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I am studying Algebraic Methods of Combinatorics and I am not able to solve the following problem.

Let $n,k,h$ be integers such that $n \geq 2k^2$ and $h \leq k-1$ (I am not sure if this condition is necessary at all). Show that there is a family $\mathcal{A} \subseteq {[n] \choose k}$ such that $|\mathcal{A}| \geq (n/2k)^h$ and $|A_i \cap A_j| \leq h-1$ for every pair of distinc subsets in $\mathcal{A}$.
[Hint: Choose a largest prime $p$ smaller than $n/k$. Fix a subset $A \subset \mathbb{F}_p$ of size $k$ and consider $A_f = \{(x,f(x)) \in A \times \mathbb{F}_p\}$ for polynomials $f \in \mathbb{F}_p[x]$ of degree at most $h-1$]

My idea was to define $\mathcal{A} = \{A_f \mid f \in \mathbb{F}_p[x] : \mathrm{deg}(f) \leq h-1 \}$, because then we have that $|\mathcal{A}| = p^h \geq (n /2k)^h$ (by using Bertrand's Postulate) as required but I don't know how to deal with the intersection size of two distinc elements of $\mathcal{A}$. Moreover, this family is a family of subsets of $[n] \times [n]$ and not $[n]$ as required. I also tried to just deal with the projection on the second coordinate of each $A_f$, but I am stuck as well with the intersection.

Lecter
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1 Answers1

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You want elements from $[n]$. What you have is a collection of pairs $(x,y)$ with $x \in [k]$ and $y \in \mathbb{F}_p$. You can map this pair to an element in $[kp] \subseteq [n]$ by Chinese remaindering. That is $h:(x,y) \to m$ where $m \equiv x$ (mod $k$) and $y \equiv f(x)$ (mod $p$). This map sends $A_f$ to a $k$-element subset of $[n]$. If $h(x,y) \in h(A_f) \cap h(A_g)$, then $f(x) \equiv g(x)$ mod $p$, so the number of such possible $x$ (and hence $(x,y)$) is at most $h-1$.

Aravind
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  • But $A$ is a random subset of $\mathbb{F}_p$, it is not necessarly ${0,1,\dots,k-1}$. So $x$ is not in $[k]$ but in $A$. – Lecter May 09 '20 at 10:53
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    The choice of $A$ is ours, so instead of letting it be arbitrary, we can let it be the first $k$ non-negative integers. – Aravind May 09 '20 at 10:56
  • So $y \in \mathbb{F}_p$ (not in $[p]$) and I think that you have to say that $h(x,y) \in h(A_f) \cap h(A_g)$ instead of the pair $(x,y)$. – Lecter May 09 '20 at 11:44
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    You're right; I have edited it. – Aravind May 09 '20 at 11:58
  • And the last question... What you did is to solve for which $x$ we have $f(x) \equiv g(x)$ (mod $p$), with $f,g$ being polynomials of degree at most $h-1$ (meaning the have at most $h-1$ real roots). Why we have at most $h-1$ solutions for this? – Lecter May 09 '20 at 13:32
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    Because it is equivalent to $f(x)-g(x) \equiv 0$ (mod $p$), and for distinct $f,g$, the LHS is a non-zero polynomial of degree at most $h-1$. – Aravind May 09 '20 at 13:38
  • Now I realise that $h$ is not really well defined, because it depends on the polynomial $f$; and so it makes no sense to say that $h(x,y)$ is in the intersection (because $A_f$ and $A_g$ would have different associated functions $h_f$ and $h_g$). – Lecter May 15 '20 at 10:55