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If an irreducible cubic polynomial with coefficients in $\mathbb Q$ has Galois group $C_3$ then, since no order $2$ symmetry lies in the Galois group no complex conjugation acts on the roots, it's roots must all be real.

What about the converse, must $3$ real roots imply $C_3$ group?

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presumably, you mean 3 irrational roots? If any root is rational the group can't be $C_3$. –  Alon Amit May 11 '11 at 20:46
    
I think they are irrational as a consequence of irreducibility. –  quanta May 11 '11 at 20:47
    
Sure, but when you say "converse" it's unclear which of the assumptions of the original situation you're still assuming. If all you know is that the 3 roots are real, you can't conclude that the polynomial is irreducible. –  Alon Amit May 11 '11 at 20:48
    
Whatever assumptions make the theorem work. –  quanta May 11 '11 at 20:50
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2 Answers

up vote 6 down vote accepted

No.

Suppose $f(x) = x^3 + ax + b$ (a linear change of variable will take it to this form). If $\alpha_1,\alpha_2,\alpha_3$ are the three roots of $f(x)$, then $$\delta = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)$$ lies in the splitting field, and its square is the discriminant of the polynomial, $\delta^2 = -4a^3 - 27b^2$.

So the splitting field contains $\mathbb{Q}(\sqrt{\Delta})$, hence the splitting field has degree $6$ over $\mathbb{Q}$ whenever the discriminant is not a rational square. In fact, the converse also holds: if the splitting field has degree $6$, then the discriminant is not a square.

On the other hand, the splitting field is real if and only if the discriminant is positive (since it equals a square).

So to find an irreducible cubic whose splitting field is completely real but has Galois group $S_3$, you just need to find an irreducible cubic whose discriminant is positive and not a square. For example, $f(x) = x^3 -4x + 1$; the only possible rational roots are $1$ and $-1$, so it is irreducible, and the discriminant is $$\Delta = -4(-4)^3 -27 = 229,$$ which is positive and not a square. So the splitting field is of degree $6$ and has Galois group $S_3$.

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No - an irreducible cubic with discriminant $D$ for which $\sqrt{D}\notin\mathbb{Q}$, which is possible even if all 3 roots are real, will have a Galois group of $S_3$. See this page.

As is worked out in the reference, if $f$ is an irreducible cubic in $\mathbb{Q}[x]$ with discriminant $D$, then the splitting field of $f$ is $K=\mathbb{Q}(\alpha,\sqrt{D})$ where $\alpha$ is any root of $f$. Because $f$ will have at least one real root, we have that $D>0$ if and only if all three roots of $f$ are real. However, there will still be an order 2 component of $\text{Gal}(K/\mathbb{Q})$ if $\sqrt{D}\notin\mathbb{Q}$.

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