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Assume that $f(x)$ is a quintic polynomial with integer coefficients and is irreducible over $\mathbb{Q}$.

If $f(x)$ has three distinct real roots and two non-real complex roots, then $f(x)$ is not solvable by radicals over $\mathbb{Q}$ since it's Galois group over $\mathbb{Q}$ is isomorphic to $S_{5}$ (which is not solvable).

This appears to be the prototypical example of a quintic polynomial with integer coefficients that is not solvable by radicals.

But what if $f(x)$ has a real root and four distinct non-real roots? Or what if $f(x)$ has five distinct real roots? In these two cases, do we need more information about $f(x)$ to determine if it's solvable by radicals?

user232456
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2 Answers2

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The polynomial $f(x)$ in general need not have three distinct real roots, and two non-real roots, in order to be not solvable by radicals. Consider, say, $f(x)=x^5+3x+3$, which is irreducible by Eisenstein. Here $f(x)$ has only one real root, because its derivative is positive everywhere, so that $f$ is an increasing function. Nevertheless the Galois group of $f$ is again $S_5$. The reason here is, that it is sufficient for $f(x)$ to have Galois group $S_5$, if $f$, modulo some prime number $p$ has cubic and quadratic factors. Note the the Galois group of an irreducible polynomial $f$ of degree $5$ is one of the following groups: $$ C_5,D_5,Fr_5,A_5,S_5. $$ Here the first three groups are solvable, and $A_5,S_5$ are not solvable. Each group really arises as Galois group of an irreducible quintic polynomial, e.g. by the following polynomials $$ f_{C_5}(t)=t^5+t^4-4t^3-3t^2+3t+1,\; f_{D_5}(t)=t^5-5t+12 $$

$$ f_{Fr_5}(t)=t^5-2,\; f_{A_5}(t)=t^5+20t+16,\; f_{S_5}(t)=t^5-4t+2. $$ However, we have many choices here, and the number of real roots is not an invariant. What we need for $f$ being not solvable by radicals is that its Galois group is either $A_5$ or $S_5$. For more on this question see also the answers here.

Community
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Dietrich Burde
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  • Since every irreducible polynomial in characteristic $0$ is separable, is it redundant to say that the roots need to be distinct? – user232456 Dec 03 '15 at 17:35
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In some cases, we can "rat out" a quintic that seems to have a favorable number of real roots and yet is in fact not solvable. We do so by counting roots in a different field.

Let's look again at $x^5+3x+3=0$. Instead of counting the usual real roots, we seek roots in $\mathbb{Z}/7\mathbb{Z}$. Clearly $x=1$ is one such root, and if we go through trials we discover that $x=5$ is another root and both roots have multiplicity one. The remaining elements are not roots.

Thus we have the factorization $(x+2)(x+6)(x^3+6x^2+3x+2)$ where the cubic factor is irreducible. So the Galois group has to include a single pairwise transposition (between the two "real" roots modulo $7$) as well as the usual $5$-cycle and thus is all of $S_5$.

Oscar Lanzi
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