Simple proof of Chebotarëv’s theorem

We give a simple proof of Chebotarëv’s theorem: Let p be a prime and ω a primitive pth root of unity. Then all minors of the matrix ( ω ij )p−1 i,j=0 are non-zero. Let p be a prime and ω a primitive pth root of unity. We write Fp for the field with p elements. In 1926, Chebotarëv proved the following theorem (see [3]): Theorem. For any sets I, J ⊆ Fp with equal cardinality, the matrix (ω )i∈I,j∈J has non-zero determinant. Independent proofs were given by Dieudonné [1], Evans and Isaacs [2], and Terence Tao [4]. Tao points out that the theorem is equivalent to the inequality |suppf |+|suppf̂ | ≥ p+1 holding for any function 0 6≡ f : Fp → C and its Fourier transform f̂ , a fact also discovered independently by András Biró. We give a very simple proof via the following two lemmas. Lemma 1 Let Ω be an indeterminate. Then we have a commutative diagram Z[Ω] → Z ↓ ↓ Z[ω] → Fp of surjective ring homomorphisms, the horizontal arrows given by Ω 7→ 1 and ω 7→ 1, and the left vertical arrow by Ω 7→ ω. The kernel of the lower horizontal arrow is the principal ideal generated by 1− ω. Proof. The vertical arrows and the upper horizontal one are obviously well defined by the text of the lemma. Let Φp(Ω) = 1 + Ω + · · · + Ω p−1 be the minimal polynomial of the algebraic integer ω. The kernel of the composite Z[Ω] → Z → Fp is the ideal (1 − Ω, p) which contains the kernel (Φp(Ω)) of the left vertical arrow, since Φp(Ω) ≡ p mod (1−Ω). Therefore, the lower horizontal arrow is well defined. Its kernel is the image of (1− Ω, p), which is (1− ω, p) = (1− ω). Indeed, p ≡ Φp(ω) = 0 mod (1− ω). Lemma 2 Let 0 6≡ g(x) ∈ Fp[x] be a polynomial of degree < p. Then the multiplicity of any element 0 6= a ∈ Fp as a root of g(x) is strictly less than the number of non-zero coefficients of g(x).