Examples in number theory

Gaussian integers. We consider the field extension K/k := ℚ(i)/ℚ. This extension is Galois with Galois group {1, σ}, satisfying σ^{2} = 1, where σ is complex conjugation (σ(i) = −i). A basis of K over k is given by 1, i; the trace is Tr(x + iy) = 2x and the norm is N(x + iy) = x^{2} + y^{2}; the degree of the extension is 2. It is straightforward to check that the ring 𝔬_{K} of integers of K is ℤ[i]. This ring is easily seen to be principal; in fact, it is even euclidian. The roots are unity are {1, i, −1, −i}, and the units are {1, i, −1, −i}.

The (global) different is easily seen to be 𝔇 = (2) = (1 + i)^{2}, and the discriminant is −4.

We now consider a few particular places.

1. The prime 2. Let 𝔭 = (2) and v be the corresponding place. The only prime of ℤ[i] lying above 𝔭 is 𝔓 = (1+i) = (1−i).

Let w be the corresponding place. We have 𝔭𝔬_{K} = 𝔓^{2}: in other words, the prime 2 ramifies in ℤ[i], and the ramification index is e = 2; the prime 𝔓 is totally and strongly ramified. The residue field is ℤ[i]/𝔓 = 𝔽_{2}, which is equal to ℤ/𝔭: in other words, the modular degree is f = 1. The local degree is ef = 2, which means that the local field K_{w} = ℚ(i)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{2}. Indeed, ℚ(i)_{w} = ℚ_{2}(i). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + iy) = 2x and N_{w}(x + iy) = x^{2} + y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{2}/𝔽_{2} is {1}. The inertia group is therefore G = {1, σ}. The decomposition field is ℚ, and the inertia field is also ℚ. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = 𝔓^{2}, witnessing again the fact that the prime in question is strongly ramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{2}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = \frac{1}{2}, |1 + i|_{w} = \frac{1}{\sqrt{2}}, |i|_{w} = 1, and ∥2∥_{w} = \frac{1}{4}, ∥1 + i∥_{w} = \frac{1}{2}, ∥i∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

2. The prime 3. Let 𝔭 = (3) and v be the corresponding place. The only prime of ℤ[i] lying above 𝔭 is 𝔓 = (3). Let w

be the corresponding place. We have 𝔭𝔬_{K} = 𝔓: in other words, the prime 3 does not ramify in ℤ[i], that is, the ramification index is e = 1. The residue field is ℤ[i]/𝔓 = 𝔽_{9}, which is quadratic over ℤ/𝔭: in other words, the modular degree is f = 2. The local degree is ef = 2, which means that the local field K_{w} = ℚ(i)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{3}. Indeed, ℚ(i)_{w} = ℚ_{3}(i). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + iy) = 2x and N_{w}(x + iy) = x^{2} + y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{9}/𝔽_{3} is {1, σ̄}. The inertia group is therefore {1}. The decomposition field is ℚ, and the inertia field is ℚ(i). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing |·|_{w} for the absolute value on K_{w} which extends |·|_{v} = |·|_{3}, and ∥·∥_{w} for the normalized absolute value on K_{w} (the module), we have |3|_{w} = \frac{1}{3}, |i|_{w} = 1, and ∥3∥_{w} = \frac{1}{9}, ∥i∥_{w} = 1; in fact, ∥·∥_{w} = |·|_{w}^{2} because 2 is the local degree.

3. The prime 5. Let 𝔭 = (5) and v be the corresponding place. There are two primes of ℤ[i] lying above 𝔭, namely

𝔓 = (1 + 2i), and 𝔓′ = (1 − 2i). Let w, w′ be the corresponding places. We have 𝔭𝔬_{K} = 𝔓𝔓′: in other words, the prime 5 splits in ℤ[i]. Since neither prime ramifies, the two ramification indices are e = e′ = 1. The residue fields are ℤ[i]/𝔓 = 𝔽_{5} and ℤ[i]/𝔓′ = 𝔽_{5}, which are both equal to ℤ/𝔭: in other words, the modular degrees are f = f′ = 1. Note that though the two residue fields are equal, the canonical maps from ℤ[i] to one and the other are not equal: in ℤ[i]/𝔓, we have i = 2, whereas in ℤ[i]/𝔓′ we have i = 3. The equality of the ramification indices and the modular degrees was necessary, since the extension K/k is Galois, so the Galois group operates transitively on the primes lying above 𝔭; in this case, σ𝔓 = 𝔓′ and σ𝔓′ = 𝔓. The local degrees are ef = e′f′ = 1, which means that the local fields K_{w} = ℚ_{w} and K_{w′} = ℚ_{w′} are equal to k_{v} = ℚ_{v} = ℚ_{5}. Indeed, the polynomial X^{2} + 1 is split over ℚ_{5} as X^{2} + 1 = (X − ...212)(X − ...233). Note that though the two local fields are equal, the canonical embeddings from ℤ[i] to one and the other are not equal. Both the local trace and the local norm are the identity on K_{w} = K_{w′} = k_{v}; however, they are not equal on K (since the embedding of K in K_{w} is not equal to that of K in K_{w′}). The decomposition group G_{𝔓} of the prime 𝔓 is G = {1}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{5}/𝔽_{5} is {1}. The inertia group is therefore {1}. The decomposition field is ℚ(i), and the inertia field is also ℚ(i). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{5}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |5|_{w} = \frac{1}{5}, |1 + 2i|_{w} = \frac{1}{5}, |1 − 2i|_{w} = 1, |i|_{w} = 1, and ∥5∥_{w} = \frac{1}{5}, ∥1 + 2i∥_{w} = \frac{1}{5}, ∥1 − 2i∥_{w} = 1, ∥i∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree. (Of course, all the final discussion applies, mutatis mutandis, to w′ and 𝔓′ instead of w and 𝔓).

