# Roots of $f(x)$.

Let $$p$$ be a prime not equal to $$2$$. Let $$f(x)$$ be an irreducible polynomial over $$\mathbb{Q}$$ of degree $$p$$ with Galois group isomorphic to the dihedral group $$D_{2p}$$. I need to show that $$f(x)$$ has all real roots or exactly one real root. (Note that $$D_{2p} = $$).

I'm very much stumped and don't even know where to begin. Can I get any aid or hints?

If $$f$$ has at one non-real root, then one $$\alpha\in G$$ is given by complex conjugation. If $$x_1,x_2$$ are two real roots, then they are fix under $$\alpha$$. By transitivity of $$G$$, there exists $$\beta\in G$$ with $$\beta(x_1)=x_2$$. Then $$\beta^{-1}\alpha\beta$$ is of order two and leaves $$x_1$$ fix. The product of two distinct reflections in $$D_{2p}$$ is a rotation (i.e., of order $$p$$), hence $$\alpha\beta^{-1}\alpha\beta$$ is of order $$p$$ and has a fixpoint. Hence it can permute only the other $$p-1$$ roots. But $$S_{p-1}$$ has no element of order $$p$$.

• I particularly like that your argument does not rely on $D_p$ acting on the roots via the same permutations it acts on the vertices of a regular $p$-gon. That is probably the only available action, but proving that could be a bit cumbersome (I think reuns does prove that in a sense). – Jyrki Lahtonen Jun 2 at 4:52

Take $$g\in Gal(f) \cong D_{2p}$$ of order $$p$$, it must permute the $$p$$ roots $$x_0,\ldots,x_{p-1}$$ of $$f$$ transitively (since otherwise it would be of order $$) so wlog $$g(x_j) = x_{j+1}$$ (with $$j$$ taken $$\bmod p$$). The dihedral group is generated by one more element $$h$$ of order $$2$$ and which satisfies $$hgh = g^{-1}$$.

Let $$h(x_j) = x_{e_j}$$, then $$g^{-1}(x_{j+1}) =x_j =h(x_{e_j})= hgh(x_{j+1})=hg(x_{e_{j+1}})=h(x_{e_{j+1}+1})$$ so $$x_{e_{j+1}+1} = x_{e_j}$$ and $$e_{j+1} = e_j-1$$ ie. $$e_j = e_0-j, h(x_j) = x_{e_0-j}$$.

If the roots are not all real we can take $$h$$ to be the complex conjugaison so that $$x_j$$ is real iff $$h(x_j) = x_j$$ iff $$2j = e_0 \bmod p$$.

• $D_{2p}$ is the group with $2p$ elements I mentioned – reuns Jun 2 at 3:58
• Of course I said $g$ has order $p$. – reuns Jun 2 at 4:03
• Nitpick about the second line. If $g$ is not a $p$-cycle it does not follow that its order is less than $p$, but it does follow that the order would be coprime to $p$. Like $S_{11}$ has elements of order $>11$. Of course, such permutations won't exist in $D_{11}$, making the point moot. Nice, of course! – Jyrki Lahtonen Jun 2 at 4:45