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A special case is the following:

Pick:

An integer $n$ that is a square;

$$H =F^{*} D F$$

a matrix with $n$ lines and $n$ columns

where

$D$ is a diagonal square matrix with $n$ lines and with integer coefficients $F$ is the Fourier matrix, with $n$ lines defined by

$$F = (1/\sqrt{n}) (s^{(i-1)(j-1)})$$

where

$$s = e^{-2 \pi i/n}$$

$*$ means conjugate-transpose.

You get

$H$ is hermitian with entries algebraic integers.

$H$ is also a circulant matrix.

Pick now:

$$U =F^{*}$$

so that

$$Q(H) \subseteq Q(U) = Q(s)$$

while

$$Q(U,D) = Q(F^{*},D) = Q(s).$$

Observe that $$Q(s)$$ is the classic extension of $Q$ containing the $n$-th roots of unity so that it has degree

$$\varphi(n)$$

over $Q$, where $\varphi$ is the Euler totient's function.

Thus,

The extension $Q(U,D)$ over $Q(H)$ has degree $d$ bounded above by $\varphi(n)$\varphi(n).$Observe that this degree$d$is substantially slower than$n !$since $$d \leq \varphi(n) < n.$$ 3 added 190 characters in body A special case is the following: Pick: An integer$n$that is a square; $$H =F^{*} D F$$ where$D$is a diagonal matrix with integer coefficients$F$is the Fourier matrix defined by $$F = (1/\sqrt{n}) (s^{(i-1)(j-1)})$$ where $$s = e^{-2 \pi i/n}$$$*$means conjugate-transpose. You get$H$is hermitian with entries algebraic integers.$H$is also a circulant matrix. Pick now: $$U =F^{*}$$ so that $$Q(H) \subseteq Q(U) = Q(s)$$ while $$Q(U,D) = Q(F^{*},D) = Q(s).$$ Observe that $$Q(s)$$ is the classic extension of$Q$containing the$n$-th roots of unity so that it has degree $$\varphi(n)$$ over$Q$, where$\varphi$is the Euler totient's function. Thus, The extension$Q(U,D)$over$Q(H)$has degree$d$bounded above by$\varphi(n)$this degree is substantially slower than$n !$since $$d \leq \varphi(n) < n$$ 2 deleted 69 characters in body Actually you can get the inclusion in A special case is the other direction !following: Pick: An integer$n$that is a square; $$H =F^{*} D F$$ where$D$is a diagonal matrix with integer coefficients$F$is the Fourier matrix defined by $$F = (1/\sqrt{n}) (s^{(i-1)(j-1)})$$ where $$s = e^{-2 \pi i/n}$$$*$means conjugate-transpose. You get$H$is hermitian with entries algebraic integers.$H$is also a circulant matrix. Pick now: $$U =F^{*}$$ so that $$Q(H) \subseteq Q(s)$$ while $$Q(U,D) = Q(F^{*},D) = Q(s)$$ so that $$Q(U,D) \subseteq Q(H) Q(s).$$ Observe that $$Q(s)$$ is the classic extension of$Q$containing the$n$-th roots of unity so that it has degree $$\varphi(n)$$ over$Q$, where$\varphi\$ is the Euler totient's function.

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