[proofplan] The Minkowski bound is produced by applying [Minkowski's Theorem](/theorems/1608) part (2b) to a carefully chosen closed convex symmetric body $B_{r,s}(t) \subset \mathbb{R}^r \times \mathbb{C}^s \cong \mathbb{R}^n$ and the lattice $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$ obtained from the Minkowski embedding. The body is tailored so that its points have controlled $\ell^1$-norm, which by the AM-GM inequality forces the product-norm $|N(\alpha)|$ to be small for any $\alpha$ landing inside it. The volume of $B_{r,s}(t)$ is computed by a two-variable recursion (one for each embedding type) whose explicit solution is $V(r,s,t) = 2^r (\pi/2)^s t^n / n!$. Setting $t$ so that $\operatorname{vol}(B_{r,s}(t)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$ and invoking [Covolume of Ideal Lattices](/theorems/1609) yields the exact constant $c_L = (4/\pi)^s (n!/n^n) |D_L|^{1/2}$. [/proofplan]

[step:Set up the Minkowski embedding and name the lattice]Let $n = [L : \mathbb{Q}]$. Let $\sigma_1, \ldots, \sigma_r : L \to \mathbb{R}$ be the real embeddings and let $\tau_1, \bar\tau_1, \ldots, \tau_s, \bar\tau_s : L \to \mathbb{C}$ be the complex embeddings, grouped into conjugate pairs, so that $n = r + 2s$. The **Minkowski embedding** is the map \begin{align*} \sigma : L &\to \mathbb{R}^r \times \mathbb{C}^s \cong \mathbb{R}^n \\ \alpha &\mapsto \bigl(\sigma_1(\alpha), \ldots, \sigma_r(\alpha),\; \tau_1(\alpha), \ldots, \tau_s(\alpha)\bigr), \end{align*} where each complex coordinate $\tau_j(\alpha) = \operatorname{Re} \tau_j(\alpha) + i \operatorname{Im} \tau_j(\alpha)$ is identified with the real pair $(\operatorname{Re} \tau_j(\alpha), \operatorname{Im} \tau_j(\alpha)) \in \mathbb{R}^2$. By [Covolume of Ideal Lattices](/theorems/1609), for any non-zero ideal $\mathfrak{a} \leq \mathcal{O}_L$ the image $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$ is a full-rank lattice with \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) = 2^{-s} |D_L|^{1/2} N(\mathfrak{a}). \end{align*}[/step]

[guided]We want to find a small non-zero element $\alpha \in \mathfrak{a}$. "Small" here means its field-theoretic norm $|N(\alpha)|$ is bounded. To bring in geometric tools, we embed $\mathfrak{a}$ into a Euclidean space where it becomes a lattice, then apply [Minkowski's Theorem](/theorems/1608). **Counting embeddings.** Every degree-$n$ number field $L$ has exactly $n$ embeddings $L \hookrightarrow \mathbb{C}$. Each is either real-valued (image in $\mathbb{R} \subset \mathbb{C}$) or genuinely complex-valued; complex embeddings come in conjugate pairs (because $L$ is the fixed field of complex conjugation on the set of $\mathbb{Q}$-embeddings). Write $r$ for the number of real embeddings and $s$ for the number of conjugate pairs, so that $n = r + 2s$. **Minkowski embedding.** We assemble these embeddings into a single map with target $\mathbb{R}^n$: \begin{align*} \sigma : L &\to \mathbb{R}^r \times \mathbb{C}^s \\ \alpha &\mapsto (\sigma_1(\alpha), \ldots, \sigma_r(\alpha), \tau_1(\alpha), \ldots, \tau_s(\alpha)). \end{align*} We pick one representative $\tau_j$ from each conjugate pair $\{\tau_j, \bar\tau_j\}$; the choice is irrelevant because the $\mathbb{R}^2$-identification $\mathbb{C} \cong \mathbb{R}^2$ swaps $z$ and $\bar z$ by a reflection, which preserves Lebesgue measure. After identifying $\mathbb{C} \cong \mathbb{R}^2$, the codomain is $\mathbb{R}^r \times \mathbb{R}^{2s} = \mathbb{R}^n$. **Covolume of $\sigma(\mathfrak{a})$.** Applying [Covolume of Ideal Lattices](/theorems/1609): the hypotheses are that $\mathfrak{a}$ is a non-zero ideal of $\mathcal{O}_L$, which holds by assumption. The conclusion gives \begin{align*} \operatorname{covol}(\sigma(\mathfrak{a})) = 2^{-s} |D_L|^{1/2} N(\mathfrak{a}). \end{align*} This is the quantitative version of the statement that $\mathfrak{a}$ embeds as a full-rank lattice in $\mathbb{R}^n$ whose fundamental parallelepiped has explicitly computable volume. Here $D_L$ is the field discriminant and $N(\mathfrak{a}) = |\mathcal{O}_L / \mathfrak{a}|$ is the ideal norm.[/guided]

