# Primitive Elements for $S_n$ Galois Extensions?

This is an offshoot of my other question two days ago. How to apply Hilbert's Irreducibilty theorem?

But it is of independent interest.

Solutions of Inverse Galois Problem for a finite group $G$ give us polynomials whose splitting field has Galois group $G$. I would like to know if there is a way of getting the minimal polynomial (degree equal to order of $G$) for a primitive element of that splitting field.

If this is too cumbersome at least is there a reasonable description of a primitive element?

For example in the case of $S_n$ can we get a recursive description something like this? give two algebraic numbers $\alpha,\beta$ one of them of degree $(n-1)!$ such that $\alpha+\beta$ is of degree $n!$ and generates an $S_n$-Galois extension? (as $S_n$ is generated by an $n$-cycle and an involution, the number of degree $(n-1)!$ above could possibly be fixed by that $n$-cycle).

Known baby example: Considering that cube roots of one and two together generate a Galois extension with $S_3$ as Galois group it is straightforward to see $\omega +\sqrt[3]{2}$ as primitive element and write down its minimal polynomial. (It turns out to be $x^6 + 3x^5 + 6x^4 - 13x^3 - 24x^2 + 33x + 121$, quite a mouthful).

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I don't understand what you're asking. – Qiaochu Yuan Jan 3 '13 at 4:28
Thanks Qiaochu: the opening paragraph was missed out during editing and I have restored it now. – P Vanchinathan Jan 3 '13 at 5:35
Suppose the grround field $F$ has at $n$ distinct elements $c_1,\ldots,c_n$ that don't sum to zero. (For example, in characteristic zero let $c_i=i$ for each $i$.) Write the extension as the Galois closure of a degree-n extension $F(x_1)$. Let $x_1,\ldots,x_n$ be the conjugates of $x_1$. Then $X := \sum_{i=1}^n c_i x_i$ is a primitive element, because the conjugates of $X$ are obtained by permuting the $c_i$, and no nontrivial permutation fixes $X$. – Noam D. Elkies Jan 3 '13 at 5:37
Moreover, one there is an explicit formula for the minimal polynomial of this element, since one can compute addition and multiplication in the ring $Q[x_1,...,x_n]$ using the obvious relations, and thus find the linear relation between $1,X,X^2,...,X^{n!}$, using polynomials. – Will Sawin Jan 3 '13 at 6:08
Thanks to Elkies, and Swain. Your answers are clear and explicit and readily usable to construct. I have now my work cut out getting that polynomial of degree 24 giving $S_4$ as Galois extension. – P Vanchinathan Jan 3 '13 at 10:09