In general an irreducible polynomial of degree $n$ over $\mathbb{Q}$ has Galois group $S_n$.
Could this be argued as follows:

A general polynomial $f(x)=a_0 + a_1 x+ \ldots + a_n x^n \in \mathbb{Q}[x]$ can be thought of as a polynomial with coefficients in $K=\mathbb{Q}(a_0,a_1,\ldots,a_n)$ where $a_i$'s are indeterminates. In the splitting field for this polynomial, say $E$, the polynomial factors as $f(x)=\prod_{i=1}^n (x-\lambda_i)$ and $a_i$'s are symmetric polynomials in $\lambda_i$. Now the Galois group of this extension $Gal(E/K)$ is $S_n$.

Is this the correct sense in which the above statement holds. If yes, then can it be made more precise. Specifically when I say "thought of" in the above statement.

If not, then what is the correct way to formalize the statement.

I do not know of any reference where this is specifically said. I just sort of believe it. Please let me know if I am wrong.

  • $\begingroup$ If I remember correctly, this is first proved in B. L. van der Waerden, Die Seltenheit der reduziblen Gleichungen und der Gleichungen mit Affekt. Monatsh. Math. Phys., 43(1):133–147, 1936. But I've not found an English translation yet. $\endgroup$ – Cave Johnson Dec 25 '16 at 10:29
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    $\begingroup$ ...It is stated that, choosing among all $n$-degree polynomials with coefficients $|a_i|<B$, the probability of its Galois field being $S_n$ tends to $1$ as $B\to\infty$. $\endgroup$ – Cave Johnson Dec 25 '16 at 10:34
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    $\begingroup$ For a generic polynomial it is easy to show that if the zeros are indeterminates (= algebraically independent transcendentals over some field $K$), then so are the coefficients, and the Galois group is $S_n$. $\endgroup$ – Jyrki Lahtonen Dec 25 '16 at 22:15

I suppose one way to say this would be the following.

For a generic polynomial $p(x)$ of degree $n$ in $\mathbb{Q}(a_0,a_1,\ldots,a_n)[x]$, there is a Zariski dense subset $A\subset \mathbb{Q}^{n+1}$ such that the specialization $f(a'_0,a_1',\ldots, a_n')$, where $a'=(a'_0,a_1',\ldots, a_n')\in A$, is irreducible in $\mathbb{Q}[x]$.

This is a consequence of Hilbert's Irreducibility Theorem.

  • $\begingroup$ that's not enough to get a galois group of $S_n$ $\endgroup$ – mercio Jan 2 '17 at 20:45
  • $\begingroup$ Why not. The Galois group remains unchanged after specialization. $\endgroup$ – user2902293 Jan 2 '17 at 20:46
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    $\begingroup$ there are irreducible polynomials of degree $n$ whose Galois group is not $S_n$ $\endgroup$ – mercio Jan 2 '17 at 20:48
  • $\begingroup$ True. But if $\mathbb{Q}(a_i)(f)/\mathbb{Q}(a_i)$ is Galois with Galois group $G$ then the specialization map from $\mathbb{Q}(a_i)\to \mathbb{Q}$ preserves the Galois group (provided the specalization polynomial is irreducible) $\endgroup$ – user2902293 Jan 2 '17 at 20:52
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    $\begingroup$ you can make it work by applying the Hilbert Irreducibility Theorem to the resolvant of, for example, $\sum k x_k$. It is a degree $n!$ polynomial over the $a_k/a_n$, in the generic case it is irreducible, and it stays irreducible after the specialization if and only if the Galois group stays $S_n$. $\endgroup$ – mercio Jan 3 '17 at 14:55

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