[step:Define the closed convex symmetric body $B_{r,s}(t)$ and verify its shape properties]For $t > 0$, define \begin{align*} B_{r,s}(t) = \left\{(y_1, \ldots, y_r, z_1, \ldots, z_s) \in \mathbb{R}^r \times \mathbb{C}^s : \sum_{i=1}^r |y_i| + 2\sum_{j=1}^s |z_j| \leq t\right\}. \end{align*} We verify that $B_{r,s}(t)$ is closed, convex, and symmetric about $\mathbf{0}$. *Closed:* the defining function $F(y, z) = \sum_i |y_i| + 2\sum_j |z_j|$ is continuous on $\mathbb{R}^r \times \mathbb{C}^s$, and $B_{r,s}(t) = F^{-1}([0, t])$ is the preimage of a closed interval, hence closed. *Convex:* for $(y, z), (y', z') \in B_{r,s}(t)$ and $\lambda \in [0, 1]$, the triangle inequality gives \begin{align*} \sum_i |\lambda y_i + (1-\lambda) y_i'| + 2\sum_j |\lambda z_j + (1-\lambda) z_j'| &\leq \lambda\Bigl(\sum_i |y_i| + 2\sum_j |z_j|\Bigr) + (1-\lambda)\Bigl(\sum_i |y_i'| + 2\sum_j |z_j'|\Bigr) \\ &\leq \lambda t + (1-\lambda) t = t. \end{align*} *Symmetric:* $F(-y, -z) = F(y, z)$, so $(y, z) \in B_{r,s}(t) \iff -(y, z) \in B_{r,s}(t)$.[/step]

[guided]**Why this body?** The choice of $B_{r,s}(t)$ is dictated by the AM-GM inequality in Step 5. We want a set whose defining inequality involves the quantity $\sum_i |y_i| + 2\sum_j |z_j|$, because that is exactly the AM-GM upper bound for $\prod_i |y_i| \cdot \prod_j |z_j|^2 = |N(\alpha)|$ (after Minkowski embedding). Minkowski's Theorem requires the body to be convex, symmetric about the origin, and either open with $\operatorname{vol} > 2^n \operatorname{covol}$ or closed with $\operatorname{vol} \geq 2^n \operatorname{covol}$; we choose the closed variant. **Closedness.** The function $F : \mathbb{R}^r \times \mathbb{C}^s \to \mathbb{R}$, $(y, z) \mapsto \sum_i |y_i| + 2\sum_j |z_j|$, is a finite sum of moduli, each of which is continuous. Preimages of closed sets under continuous maps are closed: $B_{r,s}(t) = F^{-1}([0, t])$ is closed. **Convexity.** For $(y, z), (y', z') \in B_{r,s}(t)$ and $\lambda \in [0, 1]$, we use the triangle inequality for $|\cdot|$ componentwise: \begin{align*} |\lambda y_i + (1-\lambda) y_i'| \leq \lambda |y_i| + (1-\lambda) |y_i'|, \qquad |\lambda z_j + (1-\lambda) z_j'| \leq \lambda |z_j| + (1-\lambda) |z_j'|. \end{align*} Summing with weights $1$ (real) and $2$ (complex): \begin{align*} F(\lambda(y,z) + (1-\lambda)(y',z')) \leq \lambda F(y,z) + (1-\lambda) F(y',z') \leq \lambda t + (1-\lambda)t = t, \end{align*} so the convex combination lies in $B_{r,s}(t)$. **Symmetry about $\mathbf{0}$.** $|{-y_i}| = |y_i|$ and $|{-z_j}| = |z_j|$, so $F(-y, -z) = F(y, z)$ and the condition $F \leq t$ is preserved under $(y, z) \mapsto -(y, z)$. All three hypotheses of [Minkowski's Theorem](/theorems/1608) part (2b) — closed, convex, symmetric — are satisfied.[/guided]

