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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
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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
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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
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