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Is there a quartic polynomial $p(x)\in\mathbb{Q}[x]$ irreducible with four real roots such that Galois group is ${S_4}$?

If it really exists, can someone give me a example?

ccorn
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Junior Soares
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    The most obvious are the ones with roots of the form $\pm a\pm\sqrt b$ – Lucian Jun 26 '15 at 13:13
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    @Lucian, but aren't those reducible? – lhf Jun 26 '15 at 13:21
  • Thanks @Lucian. I had starded with the roots too, but i was using $\pm \sqrt{p} \pm\sqrt{q} $ with p,q primes. In your answer 'b' should be a prime, for it does not to be reducible. – Junior Soares Jun 26 '15 at 13:58
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    @lhf: Believe it or not, some 12 hours ago, Mathematica was telling me that $x^4-6x^2+1$ $($the polynomial corresponding to $a=1$ and $b=2)$ is irreducible. Now, she's suddenly able to factor it. :-$)$ But if we change a with $\sqrt a$, where both a and b are not perfect squares, it works. – Lucian Jun 27 '15 at 00:41

1 Answers1

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As hinted in the comments to a related question and as shown in Keith Conrad's notes, a degree-$4$ polynomial from $K[X]$, where $K$ is a field with $\operatorname{char}K\not\in\{2,3\}$, has Galois group $S_4$ if

  • it is irreducible over $K$ (thus its Galois group has order divisible by $4$), and
  • its resolvent cubic is irreducible over $K$ (thus the Galois group has order divisible by $3$), and
  • its discriminant is not a square in $K$ (thus the Galois group is not a subgroup of $A_4$).

Now consider $$f(X) = X^4 - 6 X^2 + 2 X + 2$$ which changes sign between $-\infty,-1,0,+1,+\infty$ and therefore has four real roots.

$f(X)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p=2$. Therefore its Galois group has order divisible by $4$.

The resolvent cubic of $f(X)$ is $$\begin{align} g(X) &= X^3 + 6 X^2 - 8 X - 52 & g(X-2) &= X^3 - 20 X - 20 \end{align}$$ so $g(X-2)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion with $p=5$, thus $g(X)$ is irreducible over $\mathbb{Q}$ as well. The extension degree of the splitting field of $g(X)$ over $\mathbb{Q}$ is therefore divisible by $3$. By construction of the resolvent cubic, the splitting field of $g(X)$ is contained in the splitting field of $f(X)$, so the order of the Galois group of $f(X)$ must be divisible by $3$.

Again by construction, the discriminant of $f(X)$ equals the discriminant of $g(X)$ and of $g(X-2)$, which is $$\Delta=21200$$ You easily recognize the square $100$ as a factor, but the cofactor $212\equiv2\pmod{5}$ is not a square. Therefore $\Delta$ is not a square in $\mathbb{Q}$. Consequently, the Galois group of $f(X)$ is not a subgroup of $A_4$.

Summarizing the above results, the Galois group of $f(X)$ has order divisible by $3$ and $4$, yet is not a subgroup of $A_4$. This leaves $S_4$ as only possible Galois group for $f(X)$.

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ccorn
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  • Example with somewhat smaller discriminant: $f(X)=X^4-4X^2+X+1$, changing signs between $-\infty,-1,0,+1,+\infty$, irreducible $\bmod{2}$, resolvent cubic $g(X)=X^3+4X^2-4X-17$, irreducible $\bmod{3}$, $\Delta=1957$. – ccorn Jun 27 '15 at 06:36
  • Thanks. I was thinking about the same question but with a polynomial with degree 3 and $S_3$, in this case, it is not possible, right? – Junior Soares Jun 27 '15 at 12:38
  • For a degree-$3$ polynomial with real roots and Galois group $S_3$: By Keith Conrad's notes, you will find that then $f(X)$ just needs to be irreducible (thus $3$ divides the group order) and have nonsquare positive discriminant (thus it is not $A_3$, and all roots are real). Therefore, the $g(X)$ and $g(X-2)$ given above can serve as examples. – ccorn Jun 27 '15 at 15:36