4. The infinite place. Let v = ∞ be the infinite place of ℚ (it is real). The only place of ℚ(i) lying above v is a complex

place w. The local degree is 2, which means that the local field K_{w} = ℚ(i)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℝ. Indeed, ℚ(i)_{w} = ℂ. The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + iy) = 2x and N_{w}(x + iy) = x^{2} + y^{2}. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = 2, |1 + i|_{w} = \sqrt{2}, |i|_{w} = 1, and ∥2∥_{w} = 4, ∥1 + i∥_{w} = 2, ∥i∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

Witnessing the product formula, we have, e.g.:

\begin{matrix}∥i∥_{1+i} × ∥i∥_{3} × ∥i∥_{1+2i} × ∥i∥_{1−2i} × ⋯ × ∥i∥_{∞} \\ = & 1 × 1 × 1 × 1 × ⋯ × 1 = 1 \end{matrix}

\begin{matrix}∥2∥_{1+i} × ∥2∥_{3} × ∥2∥_{1+2i} × ∥2∥_{1−2i} × ⋯ × ∥2∥_{∞} \\ = & \frac{1}{4} × 1 × 1 × 1 × ⋯ × 4 = 1 \end{matrix}

\begin{matrix}∥1 + i∥_{1+i} × ∥1 + i∥_{3} × ∥1 + i∥_{1+2i} × ∥1 + i∥_{1−2i} × ⋯ × ∥1 + i∥_{∞} \\ = & \frac{1}{2} × 1 × 1 × 1 × ⋯ × 2 = 1 \end{matrix}

\begin{matrix}∥3∥_{1+i} × ∥3∥_{3} × ∥3∥_{1+2i} × ∥3∥_{1−2i} × ⋯ × ∥3∥_{∞} \\ = & 1 × \frac{1}{9} × 1 × 1 × ⋯ × 9 = 1 \end{matrix}

\begin{matrix}∥5∥_{1+i} × ∥5∥_{3} × ∥5∥_{1+2i} × ∥5∥_{1−2i} × ⋯ × ∥5∥_{∞} \\ = & 1 × 1 × \frac{1}{5} × \frac{1}{5} × ⋯ × 25 = 1 \end{matrix}

\begin{matrix}∥1 + 2i∥_{1+i} × ∥1 + 2i∥_{3} × ∥1 + 2i∥_{1+2i} × ∥1 + 2i∥_{1−2i} × ⋯ × ∥1 + 2i∥_{∞} \\ = & 1 × 1 × \frac{1}{5} × 1 × ⋯ × 5 = 1 \end{matrix}

Square root of 2. We consider the field extension K/k := ℚ(\sqrt{2})/ℚ. This extension is Galois with Galois group \{1, σ\}, satisfying σ^{2} = 1, where σ is defined by σ(\sqrt{2}) = −\sqrt{2}. A basis of K over k is given by 1, \sqrt{2}; the trace is Tr(x + y\sqrt{2}) = 2x and the norm is N(x + y\sqrt{2}) = x^{2} − 2y^{2}; the degree of the extension is 2. It is straightforward to check that the ring 𝔬_{K} of integers of K is ℤ[\sqrt{2}]. This ring is easily seen to be principal; in fact, it is even euclidian. The roots are unity are {1, −1}, and the units are generated by the fundamental unit \sqrt{2} − 1.

The (global) different is easily seen to be 𝔇 = (2\sqrt{2}) = (\sqrt{2})^{3}, and the discriminant is 8.

We now consider a few particular places.

1. The prime 2. Let 𝔭 = (2) and v be the corresponding place. The only prime of ℤ[\sqrt{2}] lying above 𝔭 is 𝔓 = (\sqrt{2}).

Let w be the corresponding place. We have 𝔭𝔬_{K} = 𝔓^{2}: in other words, the prime 2 ramifies in ℤ[\sqrt{2}], and the ramification index is e = 2; the prime 𝔓 is totally and strongly ramified. The residue field is ℤ[\sqrt{2}]/𝔓 = 𝔽_{2}, which is equal to ℤ/𝔭: in other words, the modular degree is f = 1. The local degree is ef = 2, which means that the local field K_{w} = ℚ(\sqrt{2})_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{2}. Indeed, ℚ(\sqrt{2})_{w} = ℚ_{2}(\sqrt{2}). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + y\sqrt{2}) = 2x and N_{w}(x + y\sqrt{2}) = x^{2} − 2y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{2}/𝔽_{2} is {1}. The inertia group is therefore G = {1, σ}. The decomposition field is ℚ, and the inertia field is also ℚ. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = 𝔓^{3}, witnessing again the fact that the prime in question is strongly ramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{2}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = \frac{1}{2}, |\sqrt{2}|_{w} = \frac{1}{\sqrt{2}}, and ∥2∥_{w} = \frac{1}{4}, ∥\sqrt{2}∥_{w} = \frac{1}{2}; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

2. The prime 3. Let 𝔭 = (3) and v be the corresponding place. The only prime of ℤ[\sqrt{2}] lying above 𝔭 is 𝔓 = (3). Let w be

the corresponding place. We have 𝔭𝔬_{K} = 𝔓: in other words, the prime 3 does not ramify in ℤ[\sqrt{2}], that is, the ramification index is e = 1. The residue field is ℤ[\sqrt{2}]/𝔓 = 𝔽_{9}, which is quadratic over ℤ/𝔭: in other words, the modular degree is f = 2. The local degree is ef = 2, which means that the local field K_{w} = ℚ(\sqrt{2})_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{3}. Indeed, ℚ(\sqrt{2})_{w} = ℚ_{3}(\sqrt{2}). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + y\sqrt{2}) = 2x and N_{w}(x + y\sqrt{2}) = x^{2} − 2y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{9}/𝔽_{3} is {1, σ̄}. The inertia group is therefore \{1\}. The decomposition field is ℚ, and the inertia field is ℚ(\sqrt{2}). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{3}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |3|_{w} = \frac{1}{3}, |\sqrt{2}|_{w} = 1, and ∥3∥_{w} = \frac{1}{9}, ∥\sqrt{2}∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

