Let $f = x^n + a_{n-1}x^n + \cdots + a_0$ be a monic polynomial of degree $n \geq 2$ with integer coefficients. By $\text{Gal}(f)$ we mean the Galois group over $\mathbb{Q}$ of the Galois closure of $f$. Define $H(f) = \max\{|a_i|\}$ denote the naive or box height of $f$. Hilbert's irreducibility theorem asserts that for most integer $n$-tuples $(a_0, \cdots, a_{n-1})$ with $H(f) \leq B$ say, the corresponding polynomial $f$ has symmetric group $S_n$. If we assume that one of the coefficients $a_i$ of the polynomials are fixed, then is it still true that the number of such polynomials whose galois group is not symmetric group is $o(B^{n-1})?$ Thanks in advance for any assistance.
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1$\begingroup$ What does $N$ stand for? $\endgroup$– Gerry MyersonCommented Mar 6, 2022 at 23:12
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1$\begingroup$ @GerryMyerson It has been fixed. $\endgroup$– HhhhhhhhhhhCommented Mar 7, 2022 at 5:09
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This is equivalent (by Hilbert) to asking whether the partially specialized polynomial still has symmetric Galois group (over the respective function field). This holds unless you specialized the constant coefficient to $0$.
According to Cohen, S. D., The Galois group of a polynomial with two indeterminate coefficients, Pac. J. Math. 90, 63-76 (1980), ZBL0408.12011, you may even prescribe all but two of the coefficients and still get the full symmetric group, unless you specialized such that the resulting polynomial is a polynomial in $x^r$ for some $r>1$.
