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Here is an exercise in Hungerford's Algebra, page 277 Ex.12.

Let $K$ be a subfield of real numbers and $f \in K[x]$ an irreducible quartic polynomial(of degree 4). If $f$ has exactly two roots, the Galois group of $f$ is $S_4$ or $D_4$.

If we can proof that $f$ is separable(no multiple roots in the splitting field), then it is easy to show the only possible isomophic types are $S_4$ and $D_4$. But how to show that $f$ is separable?

Plus, Can anybody show me an example of a polynomial satisfying the conditions above whose Galois group $ \cong S_4$? Thanks in advance!

No One
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  • You might look here. – rogerl Apr 28 '15 at 01:58
  • @rogerl Thanks for the reference, but does that webpage really answer my question? It seems that in the discussion of irreducible polynomials, the author assumed the separability. What if $f$ has multiple roots? The same discussion in Hungerford's book assumed separability of $f$ in the condition.(see Prop 4.11 in page 411) – No One Apr 28 '15 at 02:18

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As an example of a polynomial of this kind with Galois group $\cong S_4$ I proffer $$ f(x)=x^4+2x^2-x-3\in\Bbb{Q}[x]. $$ It has exactly two real zeros because $f(0)<0$, $\lim_{x\to\pm\infty}f(x)=+\infty$, and because it's convex - $f''(x)=12x^2+4$ is positive everywhere.

It is irreducible, because reduction modulo $2$ gives $x^4+x+1$ which is irreducible in $\Bbb{F}_2[x]$. Therefore its Galois group, when viewed as a group of permutations of its roots, is a transitivie subgroup of $S_4$, so either $V_4, C_4, D_4, A_4$ or $S_4$. Complex conjugation is an automorphism of the splitting field that acts as a $2$-cycle on the non-real roots. This rules out alternatives other than $D_4$ and $S_4$.

The final needed bit is the observation that modulo the prime $p=3$ we have the factorization $$ f(x)=x(x^3-x-1)\in\Bbb{F}_3[x]. $$ The cubic factor here is irreducible - see e.g. here. By Dedekind's theorem, see e.g. here, we then know that there is a 3-cycle in the Galois group. This leaves $S_4$ as the only alternative for the Galois group.

Pleading quilty to reverse engineering this. I started with the desired factorizations modulo $2,3$, and played with what the Chinese Remainder Theorem let's me play with to find a polynomial that fits.

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Jyrki Lahtonen
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  • Not sure whether Hungerford covers Dedekind's theorem? I used Jacobson's Basic Algebra I-II, and the result is included in vol. I, ch. 4. I also hope you are familiar with the technique of deducing irreducibility of a polynomial in $\Bbb{Z}[x]$ by reduction modulo a prime. Gauss' lemma tells us that such a polynomial is irreducible in $\Bbb{Q}[x]$, iff its irreducible in $\Bbb{Z}[x]$. – Jyrki Lahtonen Apr 28 '15 at 06:08
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Any irreducible polynomial over a field of characteristic zero is separable. More generally, any irreducible polynomial over a perfect field is separable.

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