3. The prime 7. Let 𝔭 = (7) and v be the corresponding place. There are two primes of ℤ[\sqrt{2}] lying above 𝔭, namely

𝔓 = (3 + \sqrt{2}), and 𝔓′ = (3 − \sqrt{2}). Let w, w′ be the corresponding places. We have 𝔭𝔬_{K} = 𝔓𝔓′: in other words, the prime 7 splits in ℤ[\sqrt{2}]. Since neither prime ramifies, the two ramification indices are e = e′ = 1. The residue fields are ℤ[\sqrt{2}]/𝔓 = 𝔽_{7} and ℤ[\sqrt{2}]/𝔓′ = 𝔽_{7}, which are both equal to ℤ/𝔭: in other words, the modular degrees are f = f′ = 1. Note that though the two residue fields are equal, the canonical maps from ℤ[\sqrt{2}] to one and the other are not equal: in ℤ[\sqrt{2}]/𝔓, we have \sqrt{2} = 4, whereas in ℤ[\sqrt{2}]/𝔓′ we have \sqrt{2} = 3. The equality of the ramification indices and the modular degrees was necessary, since the extension K/k is Galois, so the Galois group operates transitively on the primes lying above 𝔭; in this case, σ𝔓 = 𝔓′ and σ𝔓′ = 𝔓. The local degrees are ef = e′f′ = 1, which means that the local fields K_{w} = ℚ_{w} and K_{w′} = ℚ_{w′} are equal to k_{v} = ℚ_{v} = ℚ_{7}. Indeed, the polynomial X^{2} − 2 is split over ℚ_{7} as X^{2} − 2 = (X − ...454)(X − ...213). Note that though the two local fields are equal, the canonical embeddings from ℤ[\sqrt{2}] to one and the other are not equal. Both the local trace and the local norm are the identity on K_{w} = K_{w′} = k_{v}; however, they are not equal on K (since the embedding of K in K_{w} is not equal to that of K in K_{w′}). The decomposition group G_{𝔓} of the prime 𝔓 is G = {1}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{7}/𝔽_{7} is \{1\}. The inertia group is therefore \{1\}. The decomposition field is ℚ(\sqrt{2}), and the inertia field is also ℚ(\sqrt{2}). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing |·|_{w} for the absolute value on K_{w} which extends |·|_{v} = |·|_{7}, and ∥·∥_{w} for the normalized absolute value on K_{w} (the module), we have |7|_{w} = \frac{1}{7}, |3 + \sqrt{2}|_{w} = \frac{1}{7}, |3 − \sqrt{2}|_{w} = 1, |\sqrt{2}|_{w} = 1, and ∥7∥_{w} = \frac{1}{7}, ∥3 + \sqrt{2}∥_{w} = \frac{1}{7}, ∥3 − \sqrt{2}∥_{w} = 1, ∥\sqrt{2}∥_{w} = 1; in fact, ∥·∥_{w} = |·|_{w} because 1 is the local degree. (Of course, all the final discussion applies, mutatis mutandis, to w′ and 𝔓′ instead of w and 𝔓).

4. The infinite places. Let v = ∞ be the infinite place of ℚ (it is real). There are two infinite places, w, w′, lying above v, both

real: let w be the one of the two which is given by the canonical embedding of ℚ(\sqrt{2}) into ℝ, whereas w′ is given by the embedding σ. The local degree is 1, which means that the local fields K_{w} = ℚ(\sqrt{2})_{w} and K_{w′} = ℚ(\sqrt{2})_{w′} are both equal to k_{v} = ℚ_{v} = ℝ. Indeed, the polynomial X^{2} − 2 is split over ℝ as X^{2} −2 = (X −1.41...)(X +1.41...). Note that though the two local fields are equal, the canonical embeddings from ℤ[\sqrt{2}] to one and the other are not equal. Both the local trace and the local norm are the identity on K_{w} = K_{w′} = k_{v}; however, they are not equal on K (since the embedding of K in K_{w} is not equal to that of K in K_{w′}). Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = 2, |\sqrt{2}|_{w} = \sqrt{2}, and ∥2∥_{w} = 2, ∥\sqrt{2}∥_{w} = \sqrt{2}; in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree. (Of course, all the final discussion applies, mutatis mutandis, to w′ instead of w).

Witnessing the product formula, we have, e.g.:

\begin{matrix}∥2∥\sqrt{2} × ∥2∥_{3} × ∥2∥_{3+\sqrt{2}} × ∥2∥_{3−\sqrt{2}} × ⋯ × ∥2∥_{∞} × ∥2∥_{∞′} \\ = & \frac{1}{4} × 1 × 1 × 1 × ⋯ × 2 × 2 = 1 \end{matrix}

\begin{matrix}∥\sqrt{2}∥\sqrt{2} × ∥\sqrt{2}∥_{3} × ∥\sqrt{2}∥_{3+\sqrt{2}} × ∥\sqrt{2}∥_{3−\sqrt{2}} × ⋯ × ∥\sqrt{2}∥_{∞} × ∥\sqrt{2}∥_{∞′} \\ = & \frac{1}{2} × 1 × 1 × 1 × ⋯ × \sqrt{2} × \sqrt{2} = 1 \end{matrix}

\begin{matrix}∥3∥\sqrt{2} × ∥3∥_{3} × ∥3∥_{3+\sqrt{2}} × ∥3∥_{3−\sqrt{2}} × ⋯ × ∥3∥_{∞} × ∥3∥_{∞′} \\ = & 1 × \frac{1}{9} × 1 × 1 × ⋯ × 3 × 3 = 1 \end{matrix}

\begin{matrix}∥7∥\sqrt{2} × ∥7∥_{3} × ∥7∥_{3+\sqrt{2}} × ∥7∥_{3−\sqrt{2}} × ⋯ × ∥7∥_{∞} × ∥7∥_{∞′} \\ = & 1 × 1 × \frac{1}{7} × \frac{1}{7} × ⋯ × 7 × 7 = 1 \end{matrix}

\begin{matrix}∥3 + \sqrt{2}∥\sqrt{2} × ∥3 + \sqrt{2}∥_{3} × ∥3 + \sqrt{2}∥_{3+\sqrt{2}} × ∥3 + \sqrt{2}∥_{3−\sqrt{2}} × ⋯ × ∥3 + \sqrt{2}∥_{∞} × ∥3 + \sqrt{2}∥_{∞′} \\ = & 1 × 1 × \frac{1}{7} × 1 × ⋯ × (3 + \sqrt{2}) × (3 − \sqrt{2}) = 1 \end{matrix}

