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The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension:

Use the following skew-symmetric matrix, whose columns reduce modulo three to four of the specified nonzero tetracode words:Set $B_{+}=3I+S$, $B_{-}=-3I+S$, and define the full Euclidean lattices $L_{\pm }^0=B_{\pm }{\mathbb{Z}}^4$. Their determinants have absolute value $144$, sinceFor arbitrary positive real numbers $a_1,\dots ,a_4$, put $D=\mathrm{diag}(\sqrt{a_1},\dots ,\sqrt{a_4})$ and $L_{\pm }=D{L}_{\pm }^0$. This is the tetracode length-isospectral lattice construction. We prove equality of the length multisets by an explicit bijection, not just equality of the shortest lengths.

Modulo three, both lattices have image $\mathrm{im}S=T$. The matrix $S$ has rank two over ${\mathbb{F}}_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\mathrm{im}S=T$. Each nonzero tetracode word has exactly one zero coordinate. Every $v\in L_{\pm }^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_{\pm }$ is odd.

For $v\in L_{+}^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_i{e}_i$. The coordinate reflection bijection for tetracode lattices will show that $w\in L_{-}^0$. First, $w\equiv v(\mathrm{mod}3)$, so $Sw\equiv 0(\mathrm{mod}3)$. Second, positive-lattice membership implies $Sv\equiv 3v\equiv -v(\mathrm{mod}4)$. HenceThe expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_i{S}_{ji}$ even, by the common parity and $S_{ji}=\pm 1$. Thus $(I-S)w\equiv 0(\mathrm{mod}4)$. Together the congruences modulo three and four say $(-3I-S)w\equiv 0(\mathrm{mod}12)$, which is precisely $B_{-}^{-1}w\in {\mathbb{Z}}^4$.

Conversely, if $v\in L_{-}^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v(\mathrm{mod}4)$. ThusIts $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_{+}^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an involution between the two lattices. It preserves every $|v_j|$ and therefore preserves $\Vert Dv{\Vert }^2={\sum }_j{a}_j{v}_j^2$. The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.

To prove a genuine failure of isometry, chooseThese four numbers are linearly independent over $\mathbb{Q}$: changing the signs of the independent square roots in the multiquadratic field $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ isolates each coefficient of a rational relation. If an orthogonal transformation $O$ maps $L_{+}$ to $L_{-}$, thenand $P^T\mathrm{diag}(a_j)P=\mathrm{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2={\delta }_{ji}$. Consequently $P$ is diagonal with entries $\pm 1$.

Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero tetracode words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_{+}^0=L_{-}^0$. ButThis contradiction proves $L_{+}$ and $L_{-}$ are not isometric. Allowing affine isometries changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.

For the flat tori, take period lattices equal to the dual lattices:The standard spectrum of a flat torus lemma states that its characters of a real torus are indexed by the dual of its period lattice. Here they are $e^{2\pi i⟨v,x⟩}$ for $v\in L_{\pm }$, with Laplacian eigenvalue $4{\pi }^2\Vert v{\Vert }^2$. These characters give a complete orthogonal basis, so our length-preserving bijection proves that $X_{+}$ and $X_{-}$ are isospectral manifolds. The second standard lemma is that an isometry of flat tori lifts to an affine isometry of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their dual lattices to one another. This would give the impossible isometry $L_{+}\cong L_{-}$. Hence $X_{+}$ and $X_{-}$ are isospectral non-isometric flat four-dimensional tori.