Intuitively, the Galois group (of a splitting field over $\mathbb Q$ of) a polynomial $f\in\mathbb Q[X]$ taken at random is most probably the full permutation group on the roots of $f$. This intuition can be made precise as follows.

Elementary: For inegers $d \geq 1$ and $N \geq 1$, consider the set $P_{d,N}$ of polynomials in one variable of degree $\leq d$ whose coefficients have absolute value $\leq N$. The set $P_{d,N}$ is finite, and we can consider the subset $Q_{d,N}$ of those $f\in P_{d,N}$ whose Galois group is the full symmetric group $S_d$. Then, is it true that the ratio $|Q_{d,N}|/|P_{d,N}|$ goes to $1$ as $N$ goes to infinity? If so, are there good estimates for the speed of convergence?

Less elementary, but still the same: Let $k$ be a number field. For an integer $d\geq 1$, indentify $k$--points of the $d$--dimensional projective space $\mathbb P^d$ with nonzero polynomials of degree $\leq d$ over $k$ up to scalars. This way, we can speak about the Galois group of a point $f\in \mathbb P^d(k)$. For an integer $N$, let $p_{d,N}$ be the number of points of $P^d(k)$ of height $\leq N$, and let $q_{d,N}$ be the number of points of height $\leq N$ and with Galois group $S_d$. Then, is it true that $$\frac{q_{d,N}}{p_{d,N}}\to 1$$ as $N$ goes to infinity? And again, are there good estimates for the speed of convergence?

Remark: The subset of $\mathbb Q^{d+1} = \mathbb A^{d+1}(\mathbb Q)$ of those elements $(a_0,\ldots, a_d)$ such that the Galois group of the polynomial $f = a_0+a_1X+\cdots+a_dX^d$ is $S_d$ is Zariski dense in $\mathbb A^{d+1}$. This is so because the Galois group of the generic polynomial $f = t_0+t_1X+\cdots+t_dX^d$ with indeterminates $t_0,\ldots, t_d$ over $\mathbb Q(t_0,\ldots, t_d)$ is $S_d$ and the fact that $\mathbb Q$ (or any number field) is a Hilbertian field.

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    $\begingroup$ Yes to everything (!). A nice reference is J.-P. Serre Lectures on the Mordell-Weil theorem, Aspects of Mathematics, E 15, Vieweg which includes estimates of the error term. $\endgroup$ Mar 14, 2011 at 9:10
  • $\begingroup$ Out of interest, Rainer Dietmann has handled the case of degree four and improved on Gallagher's bounds: Probabilistic Galois theory for quartic polynomials , Glasgow Math. J. 48 (2006), 553--556. $\endgroup$ Mar 14, 2011 at 10:17
  • $\begingroup$ See also mathoverflow.net/questions/28453/… $\endgroup$ Mar 14, 2011 at 12:45
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    $\begingroup$ But perhaps worth noting -- if you order quartic fields by discriminant, instead of quartic polynomials by height, it is not the case that almost all have Galois group S_4; instead, a positive proportion have S_4, a positive proportion have D_4, and all the other groups occur with zero proportion. $\endgroup$
    – JSE
    Mar 22, 2011 at 1:08

1 Answer 1


As Torsten remarks, the answer is yes and the result is well-known. P. X. Gallagher proved the following: If $E_d(N)$ denotes the set of degree $d$ polynomials with coefficients $|a_i|\le N$ with Galois group not equal to $S_d$, then $$ |E_d(N)|\ll N^{d-1/2}\log N $$This bound gives you what you want. There is a discussion and a list of references in this sci.math post.


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