proof of Cauchy-Davenport theorem
There is a proof, essentially from here http://www.math.tau.ac.il/ nogaa/PDFS/annr3.pdf(Imre Ruzsa et al.):
Let |A|=k and |B|=l and |A+B|=n.
If n≥k+l-1, the assertion is true, now assume n<k+l-1.
Form the polynomial f(x,y)=∏c∈A+B(x+y-c)=∑i+j≤nfij*xi*yj.
The sum is over i,j with i+j≤n, because there are n factors in the product.
Since Zp is a field, there are polynomials gi of degree <k and hj of degree <l such that gi(x)=xi for all x∈A and hj(y)=yj for all y∈B. Define a polynomial p by p(x,y)=∑i<k,j<lfij*xi*yj+∑i≥k,j≤n-ifij*gi(x)*yj+∑j≥l,i≤n-jfij*xi*hj(y).
This polynomial coincides with f(x,y) for all x in A and y in B, for these x,y we have, however, f(x,y)=0. The polynomial p(x,y) is of degree <k in x and of degree <l in y. Let x∈A, then p(x,y)=∑pj(x)*yj is zero for all y∈B, and all coefficients must be zero. Finally, pj(x) is zero for all x∈A, and all coefficients pij of p(x,y)=∑pij*xi*yj must be zero as elements of Zp.
Should the assertion of the theorem be false, then there are numbers u,v with u+v=n<p and u<k and v<l.
But the monomial xu*yv does not appear in the second and third sum, because for j=v we have i≤n-j=u<k, and for i=u we have j≤n-i=v<l.
Then puv=0 is modulop equal to fuv, this is equal to the binomial coefficient (nv), which is not divisible by p for n<p, a contradiction
.
The Cauchy-Davenport theorem
is proved.
Title | proof of Cauchy-Davenport theorem |
---|---|
Canonical name | ProofOfCauchyDavenportTheorem |
Date of creation | 2013-03-22 14:34:49 |
Last modified on | 2013-03-22 14:34:49 |
Owner | Wolfgang (5320) |
Last modified by | Wolfgang (5320) |
Numerical id | 26 |
Author | Wolfgang (5320) |
Entry type | Proof |
Classification | msc 11B05 |