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I'm working with this problem: Let L/K be a Galois extension with Galois group $S_4.$ Then L is the splitting field of a monic degree 4 irreducible polynomial over K. Char(K)=0.

My method is since [L:K] is finite, there is a primitive element $a. $ Then $a $ has an irrducible polynomial g(x) in K[x]. Since L/K is Galois and irreducible polynomial g(x) has a root $a $ in L, then it splits in L[x]. Then degree of $g(x) = [K(a):K]=[L:K]=|S_4|=24$, not 4. I'm confused. What's wrong? Thank you!

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  • $\begingroup$ Since $a $ is a primitive element in L over K, $K(a)=L.$ $\endgroup$ – user229305 Apr 7 '15 at 1:01
  • $\begingroup$ Since [L:K] is finite, then $a$ is an algebraic element over K, then it is a root of some $g$. $\endgroup$ – user229305 Apr 7 '15 at 1:05
  • $\begingroup$ @Eoin: $g(x)$ is, by definition, the minimal polynomial of $a$, so $g(a) = 0$. OP: Yes, the degree of $g(x)$ is 24; nothing is incorrect there. But if you are in search of a degree 4 polynomial whose splitting field is $L$, the primitive element will not give it to you directly. $\endgroup$ – Brandon Carter Apr 7 '15 at 1:12
  • $\begingroup$ I name the minimal polynomial of $a$ g. Since g has a root in K, then g splits in L[x]. $\endgroup$ – user229305 Apr 7 '15 at 1:14
  • $\begingroup$ The degree of the splitting field of a polynomial is the order of its Galois group, it is not the degree of the irreducible polynomial. More generally, for an irreducible polynomial of degree $n$, its splitting field has degree a divisor of $n!$. $\endgroup$ – Bernard Apr 7 '15 at 1:14
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You know that your Galois group is (isomorphic to) $\mathcal S_4$. You want an element $\beta$ whose minimal polynomial $f$ is of degree $4$, and you want $L$ to be the splitting field of $f$. In other words, $\bigl[K(\beta):K\bigr]=4$. So you want a subgroup $\mathcal H\subset\mathcal S_4$ of index $4$, such that the only subgroup $\mathcal N$ of $\mathcal H$ that’s normal in the big group is $\mathcal N=\{e\}$.

Easy: it’s the most obvious subgroup of index $4$ that you can think of.

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Nothing is wrong, per se. As put forth in the comments, if you have that $\mathbb{L}$ is a splitting field over $\mathbb{F}$ of two polynomials $f(x), g(x)$, then it is not guaranteed that $\deg(f) = \deg(g)$. An example would be the splitting field of $x^3-2$ over $\mathbb{Q}$. This is a degree $6$ extension, and it's not too hard to see that it's also the splitting field of $x^6+108$, the minimal polynomial for the primitive element $i\sqrt{3}\sqrt[3]{2} = 2(-1/2 + i\sqrt{3}/2)\sqrt[3]{2}+ \sqrt[3]{2}$.

In your case, you need an element that is not the primitive generator to be a root of your desired degree four polynomial. My hint is to consider how the degree of an element's minimal polynomial relates to the size of its stabilizer group under the Galois action. (You will probably also need to use the correspondence between subgroups of $S_4$ and the intermediate fields for your extension)

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