Is it true that $p\mid [\mathbb Q(\cos{\frac{\pi}{p^2}}):\mathbb Q]$?

Question

Is it true that $$p\mid [\mathbb Q(\cos{\frac{\pi}{p^2}}):\mathbb Q],\forall p\in \mathbb P,$$ where $[K:\mathbb Q]$ is the degree of a field extension ?

Answer 1

This is a slight variant of anon's answer, together with additional information about the minimal polynomial over ${\bf Q}$. We want to compute $[{\bf Q}(\beta):{\bf Q}]$ where $\beta=\cos(\frac{\pi}{m})$, with $m$ a positive integer. The moral of Galois theory is that

when studying questions about field extensions, it is often helpful to translate them into questions about groups.

To achieve such a translation, we need to fit the extension ${\bf Q}(\beta)/{\bf Q}$ inside a Galois extension. We do this by writing $\zeta:=\cos(\frac{\pi}m) + i\sin(\frac{\pi}m)$, and noting that $\zeta^{-1}=\cos(\frac{\pi}m) - i\sin(\frac{\pi}m)$, so that $\zeta+\zeta^{-1}=2\beta$. This shows that ${\bf Q}(\beta)$ is a subfield of ${\bf Q}(\zeta)$; in fact, it shows that $[{\bf Q}(\zeta):{\bf Q}(\beta)]\le 2$, since $\zeta$ is a root of the degree-$2$ polynomial $X^2+1-2\beta X$ whose coefficients are in ${\bf Q}(\beta)$. Now, the advantage of introducing $\zeta$ is that we know from Galois theory that the extension ${\bf Q}(\zeta)/{\bf Q}$ is Galois. Writing $G$ for its Galois group, there is an isomorphism $$ \rho:({\bf Z}/2m{\bf Z})^\times\rightarrow G$$ defined by $$ \rho(k):\zeta\mapsto \zeta^k.$$ Since ${\bf Q}(\beta)$ is an intermediate field between ${\bf Q}$ and ${\bf Q}(\zeta)$, and ${\bf Q}(\zeta)/{\bf Q}$ is Galois, it follows that ${\bf Q}(\zeta)/{\bf Q}(\beta)$ is Galois as well, say with Galois group $H$. Moreover, $[{\bf Q}(\zeta):{\bf Q}(\beta)]=\#H$, so we already know that $\#H\le 2$. But we can see that $\#H\ge 2$ as well, since $\rho(-1)$ interchanges $\zeta$ and $\zeta^{-1}$, so $\rho(-1)$ fixes $\zeta+\zeta^{-1}=2\beta$. Thus $2\le \#H\le 2$, so $\#H=2$. Finally, since $$[{\bf Q}(\zeta):{\bf Q}] = [{\bf Q}(\zeta):{\bf Q}(\beta)]\cdot [{\bf Q}(\beta):{\bf Q}],$$ we find that $$[{\bf Q}(\beta):{\bf Q}] = \frac{[{\bf Q}(\zeta):{\bf Q}]}{[{\bf Q}(\zeta):{\bf Q}(\beta)]} = \frac{\#G}{\#H} = \frac{\#({\bf Z}/2m{\bf Z})}{2} = \frac{\varphi(2m)}2.$$ Finally, the question asked about the case that $m=p^2$ where $p$ is prime. If $p$ is odd then $\varphi(2m)=p^2-p$, so that indeed $[{\bf Q}(\beta):{\bf Q}]=p\frac{p-1}2$ is divisible by $p$. If $p=2$ then $\varphi(2m)=2$, so $[{\bf Q}(\beta):{\bf Q}]=1$ (which can also be seen directly, since $\beta=\cos(\frac{\pi}2)=0$ is clearly a rational number).

