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Let $P_n$ be the set of all polynomials over $\mathbb{Z}_2$ and of degree less than $n$. What is the rate of growth of $|\{x y : x,y \in P_n \}|$? I'm looking for an answer like "$\Theta(2^{3n/2})$" or "$\Theta(n2^n)$"; an answer does not need to have details on the leading constant or any lower-order terms.

The motivation is for proving regularity properties of hash function that use multiplication of binary polynomials. Regularity is roughly the property that the pre-image of any value in the range should not be too large. Consider the hash function $f_h : P_{2n} \times P_n \to P_{2n}$, parametrized by $0 \neq h \in P_n$ and defined by

$$ f_h(a,b) = a + hb $$

I believe that the smaller the size of the set $\{x y : x,y \in P_n \}$, the more regular $f_h$.

Certainly, the set has size at least $|P_n| = 2^n$. Also, there are no duplicates in $\{(p + x^{n-1}) x^i : p \in P_{n-1}, i \in 0 \dots n-1\}$, which has size $n2^{n-1}$, so that's a (weak-looking) lower bound.

For an upper bound, from another answer on this site, I learned that the number of irreducible polynomials in $P_{2n-1}$ is $\sim2^{2n-2}/(2n-2)$, and only $\sim 2^{n-1}/(n-1)$ are of degree less than $n$. So $\sim(2^{2n-2}/(2n-2)-2^{n-1}/(n-1))$ are of degree $\in [n..2n-1)$. These can be excluded from belonging to $\{x y : x,y \in P_n \}$, but that's also a weak upper bound.

jbapple
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    I think the lower bound can be improved to $\Omega(2^{2n}/n^2)$, since the product of two irreducible degree-$n$ polynomials is unique. This brings it within a factor of $O(n^2)$ of the upper bound, no? – Erick Wong Feb 21 '16 at 00:33

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