Cube root of unity. We consider the field extension K/k := ℚ(ω)/ℚ, where ω = e^{2iπ/3}. This extension is Galois with Galois group {1, σ}, satisfying σ^{2} = 1, where σ is complex conjugation, σ(ω) = ω^{2} = −1 − ω. A basis of K over k is given by 1, ω; the trace is Tr(x + yω) = 2x − y and the norm is N(x + yω) = x^{2} − xy + y^{2}; the degree of the extension is 2. It is straightforward to check that the ring 𝔬_{K} of integers of K is ℤ[ω]. This ring is easily seen to be principal; in fact, it is even euclidian. The roots are unity are {1, ω, ω^{2}}, and the units are {1, ω, ω^{2}}.

The (global) different is easily seen to be 𝔇 = (1 − ω) = (1 + 2ω), and the discriminant is −3.

We now consider a few particular places.

1. The prime 2. Let 𝔭 = (2) and v be the corresponding place. The only prime of ℤ[ω] lying above 𝔭 is 𝔓 = (2). Let w

be the corresponding place. We have 𝔭𝔬_{K} = 𝔓: in other words, the prime 2 does not ramify in ℤ[ω], that is, the ramification index is e = 1. The residue field is ℤ[ω]/𝔓 = 𝔽_{4}, which is quadratic over ℤ/𝔭: in other words, the modular degree is f = 2. The local degree is ef = 2, which means that the local field K_{w} = ℚ(ω)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{2}. Indeed, ℚ(ω)_{w} = ℚ_{2}(ω) (where ω is a root over ℚ_{2} of the irreducible polynomial X^{2} + X + 1). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + yω) = 2x − y and N_{w}(x + yω) = x^{2} − xy + y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{4}/𝔽_{2} is {1, σ̄}. The inertia group is therefore \{1\}. The decomposition field is ℚ, and the inertia field is ℚ(\sqrt{2}). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{2}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = \frac{1}{2}, |ω|_{w} = 1, and ∥2∥_{w} = \frac{1}{4}, ∥ω∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

2. The prime 3. Let 𝔭 = (3) and v be the corresponding place. The only prime of ℤ[ω] lying above 𝔭 is 𝔓 = (1 − ω) =

(1 + 2ω). Let w be the corresponding place. We have 𝔭𝔬_{K} = 𝔓^{2}: in other words, the prime 3 ramifies in ℤ[ω], and the ramification index is e = 2; the prime 𝔓 is totally and tamely ramified. The residue field is ℤ[ω]/𝔓 = 𝔽_{3}, which is equal to ℤ/𝔭: in other words, the modular degree is f = 1. The local degree is ef = 2, which means that the local field K_{w} = ℚ(ω)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℚ_{3}. Indeed, ℚ(ω)_{w} = ℚ_{3}(ω) (where ω is a root over ℚ_{3} of the irreducible polynomial X^{2} + X + 1). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + yω) = 2x − y and N_{w}(x + yω) = x^{2} − xy + y^{2}. The decomposition group G_{𝔓} of the prime 𝔓 is G = {1, σ}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{3}/𝔽_{3} is {1}. The inertia group is therefore G = {1, σ}. The decomposition field is ℚ, and the inertia field is also ℚ. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = 𝔓, witnessing again the fact that the prime in question is tamely ramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{3}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |3|_{w} = \frac{1}{3}, |1 − ω|_{w} = \frac{1}{\sqrt{3}}, |ω|_{w} = 1, and ∥3∥_{w} = \frac{1}{9}, ∥1 − ω∥_{w} = \frac{1}{3}, ∥ω∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

3. The prime 7. Let 𝔭 = (7) and v be the corresponding place. There are two primes of ℤ[ω] lying above 𝔭, namely

𝔓 = (2 − ω), and 𝔓′ = (3 + ω). Let w, w′ be the corresponding places. We have 𝔭𝔬_{K} = 𝔓𝔓′: in other words, the prime 7 splits in ℤ[ω]. Since neither prime ramifies, the two ramification indices are e = e′ = 1. The residue fields are ℤ[ω]/𝔓 = 𝔽_{7} and ℤ[ω]/𝔓′ = 𝔽_{7}, which are both equal to ℤ/𝔭: in other words, the modular degrees are f = f′ = 1. Note that though the two residue fields are equal, the canonical maps from ℤ[ω] to one and the other are not equal: in ℤ[ω]/𝔓, we have ω = 2, whereas in ℤ[ω]/𝔓′ we have ω = 4. The equality of the ramification indices and the modular degrees was necessary, since the extension K/k is Galois, so the Galois group operates transitively on the primes lying above 𝔭; in this case, σ𝔓 = 𝔓′ and σ𝔓′ = 𝔓. The local degrees are ef = e′f′ = 1, which means that the local fields K_{w} = ℚ_{w} and K_{w′} = ℚ_{w′} are equal to k_{v} = ℚ_{v} = ℚ_{7}. Indeed, the polynomial X^{2} + X + 1 is split over ℚ_{7} as X^{2} + X + 1 = (X − ..642)(X − ...024). Note that though the two local fields are equal, the canonical embeddings from ℤ[ω] to one and the other are not equal. Both the local trace and the local norm are the identity on K_{w} = K_{w′} = k_{v}; however, they are not equal on K (since the embedding of K in K_{w} is not equal to that of K in K_{w′}). The decomposition group G_{𝔓} of the prime 𝔓 is G = {1}; the galois group Ḡ_{𝔓} of the residue field extension 𝔽_{7}/𝔽_{7} is {1}. The inertia group is therefore {1}. The decomposition field is ℚ(ω), and the inertia field is also ℚ(ω). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{7}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |7|_{w} = \frac{1}{7}, |2 − ω|_{w} = \frac{1}{7}, |3 + ω|_{w} = 1, |ω|_{w} = 1, and ∥7∥_{w} = \frac{1}{7}, ∥2 − ω∥_{w} = \frac{1}{7}, ∥3 + ω∥_{w} = 1, ∥ω∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree. (Of course, all the final discussion applies, mutatis mutandis, to w′ and 𝔓′ instead of w and 𝔓).