We can go further and obtain information about the minimal polynomial of $\beta$ over ${\bf Q}$. I do this in part to explain the empirical observation made in the comments to this other question, according to which the minimal polynomial is related to a Chebyshev polynomial. By Galois theory, we know that the roots of the minimal polynomial of $\beta$ over ${\bf Q}$ are the elements $\sigma(\beta)$, where $\sigma$ ranges over a set of representatives for the left cosets of $H$ in $G$. Since $H$ is generated by $\rho(-1)$, this means $\sigma$ ranges over the set $\rho(S)$ where $S$ is a subset of $({\bf Z}/2m{\bf Z})^\times$ which contains precisely one element of $\{k,-k\}$ for every $k\in ({\bf Z}/2m{\bf Z})^\times$. One such $S$ is the set of elements of $({\bf Z}/2m{\bf Z})^\times$ corresponding to integers between $1$ and $m$ which are coprime to $2m$. Thus, the roots of the minimal polynomial of $\beta$ over ${\bf Q}$ (which are the same thing as the conjugates of $\beta$ over ${\bf Q}$) are precisely the elements $(\rho(k))(\beta)$, where $1\le k\le m$ and $\gcd(k,2m)=1$. Since $\rho(k)$ maps $\zeta\mapsto \zeta^k$, it also maps $\zeta+\zeta^{-1}\mapsto \zeta^k+\zeta^{-k}$. Since $\beta=\frac12(\zeta+\zeta^{-1})$, it follows that $\rho(k)$ maps $\beta$ to $\frac12(\zeta^k+\zeta^{-k})$, which equals $\cos \frac{\pi k}m$. Thus, the minimal polynomial of $\beta$ is $$ \prod_{\substack{1\le k\le m \\ \gcd(k,2m)=1}} (X-\cos\frac{\pi k}m).$$ To write this polynomial in a simpler way, recall the Chebyshev polynomial (of the first kind) $T_n(X)$, which is defined by the functional equation $$ T_n(\cos\theta)=\cos(n\theta).$$ Here $T_n(X)$ is a degree-$n$ polynomial with integer coefficients. The functional equation implies that $$ T_m(\cos\frac{\pi k}m) = \cos(\pi k),$$ which equals $-1$ when $k$ is odd. Thus, $T_m(X)$ is a degree-$m$ polynomial with integer coefficients whose roots include all the numbers $\cos(\frac{\pi k}m)$ where $1\le k\le m$ and $k$ is odd. In particular, $T_m(\beta)=-1$, so $T_m(X)+1$ is a multiple of the minimal polynomial of $\beta$ over ${\bf Q}$. We already computed the degree of this minimal polynomial, namely $\varphi(2m)/2$. So we need to identify a factor of $T_m(X)+1$ having this degree (and having $\beta$ as a root). First note that $T_m(X)+1$ has repeated roots whenever $m>1$. To see this, note that the functional equation defining $T_m(X)$ implies that $$ T_m\left(\frac{Y+Y^{-1}}2\right) = \frac{Y^m+Y^{-m}}2 $$ is true whenever $Y$ is a complex number of absolute value $1$. But both sides of this equation are rational functions in $Y$, so their difference is a rational function in $Y$ which vanishes for infinitely many values of $Y$. Hence the numerator of this difference is a polynomial in $Y$ which has infinitely many roots, so this numerator must be zero, whence the displayed equation is an equality of rational functions. In order to factor $T_m(X)+1$, write $X=(Y+Y^{-1})/2$ so that $$ T_m(X)+1 = T_m\left(\frac{Y+Y^{-1}}2\right) +1= \frac{Y^m+Y^{-m}}2+1 = \frac12(Y^{m/2}+Y^{-m/2})^2. $$ If $m$ is even then we compute $$ T_m(X)+1 = 2\left(\frac{Y^{m/2}+Y^{-m/2}}2\right)^2 = 2\left(T_{m/2}\left(\frac{Y+Y^{-1}}2\right)\right)^2 = 2(T_{m/2}(X))^2, $$ so that the minimal polynomial of $\beta$ over ${\bf Q}$ divides $T_{m/2}(X)$. If $m$ is odd then we have to work harder to obtain this sort of conclusion, since $Y^{m/2}+Y^{-m/2}$ is not a function of $X$. In this case, write $$ T_m(X)+1 = \frac12(Y^{1/2}+Y^{-1/2})^2\cdot\left(\frac{Y^{m/2}+Y^{-m/2}}{Y^{1/2}+Y^{-1/2}}\right)^2.$$ The first factor is nice: $$\frac12(Y^{1/2}+Y^{-1/2})^2 = \frac12(Y+2+Y^{-1}) = X+1.$$ To make the second factor nicer, multiply numerator and denominator by $Y^{1/2}-Y^{-1/2}$ to get $$\frac{Y^{m/2}+Y^{-m/2}}{Y^{1/2}+Y^{-1/2}} = \frac{Y^{(m+1)/2}-Y^{(m-1)/2}+Y^{(1-m)/2}-Y^{(-1-m)/2}}{Y-Y^{-1}}.$$ Now recall that the Chebyshev polynomial of the second kind, $U_n(X)$, is a degree-$n$ polynomial with integer coefficients which satisfies $$ U_n(\cos\theta) = \frac{\sin((n+1)\theta)}{\sin\theta}.$$ As above, this equation implies that $$ U_n\left(\frac{Y+Y^{-1}}2\right) = \frac{Y^{n+1}-Y^{-n-1}}{Y-Y^{-1}}.$$ Thus, when $m$ is odd and $X=(Y+Y^{-1})/2$, we have $$T_m(X)+1 = (X+1)\cdot \left(U_{(m-1)/2}(X) - U_{(m-3)/2}(X)\right)^2.$$ Since $\beta\ne -1$ when $m>1$, it follows that if $m$ odd and $m>1$ then the minimal polynomial of $\beta$ over ${\bf Q}$ divides $U_{(m-1)/2}(X)-U_{(m-3)/2}(X)$. Finally, if $m$ is an odd prime then the minimal polynomial of $\beta$ has degree $\varphi(2m)/2=(m-1)/2$, so it must be a constant times $U_{(m-1)/2}(X)-U_{(m-3)/2}(X)$. Analogous expressions for other values of $m$ are more complicated.

