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I am reading an article about Galois groups. The article states that:

It can also be shown that for each degree $d$ there exist polynomials whose Galois group is the fully symmetric group $S_d$.

I think that the Galois group of a quadratic is isomorphic to $S_2$ if the roots are not rational.

I think that the Galois group of $x^3 - a$ is isomorphic to $S_3$ if $a$ is not a perfect cube.

Is it difficult to find an example of a degree $4$ polynomial whose Galois group is isomorphic to $S_4$? I am reading a text book and so far most of the examples have been for Galois groups of "small" order.

I have not progressed to the point where I can appreciate a proof of the statement made in the article but I would like to see an example of such a polynomial of degree $4$ or hear from someone knowledgeable that this is a difficult and/or tedious task.

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2 Answers 2

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This isn't too hard to do, but the below method requires a little knowledge of the cubic resolvent $h$ of a quartic polynomial $g$, namely that $\operatorname{Gal}(h) \leq \operatorname{Gal}(g)$ and that the discriminants of $g$ and $h$ (essentially) coincide. (I can expand about this some if it would be helpful.)

If a polynomial $g$ is irreducible (which is a necessary condition for $\operatorname{Gal}(g) \cong S_{\deg g}$, and which we thus henceforth assume), then its Galois group acts transitively. The only transitive subgroups of $S_4$ are (up to conjugacy) $S_4, A_4, D_8, \Bbb Z_4, \Bbb Z_2 \times \Bbb Z_2$. We'll use the fact that the only groups among these whose order is divisible by $6$ are $S_4$ and $A_4$.

At least when the character of the underlying field $\Bbb F$ is not $3$, we may for simplicity of the below formulae make a linear change of variables so that $g$ has zero coefficient in its $x^3$ term and write (after dividing by the leading coefficient) $$g(x) = x^4 + p x^2 + q x + r,$$ its resolvent cubic is $$h(x) = x^3 - 2 p x^2 + (p^2 - 4 r) x + q^2,$$ and the discriminants of $g$ and $h$ coincide (perhaps up to an overall nonzero multiplicative constant), and are $$D = 16 p^4 r − 4 p^3 q^2 − 128 p^2 r^2 + 144 p q^2 r − 27 q^4 + 256 r^3.$$ If $h$ is irreducible and its discriminant $D$ is not a square, then (1) $\operatorname{Gal}(h) \leq G := \operatorname{Gal}(g)$ is $S_3$, so $G$ has order divisible by $6$ and hence by the above is $S_4$ or $A_4$, and (2) since $D$ (the discriminant of $g$) is not a square, $G \not\leq A_4$ and hence $G \cong S_4$. For $g$ and $h$ to be irreducible, they must have nonzero constant terms and hence $q, r \neq 0$. For $p = 0$, the above formulae simplify to $$h(x) = x^3 - 4 r x + q^2$$ and $$D = - 27 q^4 + 256 r^3,$$ so to find an example we can search for $q, r$ for which $h$ is irreducible and $D$ is nonsquare. For $\Bbb F = \Bbb Q$ the simple choice $q = r = 1$ satisfies these criteria (the first by the Rational Root Test), so, we have for example, that $$\operatorname{Gal}(x^4 + x + 1) \cong S_4 .$$

See these notes for more details (using the same notation). Also, note that these sorts of examples are generic in that, in a sense that can be made precise, most irreducible polynomials of degree $n$ in $\Bbb Q[x]$ have Galois group $S_n$.

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Consider $F=\Bbb C(x_1,x_2,\dots,x_n)$ where the $x_i$'s are indeterminates (variables). A group of automorphisms can be determined by how they permute the $x_i$'s. So that group of automorphisms is $S_n$. Now take the fixed field $K$ of those automorphisms. Then Gal$(F/K)$ should be $S_n$ by Artin's theorem.

The polynomial is then $P(Y)=(Y-x_1)\cdots(Y-x_n)$ which if you multiply out the coefficients are the symmetric polynomials in the $x_i$'s.

If you want it as an extension of $\Bbb Q$ then that's harder.

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