4. The infinite place. Let v = ∞ be the infinite place of ℚ (it is real). The only place of ℚ(ω) lying above v is a complex

place w. The local degree is 2, which means that the local field K_{w} = ℚ(ω)_{w} is of degree 2 over k_{v} = ℚ_{v} = ℝ. Indeed, ℚ(i)_{w} = ℂ. The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x+yω) = 2x−y and N_{w}(x + yω) = x^{2} − xy + y^{2}. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = 2, |ω|_{w} = 1, and ∥2∥_{w} = 4, ∥ω∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

Witnessing the product formula, we have, e.g.:

\begin{matrix}∥ω∥_{2} × ∥ω∥_{1−ω} × ∥ω∥_{2−ω} × ∥ω∥_{3+ω} × ⋯ × ∥ω∥_{∞} \\ = & 1 × 1 × 1 × 1 × ⋯ × 1 = 1 \end{matrix}

\begin{matrix}∥2∥_{2} × ∥2∥_{1−ω} × ∥2∥_{2−ω} × ∥2∥_{3+ω} × ⋯ × ∥2∥_{∞} \\ = & \frac{1}{4} × 1 × 1 × 1 × ⋯ × 4 = 1 \end{matrix}

\begin{matrix}∥3∥_{2} × ∥3∥_{1−ω} × ∥3∥_{2−ω} × ∥3∥_{3+ω} × ⋯ × ∥3∥_{∞} \\ = & 1 × \frac{1}{9} × 1 × 1 × ⋯ × 9 = 1 \end{matrix}

\begin{matrix}∥1 − ω∥_{2} × ∥1 − ω∥_{1−ω} × ∥1 − ω∥_{2−ω} × ∥1 − ω∥_{3+ω} × ⋯ × ∥1 − ω∥_{∞} \\ = & 1 × \frac{1}{3} × 1 × 1 × ⋯ × 3 = 1 \end{matrix}

\begin{matrix}∥7∥_{2} × ∥7∥_{1−ω} × ∥7∥_{2−ω} × ∥7∥_{3+ω} × ⋯ × ∥7∥_{∞} \\ = & 1 × 1 × \frac{1}{7} × \frac{1}{7} × ⋯ × 49 = 1 \end{matrix}

\begin{matrix}∥2 − ω∥_{2} × ∥2 − ω∥_{1−ω} × ∥2 − ω∥_{2−ω} × ∥2 − ω∥_{3+ω} × ⋯ × ∥2 − ω∥_{∞} \\ = & 1 × 1 × \frac{1}{7} × 1 × ⋯ × 7 = 1 \end{matrix}

Cube root of 2. We consider the field extension K/k := ℚ( ^{3} \sqrt{2})/ℚ. This extension is not Galois and the Galois group \{1\}. A basis of K over k is given by 1, ^{3} \sqrt{2}, ^{3} \sqrt{4}; the trace is Tr(x + y ^{3} \sqrt{2} + z ^{3} \sqrt{4}) = 3x and the norm is N(x + y ^{3} \sqrt{2} + z ^{3} \sqrt{4}) = x^{3} + 2y^{3} + 4z^{3} − 6xyz; the degree of the extension is 3. It is straightforward (but painful) to check that the ring 𝔬_{K} of integers of K is ℤ[ ^{3} \sqrt{2}]. The roots are unity are \{1, −1\}, and the units are generated by the fundamental unit ^{3} \sqrt{2} − 1.

The (global) different is easily seen to be 𝔇 = (3 ^{3} \sqrt{4}), and the discriminant is −108 = −2^{2} × 3^{3}.

We now consider a few particular places.

1. The prime 2. Let 𝔭 = (2) and v be the corresponding place. The only prime of ℤ[ ^{3} \sqrt{2}] lying above 𝔭 is 𝔓 = ( ^{3} \sqrt{2}).

Let w be the corresponding place. We have 𝔭𝔬_{K} = 𝔓^{3}: in other words, the prime 2 ramifies in ℤ[ ^{3} \sqrt{2}], and the ramification index is e = 3; the prime 𝔓 is totally and tamely ramified. The residue field is ℤ[ ^{3} \sqrt{2}]/𝔓 = 𝔽_{2}, which is equal to ℤ/𝔭: in other words, the modular degree is f = 1. The local degree is ef = 3, which means that the local field K_{w} = ℚ( ^{3} \sqrt{2})_{w} is of degree 3 over k_{v} = ℚ_{v} = ℚ_{2}. Indeed, ℚ( ^{3} \sqrt{2})_{w} = ℚ_{2}( ^{3} \sqrt{2}). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x+y ^{3} \sqrt{2}+z ^{3} \sqrt{4}) = 3x and N_{w}(x+y ^{3} \sqrt{2}+z ^{3} \sqrt{4}) = x^{3} +2y^{3} +4z^{3} −6xyz. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = 𝔓^{2}, witnessing again the fact that the prime in question is tamely ramified. Writing |·|_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{2}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |2|_{w} = \frac{1}{2}, | ^{3} \sqrt{2}|_{w} = \frac{1}{\sqrt[3]{2}}, and ∥2∥_{w} = \frac{1}{8}, ∥ ^{3} \sqrt{2}∥_{w} = \frac{1}{2}; in fact, ∥ · ∥_{w} = | · |_{w}^{3} because 3 is the local degree.