In the earlier question, it was observed empirically that, if $\zeta$ is a primitive $13$-th root of unity, then the minimal polynomial of $\zeta+\zeta^{-1}$ over ${\bf Q}$ is a polynomial $f(X)$ which satisfies $$T_{13}\left(\frac{X}2\right)-1=\frac{X-2}2\cdot f(X)^2.$$ The above argument explains this observation, and shows how it generalizes to other roots of unity. For, if we write $\alpha:=\zeta+\zeta^{-1}$ where $\zeta$ is a primitive $m$-th root of unity, then $$ T_m\left(\frac{\alpha}2\right) = \frac{\zeta^m+\zeta^{-m}}2 = 1.$$ Thus $\alpha$ is a root of $T_m(X/2)-1$, so the minimal polynomial of $\alpha$ over ${\bf Q}$ is a factor of $T_m(X/2)-1$. Now write $X=Z+Z^{-1}$, so that $$ T_m\left(\frac{X}2\right)-1 = \frac{Z^m+Z^{-m}}2-1 = \frac12(Z^{m/2}-Z^{-m/2})^2.$$ If $m=2n$ is even then $\alpha=\cos(\pi k/n)$ for some integer $k$ which is coprime to $2n$, and we analyzed this case in the previous paragraph. If $m$ is odd then $$ \frac{Z^{m/2}-Z^{-m/2}}{Z^{1/2}-Z^{-1/2}} = \frac{(Z^{m/2}-Z^{-m/2})(Z^{1/2}+Z^{-1/2})}{(Z^{1/2}-Z^{-1/2})(Z^{1/2}+Z^{-1/2})} = \frac{Z^{(m+1)/2}+Z^{(m-1)/2}-Z^{(1-m)/2}-Z^{-1-m)/2}}{Z-Z^{-1}} = U_{(m-1)/2}\left(\frac{X}2\right) + U_{(m-3)/2}\left(\frac{X}2\right),$$ so that $$T_m\left(\frac{X}2\right)-1 = \frac12(Z^{1/2}-Z^{-1/2})^2\cdot\left(U_{(m-1)/2}\left(\frac{X}2\right) + U_{(m-3)/2}\left(\frac{X}2\right)\right)^2 = \frac12(X-2)\cdot\left(U_{(m-1)/2}\left(\frac{X}2\right) + U_{(m-3)/2}\left(\frac{X}2\right)\right)^2.$$ If $m>1$ then $\alpha\ne 2$, so if $m$ is odd and $m>1$ then the minimal polynomial of $\alpha$ over ${\bf Q}$ divides $$U_{(m-1)/2}\left(\frac{X}2\right) + U_{(m-3)/2}\left(\frac{X}2\right).$$ Finally, if $m$ is an odd prime then the first paragraph of this answer shows that the minimal polynomial of $\alpha$ over ${\bf Q}$ has degree $\varphi(m)/2=(m-1)/2$, which equals the degree of the polynomial displayed above, so the minimal polynomial is a constant times the displayed polynomial. One can easily check that the displayed polynomial is monic, so it equals the minimal polynomial of $\alpha$ over ${\bf Q}$. Thus the empirically observed factorization for $m=13$ generalizes at once to any odd prime, and is a special case of a more general result valid for all odd $m>1$.

Answer 2

To quote my previous answer on the Galois machinery behind all of this:

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Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then

$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,

$\quad \rm(x-\beta)$ all coprime, $\rm \beta\in G\alpha$, implies $\rm f(x):=\prod\limits_{\beta\in G\alpha}(x-\beta)\mid m$ in $\rm L[x]$,

$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,

$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.

Therefore the zeros of $\rm\alpha$'s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.

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Now, $\alpha=2\cos(2\pi\frac{n}{m})=\zeta^n+\zeta^{-n}$ where $\zeta=e^{2\pi i/m}$ is a primitive $m$th root of unity (we assume $n/m$ is reduced). As ${\rm Gal}({\bf Q}(\zeta)/{\bf Q})\cong({\bf Z}/m{\bf Z})^\times=U(m)$, the Galois conjugates of $\alpha$ are $\zeta^\sigma+\zeta^{-\sigma}$ as $[\sigma]$ ranges over $U(m)$. For each $[\sigma]\in U(m)$, $[\sigma]\alpha=[-\sigma]\alpha$ by symmetry. Finally, $\cos$ is injective on the interval $[0,\pi]$ so $\{\cos(2\pi r/m):0\le r\le m/2,[r]\in U(m)\}$ is a complete set of conjugates without any repetitions. As $[K(\alpha):K]=\deg{\rm minpoly}_{\alpha,K}$, we therefore have

$$[{\bf Q}(\cos2\pi\frac{n}{m}):{\bf Q}]=\begin{cases}\varphi(m)/2 & m>2 \\ 1 & m=2 \end{cases} $$

since $m/2\in{\bf N}$ and $(m/2,m)=1$ iff $m=2$. Your case corresponds to $n=1$, $m=2p^2$.

More information: Galois theory and Cyclotomic fields.