Let $f(x)\in\mathbb{Z}[x]$ be a polynomial of degree $d$ and naive height (maximum of the absolute values of the coefficients) at most $H$. Is there anything known about the number of prime factors of the discriminant of $f$? Particularly upper bounds?


[This](https://mathoverflow.net/questions/11880/what-primes-divide-the-discriminant-of-a-polynomial) question is similar, but more restrictive, in that it asks about palindromic polynomials and asks for the prime factors explicitly.



Number of prime factors of a polynomial discriminant
The discriminant is $O(H^{2(d-1)})$ so its number of prime factors is bounded by $$ (1 + o(1)) \frac{ \log (H^{2(d-1)})}{\log \log (H^{2(d-1)})} = (2(d-1) + o(1)) \frac{ \log H}{\log \log H}$$ by the asymptotics for primorials (a consequence of the prime number theorem).


This is not too far from optimal (in the worst case) since the for any set of primes whose product is less than $H$, we can choose a polynomial of height $\leq H$ whose discriminant is divisible for all those primes by the Chinese remainder theorem, and such a set of primes can have size $(1+o(1))\frac{\log H}{\log \log H}$ by the same primorial asymptotics, so the upper bound is sharp (in the worst case) to within a factor of $2(d-1)$.


However, based on your comments, you may be interested in the average case also. For each $p<H$, the fraction of polynomials for which $p$ divides the discriminant is approximately $\frac{1}{p}$, so the average number of prime factors of the discriminant less than $H$ is approximately $\sum\_{p<H} \frac{1}{p} \approx \log \log H$, and the maximum number of prime factors of size $\geq H$ is $2(d-1)$ by the $O(H^{2(d-1)})$ bound for the discriminant, so the average number of prime factors is $\log \log H + O\_d(1)$.