2. The prime 3. Let 𝔭 = (3) and v be the corresponding place. The only prime of ℤ[ ^{3} \sqrt{2}] lying above 𝔭 is 𝔓 = (1 + ^{3} \sqrt{2}).

Let w be the corresponding place. We have 𝔭𝔬_{K} = 𝔓^{3}: in other words, the prime 3 ramifies in ℤ[ ^{3} \sqrt{2}], and the ramification index is e = 3; the prime 𝔓 is totally and strongly ramified. The residue field is ℤ[ ^{3} \sqrt{2}]/𝔓 = 𝔽_{2}, which is equal to ℤ/𝔭: in other words, the modular degree is f = 1. The local degree is ef = 3, which means that the local field K_{w} = ℚ( ^{3} \sqrt{2})_{w} is of degree 3 over k_{v} = ℚ_{v} = ℚ_{3}. Indeed, ℚ( ^{3} \sqrt{2})_{w} = ℚ_{2}( ^{3} \sqrt{2}). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x+y ^{3} \sqrt{2}+z ^{3} \sqrt{4}) = 3x and N_{w}(x+y ^{3} \sqrt{2}+z ^{3} \sqrt{4}) = x^{3} +2y^{3} +4z^{3} −6xyz. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = 𝔓^{3}, witnessing again the fact that the prime in question is strongly ramified. Writing | · |_{w} for the absolute value on K_{w} which extends |·|_{v} = |·|_{2}, and ∥·∥_{w} for the normalized absolute value on K_{w} (the module), we have |3|_{w} = \frac{1}{3}, |1 + ^{3} \sqrt{2}|_{w} = \frac{1}{\sqrt[3]{3}}, and ∥2∥_{w} = \frac{1}{8}, ∥1 + ^{3} \sqrt{2}∥_{w} = \frac{1}{3}; in fact, ∥ · ∥_{w} = | · |_{w}^{3} because 3 is the local degree.

3. The prime 5. Let 𝔭 = (7) and v be the corresponding place. There are two primes of ℤ[ ^{3} \sqrt{2}] lying above 𝔭, namely

𝔓 = (1 + ^{3} \sqrt{4}), and 𝔓′ = (1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}). Let w, w′ be the corresponding places. We have 𝔭𝔬_{K} = 𝔓𝔓′: in other words, the prime 5 splits, but not completely, in ℤ[ ^{3} \sqrt{2}]. Since neither prime ramifies, the two ramification indices are e = e′ = 1. The residue fields are ℤ[ ^{3} \sqrt{2}]/𝔓 = 𝔽_{5} and ℤ[ ^{3} \sqrt{2}]/𝔓′ = 𝔽_{25}, which respectively equal to, and quadratic over, ℤ/𝔭: in other words, the modular degrees are f = 1 and f′ = 2. The local degrees are ef = 1 and e′f′ = 2, which means that the local fields K_{w} = ℚ_{w} and K_{w′} = ℚ_{w′} are respecively of degree 1 and 2 over k_{v} = ℚ_{v} = ℚ_{5}. Indeed, the polynomial X^{3} − 2 is decomposes over ℚ_{5} as X^{3} −2 = (X −...203)(X^{2} +...203X +...214); in fact, K_{w′} = ℚ_{5}(\sqrt{...243}) (because ...243 is the discriminant of the polynomial X^{2}+...203X+...214). Both the local trace and the local norm are the identity on K_{w} = k_{v}; on K_{w′}, the trace is Tr(x+y\sqrt{...243}) = 2x and the norm is N(x+y\sqrt{...243}) = x^{2} −...243y^{2}. The local differents (that is, the 𝔓- and 𝔓′-components of the different) are 𝔇_{𝔓} = 𝔇_{𝔓′} = (1), witnessing again the fact that the primes in question are unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{5}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), and | · |_{w′} and ∥ · ∥_{w′} for the corresponding objects for K_{w′}, we have |5|_{w} = \frac{1}{5}, |1 + ^{3} \sqrt{4}|_{w} = \frac{1}{5}, |1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}|_{w} = 1, and ∥5∥_{w} = \frac{1}{5}, ∥1 + ^{3} \sqrt{4}∥_{w} = \frac{1}{5}, ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{w} = 1, in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree; on the other hand, we have |5|_{w′} = \frac{1}{5}, |1 + ^{3} \sqrt{4}|_{w′} = 1, |1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}|_{w′} = \frac{1}{5}, and ∥5∥_{w′} = \frac{1}{25}, ∥1 + ^{3} \sqrt{4}∥_{w′} = 1, ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{w′} = \frac{1}{25}, in fact, ∥ · ∥_{w′} = | · |_{w′}^{2} because 2 is the local degree.

4. The prime 7. Let 𝔭 = (7) and v be the corresponding place. The only prime of ℤ[ ^{3} \sqrt{2}] lying above 𝔭 is 𝔓 = (7). Let w

be the corresponding place. We have 𝔭𝔬_{K} = 𝔓: in other words, the prime 7 does not ramify in ℤ[ ^{3} \sqrt{2}], that is, the ramification index is e = 1. The residue field is ℤ[ ^{3} \sqrt{2}]/𝔓 = 𝔽_{343}, which is cubic over ℤ/𝔭: in other words, the modular degree is f = 3. The local degree is ef = 3, which means that the local field K_{w} = ℚ( ^{3} \sqrt{2})_{w} is of degree 3 over k_{v} = ℚ_{v} = ℚ_{7}. Indeed, ℚ( ^{3} \sqrt{2})_{w} = ℚ_{7}( ^{3} \sqrt{2}). The local trace and norm extend the global trace and norm; in fact, Tr_{w}(x + y ^{3} \sqrt{2} + z ^{3} \sqrt{4}) = 3x and N_{w}(x + y ^{3} \sqrt{2} + z ^{3} \sqrt{4}) = x^{3} + 2y^{2} + 4z^{2} − 6xyz. The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |_{7}, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), we have |7|_{w} = \frac{1}{7}, | ^{3} \sqrt{2}|_{w} = 1, and ∥7∥_{w} = \frac{1}{343}, ∥ ^{3} \sqrt{2}∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w}^{3} because 3 is the local degree.

