝑘≡ 1 (mod 𝑝)where𝑝is an odd prime and1 ≤ 𝑘 ≤ 𝑝 − 2 (b) 𝑥2≡ −1 (mod 𝑝), where𝑝is a prime of the form4𝑘 + 1.

10. Let𝑝 > 3be a prime,(𝑎0, 𝑝) = (𝑎1, 𝑝) = (𝑎𝑝−2, 𝑝) = 1and 𝑓 = 𝑎0 + 𝑎1𝑥 + ⋯ + 𝑎𝑝−3𝑥𝑝−3+ 𝑎𝑝−2𝑥𝑝−2

𝑔 = 𝑎1 + 𝑎2𝑥 + ⋯ + 𝑎𝑝−2𝑥𝑝−3+ 𝑎0𝑥𝑝−2 ℎ = 𝑎𝑝−2 + 𝑎𝑝−3𝑥 + ⋯ + 𝑎1𝑥𝑝−3+ 𝑎0𝑥𝑝−2. Prove that the congruences

𝑓(𝑥) ≡ 0 (mod 𝑝) , 𝑔(𝑥) ≡ 0 (mod 𝑝) , and ℎ(𝑥) ≡ 0 (mod 𝑝) have the same number of solutions.

11. Let𝑔 = 𝑏0+𝑏1𝑥+⋯+𝑏𝑛𝑥𝑛be an arbitrary polynomial with integer coefficients. To determine the number of solutions of𝑔(𝑥) ≡ 0 (mod 𝑝)in the case𝑛 > 𝑝−2and/or 𝑏0≡ 0 (mod 𝑝), how can we reduce the problem to the Kőnig–Rados Theorem?