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From the theory of numbers we have the

Proposition:
If $\mathfrak{a}$ and $\mathfrak{b}$ are mutually prime, then the density of primes congruent to $\mathfrak{b}$ modulo $\mathfrak{a}$ in the set of all primes is the reciprocal of $\phi (\mathfrak{a})$ where $\phi$ denotes the Euler function.

And it can be shown that every polynomial with integer coefficients has infinitely many prime divisors where by a prime divisor of a polynomial we understand a prime which divides the value of that polynomial at some integer $\mathfrak{n}$; it is natural then for one to ask whether or not there is a similar result to the density theorem above, i.e. the

Conjecture:
If $\mathfrak{f}$ belongs to $\mathbb{Z} [x]$, then the density of primes which divide some values of $\mathfrak{f}$ at integers in the set of all primes is uniquely determined by its coefficients somehow.

I haven't heard of any result like this, and it is desirable to have some sources to search, if there exists,; in any case, thanks for paying attention.

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Thanks for the edit, @Eric Narslund –  awllower Apr 1 '11 at 5:01

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up vote 2 down vote accepted

Of course, your "conjecture" is trivially true, because the set of prime divisors of $f(n)$ as $n$ ranges over the integers is determined by $f$. I guess what you meant to ask was "does this set have a density and can it be computed".

The answer to both questions is yes and the relevant result is called Chebotarev's density theorem. First, note that if $f=g\cdot h$, then the set of prime divisors of $f(n),\;n\in\mathbb{Z}$ is the union of the corresponding sets for $g$ and $h$. So we may without loss of generality consider irreducible $f$. For such an $f$ of degree $n$, let $G$ be the Galois group of its splitting field, thought of as a subgroup of $S_n$, and let $C$ be the subset of $G$ consisting of all fixed point-free elements. Then the set of primes that divide some $f(n)$ has a natural density and it is equal to $|C|/|G|$. Note that this is the set of primes modulo which $f$ does not attain a linear factor. Similar results hold for any given splitting pattern of $f$ modulo $p$, as the wikipedia page explains.

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Thanks @Alex Bartel: It does explain many things to me. –  awllower Apr 1 '11 at 5:00

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