5. The prime 31.

Let 𝔭 = (31) and v be the corresponding place. There are three primes of ℤ[ ^{3} \sqrt{2}] lying above 𝔭, namely 𝔓 = (3+ ^{3} \sqrt{4}), 𝔓′ = (1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}), and 𝔓′′ = (1−2 ^{3} \sqrt{4}). Let w, w′, w′′ be the corresponding places. We have 𝔭𝔬_{K} = 𝔓𝔓′𝔓′′: in other words, the prime 31 splits completely in ℤ[ ^{3} \sqrt{2}]. Since none of the primes ramifies, the ramification indices are e = e′ = e′′ = 1. The residue fields are ℤ[ ^{3} \sqrt{2}]/𝔓 = 𝔽_{31}, ℤ[ ^{3} \sqrt{2}]/𝔓′ = 𝔽_{31}, and ℤ[ ^{3} \sqrt{2}]/𝔓′′ = 𝔽_{31}, which are all equal to ℤ/𝔭: in other words, the modular degrees are f = f′ = f′′ = 1. Note that though the two residue fields are equal, the canonical maps from ℤ[ ^{3} \sqrt{2}] to one and the other are not equal: in ℤ[ ^{3} \sqrt{2}]/𝔓, we have ^{3} \sqrt{2} = 20, whereas in ℤ[ ^{3} \sqrt{2}]/𝔓′ we have ^{3} \sqrt{2} = 7, and in ℤ[ ^{3} \sqrt{2}]/𝔓′′ we have ^{3} \sqrt{2} = 4. The local degrees are ef = e′f′ = e′′f′′ = 1, which means that the local fields K_{w} = ℚ_{w}, K_{w′} = ℚ_{w′} and K_{w′′} = ℚ_{w′′} are equal to k_{v} = ℚ_{v} = ℚ_{31}. Indeed, the polynomial X^{2} − 2 is split over ℚ_{31} as X^{2} − 2 = (X − ...8 : 20)(X − ...13 : 7)(X − ...9 : 4). Note that though the three local fields are equal, the canonical embeddings from ℤ[ ^{3} \sqrt{2}] to one and the others are not equal. Both the local trace and the local norm are the identity on K_{w} = K_{w′} = K_{w′′} = k_{v}; however, they are not equal on K (since the embedding of K in K_{w} is not equal to that of K in K_{w′}). The local different (that is, the 𝔓-component of the different) is 𝔇_{𝔓} = (1), witnessing again the fact that the prime in question is unramified. Writing | · |_{w} for the absolute value on K_{w} which extends |·|_{v} = |·|_{31}, and ∥·∥_{w} for the normalized absolute value on K_{w} (the module), we have |31|_{w} = \frac{1}{31}, |3 + ^{3} \sqrt{4}|_{w} = \frac{1}{31}, |1 − 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}|_{w} = 1, |1 − 2 ^{3} \sqrt{4}|_{w} = 1, and ∥31∥_{w} = \frac{1}{31}, ∥3 + ^{3} \sqrt{4}∥_{w} = \frac{1}{31}, ∥1 − 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{w} = 1, ∥1 − 2 ^{3} \sqrt{4}∥_{w} = 1; in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree. (Of course, all the final discussion applies, mutatis mutandis, to w′ and 𝔓′ or w′′ and 𝔓′′ instead of w and 𝔓).

6. The infinite places. Let v = ∞ be the infinite place of ℚ (it is real). There are two infinite places, w, w′, lying above v, one

real and one complex: let w be the real one, which is given by the canonical embedding of ℚ(\sqrt{2}) into ℝ, and let w′ be that which is given by the embedding ^{3} \sqrt{2} ↦ e^{2iπ/3} ^{3} \sqrt{2}. The local degrees of w, w′ are 1 and 2 respectively, which means that the local fields K_{w} = ℚ( ^{3} \sqrt{2})_{w} and K_{w′} = ℚ( ^{3} \sqrt{2})_{w′} are respectively equal to and quadratic over k_{v} = ℚ_{v} = ℝ; in fact, K_{w′} = ℂ. Indeed, the polynomial X^{2} − 2 decomposes over ℝ as X^{2} − 2 = (X − 1.25...)(X + 1.25... + 1.58...). Both the local trace and the local norm are the identity on K_{w} = k_{v}; on K_{w′}, they are the ordinary trace and norm of ℂ over ℝ. Writing | · |_{w} for the absolute value on K_{w} which extends | · |_{v} = | · |, and ∥ · ∥_{w} for the normalized absolute value on K_{w} (the module), and | · |_{w′} and ∥ · ∥_{w′} for the corresponding objects for K_{w′}, we have |2|_{w} = 2, | ^{3} \sqrt{2}|_{w} = ^{3} \sqrt{2}, and ∥2∥_{w} = 2, ∥\sqrt{2}∥_{w} = \sqrt{2}; in fact, ∥ · ∥_{w} = | · |_{w} because 1 is the local degree; on the other hand, we have |2|_{w′} = 2, | ^{3} \sqrt{2}|_{w′} = ^{3} \sqrt{2}, and ∥2∥_{w′} = 4, ∥ ^{3} \sqrt{2}∥_{w′} = ^{3} \sqrt{4}; in fact, ∥ · ∥_{w} = | · |_{w}^{2} because 2 is the local degree.

