# Number of real roots of irreducible polynomials that are solvable by radicals

Let $$n \geq 3$$ be a natural number. Define the set $$X_n$$ as the set of natural numbers that appear as the number of real roots an irreducible polynomial of degree $$n$$ over $$\mathbb{Q}$$ which is solvable by radicals can have.

Example: In case $$n=p$$ is a prime, we have $$X_p= \{1,p \}$$.

Is $$X_n$$ or the cardinality $$|X_n|$$ also known for other values?

It would be interesting to see the beginning of the sequence $$a_n=|X_n|$$ for small values of $$n$$, maybe it appears in the oeis.

• what (trivial or nontrivial) things do you know about $X_n$ in general? – Arno Fehm May 10 '20 at 20:20
• @ArnoFehm I dont really know much about $X_n$ except for the stated result for prime numbers. – Mare May 10 '20 at 20:32
• You certainly knew $X_4=\{0,2,4\}$, so is $X_6$ the first case you don't know? – YCor May 10 '20 at 20:36
• @YCor Thanks, yes 6 is the first non-trivial case it seems. – Mare May 10 '20 at 20:38
• GAP might provide for small $n$ the list of $k$ such that $S_n$ has a transitive solvable subgroup that contains a product of $k$ disjoint transpositions (which is a necessary condition for the existence of such a polynomial with $n-2k$ real roots). Nevertheless for $n=6,8$ this yields no obstruction. – YCor May 10 '20 at 20:47

Klueners and Malle have a database of number fields of degree $$\leq 19$$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $$n$$ with $$4 \leq n \leq 18$$, $$X_{n} = \{ k : 0 \leq k \leq n \text{ and } k \equiv n \pmod{2} \}$$.
EDIT: If $$n = 2k$$ is even, then $$\mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/k\mathbb{Z}$$ contains elements of order $$2$$ with any desired even number of fixed points. It seems possible that one can obtain the same thing for $$n$$'s that are multiples of $$3$$. The smallest $$n$$ that does not fall into this category is $$n = 25$$, and using the classification of transitive subgroups of $$S_{25}$$, one finds that $$X_{25} \subseteq \{ 1, 5, 9, 13, 17, 21, 25 \}$$.
• So quite surprisingly for primes it tends to be very small and for composite it's as large as possible... until 19. For every such $n$ and $k$ with $n-k$ even you manually checked inside the list the polynomials until finding one with $k$ real roots? or you found a quicker way? – YCor May 10 '20 at 21:12
Loewy’s theorem. Let $$K$$ be a real number ﬁeld and $$f(X)$$ an irreducible polynomial in $$K[X]$$ of odd degree $$n$$. If $$p$$ is the smallest prime divisor of $$n$$ and the Galois group of $$f(X)$$ over $$K$$ is solvable, then $$r(f) = 1$$ or $$n$$ or satisﬁes the inequalities $$p ≤ r(f)≤ n−p +1.$$