Witnessing the product formula, we have, e.g.:

∥ ^{3} \sqrt{2}∥\sqrt[3]{2} × ∥ ^{3} \sqrt{2}∥_{1+ ^{3} \sqrt{2}} × ∥ ^{3} \sqrt{2}∥_{1+ ^{3} \sqrt{4}} × ∥ ^{3} \sqrt{2}∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×
∥ ^{3} \sqrt{2}∥_{7} × ∥ ^{3} \sqrt{2}∥_{3+ ^{3} \sqrt{4}} × ∥ ^{3} \sqrt{2}∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥ ^{3} \sqrt{2}∥_{1−2 ^{3} \sqrt{4}} × ⋯
× ∥ ^{3} \sqrt{2}∥_{∞} × ∥ ^{3} \sqrt{2}∥_{∞′}
=\frac{1}{2} × 1 × 1 × 1 × 1 × 1 × 1 × 1 × ⋯ × ^{3} \sqrt{2} × ^{3} \sqrt{4} = 1
∥2∥_{2} × ∥2∥_{1+2} × ∥2∥_{1+ ^{3} \sqrt{4}} × ∥2∥_{1+22− ^{3} \sqrt{4}}×
∥2∥_{7} × ∥2∥_{3+ ^{3} \sqrt{4}} × ∥2∥_{1−22− ^{3} \sqrt{4}} × ∥2∥_{1−2 ^{3} \sqrt{4}} × ⋯
× ∥2∥_{∞} × ∥2∥_{∞′}
=\frac{1}{8} × 1 × 1 × 1 × 1 × 1 × 1 × 1 × ⋯ × 2 × 4 = 1

∥1 + ^{3} \sqrt{2}∥\sqrt[3]{2} × ∥1 + ^{3} \sqrt{2}∥_{1+ ^{3} \sqrt{2}} × ∥1 + ^{3} \sqrt{2}∥_{1+ ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{2}∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×

∥1 + ^{3} \sqrt{2}∥_{7} × ∥1 + ^{3} \sqrt{2}∥_{3+ ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{2}∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{2}∥_{1−2 ^{3} \sqrt{4}} × ⋯

× ∥1 + ^{3} \sqrt{2}∥_{∞} × ∥1 + ^{3} \sqrt{2}∥_{∞′}

=1 × \frac{1}{3} × 1 × 1 × 1 × 1 × 1 × 1 × ⋯ × (1 + ^{3} \sqrt{2}) × (1 − ^{3} \sqrt{2} + ^{3} \sqrt{4}) = 1
∥3∥\sqrt[3]{2} × ∥3∥_{1+ ^{3} \sqrt{2}} × ∥3∥_{1+ ^{3} \sqrt{4}} × ∥3∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×
∥3∥_{7} × ∥3∥_{3+ ^{3} \sqrt{4}} × ∥3∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥3∥_{1−2 ^{3} \sqrt{4}} × ⋯
× ∥3∥_{∞} × ∥3∥_{∞′}
=1 × \frac{1}{27} × 1 × 1 × 1 × 1 × 1 × 1 × ⋯ × 3 × 9 = 1

∥1 + ^{3} \sqrt{4}∥\sqrt[3]{2} × ∥1 + ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{2}} × ∥1 + ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{4}∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×

∥1 + ^{3} \sqrt{4}∥_{7} × ∥1 + ^{3} \sqrt{4}∥_{3+ ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥1 + ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{4}} × ⋯

× ∥1 + ^{3} \sqrt{4}∥_{∞} × ∥1 + ^{3} \sqrt{4}∥_{∞′}

=1 × 1 × \frac{1}{5} × 1 × 1 × 1 × 1 × 1 × ⋯ × (1 + ^{3} \sqrt{4}) × (1 + 2 ^{3} \sqrt{4} − ^{3} \sqrt{4}) = 1

∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥\sqrt[3]{2} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{2}} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{4}} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×

∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{7} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{3+ ^{3} \sqrt{4}} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{4}} × ⋯

× ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{∞} × ∥1 + 2 ^{3} \sqrt{2} − ^{3} \sqrt{4}∥_{∞′}

=1 × 1 × 1 × \frac{1}{25} × 1 × 1 × 1 × 1 × ⋯ × (1 + 2 ^{3} \sqrt{4} − ^{3} \sqrt{4}) × (5 + 5 ^{3} \sqrt{4}) = 1
∥7∥\sqrt[3]{2} × ∥7∥_{1+ ^{3} \sqrt{2}} × ∥7∥_{1+ ^{3} \sqrt{4}} × ∥7∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×
∥7∥_{7} × ∥7∥_{3+ ^{3} \sqrt{4}} × ∥7∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥7∥_{1−2 ^{3} \sqrt{4}} × ⋯
× ∥7∥_{∞} × ∥7∥_{∞′}
=1 × 1 × 1 × 1 × \frac{1}{343} × 1 × 1 × 1 × ⋯ × 7 × 49 = 1

∥3 + ^{3} \sqrt{4}∥\sqrt[3]{2} × ∥3 + ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{2}} × ∥3 + ^{3} \sqrt{4}∥_{1+ ^{3} \sqrt{4}} × ∥3 + ^{3} \sqrt{4}∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×

∥3 + ^{3} \sqrt{4}∥_{7} × ∥3 + ^{3} \sqrt{4}∥_{3+ ^{3} \sqrt{4}} × ∥3 + ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥3 + ^{3} \sqrt{4}∥_{1−2 ^{3} \sqrt{4}} × ⋯

× ∥3 + ^{3} \sqrt{4}∥_{∞} × ∥3 + ^{3} \sqrt{4}∥_{∞′}

=1 × 1 × 1 × 1 × 1 × \frac{1}{31} × 1 × 1 × ⋯ × (3 + ^{3} \sqrt{4}) × (9 + 2 ^{3} \sqrt{2} − 3 ^{3} \sqrt{4}) = 1
∥31∥\sqrt[3]{2} × ∥31∥_{1+ ^{3} \sqrt{2}} × ∥31∥_{1+ ^{3} \sqrt{4}} × ∥31∥_{1+2 ^{3} \sqrt{2}− ^{3} \sqrt{4}}×
∥31∥_{7} × ∥31∥_{3+ ^{3} \sqrt{4}} × ∥31∥_{1−2 ^{3} \sqrt{2}− ^{3} \sqrt{4}} × ∥31∥_{1−2 ^{3} \sqrt{4}} × ⋯
× ∥31∥_{∞} × ∥31∥_{∞′}
=1 × 1 × 1 × 1 × 1 × \frac{1}{31} × \frac{1}{31} × \frac{1}{31} × ⋯ × 31 × 962 = 1