[step:Compute the volume $V(r,s,t) = \operatorname{vol}(B_{r,s}(t))$ by induction on $n$]Write $V(r, s, t) := \operatorname{vol}(B_{r,s}(t))$, where the volume is the Lebesgue measure $\mathcal{L}^n$ on $\mathbb{R}^n$ after identifying $\mathbb{C}^s \cong \mathbb{R}^{2s}$. We claim \begin{align*} V(r, s, t) = 2^r \left(\frac{\pi}{2}\right)^s \frac{t^n}{n!}, \qquad n = r + 2s. \end{align*} [claim:Base Cases] $V(1, 0, t) = 2t$ and $V(0, 1, t) = \pi t^2 / 4$. [/claim] [proof] For $(r, s) = (1, 0)$: $B_{1,0}(t) = \{y_1 \in \mathbb{R} : |y_1| \leq t\} = [-t, t]$, whose one-dimensional Lebesgue measure is $2t$. The formula $2^1 (\pi/2)^0 t^1/1! = 2t$ agrees. For $(r, s) = (0, 1)$: $B_{0,1}(t) = \{z_1 \in \mathbb{C} : 2|z_1| \leq t\}$ is the closed disc of radius $t/2$ in $\mathbb{R}^2 \cong \mathbb{C}$, with area $\pi (t/2)^2 = \pi t^2 / 4$. The formula $2^0 (\pi/2)^1 t^2 / 2! = \pi t^2 / 4$ agrees. [/proof] [claim:Real Coordinate Recursion] For $r \geq 0$, $s \geq 0$, and $t > 0$, \begin{align*} V(r+1, s, t) = 2 \int_0^t V(r, s, t - u)\, d\mathcal{L}^1(u). \end{align*} [/claim] [proof] By Tonelli's theorem on the non-negative indicator $\mathbb{1}_{B_{r+1, s}(t)}$, \begin{align*} V(r+1, s, t) = \int_{-t}^t \operatorname{vol}\bigl(\{(y_2, \ldots, z_s) : F(y_2, \ldots, z_s) \leq t - |y_1|\}\bigr) \, d\mathcal{L}^1(y_1) = \int_{-t}^t V(r, s, t - |y_1|)\, d\mathcal{L}^1(y_1). \end{align*} The integrand is symmetric in $y_1 \mapsto -y_1$, so the integral equals $2 \int_0^t V(r, s, t - u)\, d\mathcal{L}^1(u)$ after the substitution $u = y_1$. [/proof] [claim:Complex Coordinate Recursion] For $r \geq 0$, $s \geq 0$, and $t > 0$, \begin{align*} V(r, s+1, t) = 2\pi \int_0^{t/2} V(r, s, t - 2\rho)\, \rho\, d\mathcal{L}^1(\rho). \end{align*} [/claim] [proof] By Tonelli on $\mathbb{1}_{B_{r, s+1}(t)}$, integrating out the first complex coordinate $z_1 = \rho e^{i\theta}$: \begin{align*} V(r, s+1, t) = \int_{\{z_1 : 2|z_1| \leq t\}} V(r, s, t - 2|z_1|)\, d\mathcal{L}^2(z_1). \end{align*} We apply the polar coordinates formula for two-dimensional Lebesgue measure: under $z_1 = \rho e^{i\theta}$, $d\mathcal{L}^2(z_1) = \rho\, d\mathcal{L}^1(\theta)\, d\mathcal{L}^1(\rho)$, and the domain $\{|z_1| \leq t/2\}$ becomes $[0, t/2] \times [0, 2\pi)$: \begin{align*} V(r, s+1, t) = \int_0^{t/2} \int_0^{2\pi} V(r, s, t - 2\rho)\, \rho\, d\mathcal{L}^1(\theta)\, d\mathcal{L}^1(\rho) = 2\pi \int_0^{t/2} V(r, s, t - 2\rho)\, \rho\, d\mathcal{L}^1(\rho). \end{align*} The $\theta$-integral is a constant $V(r, s, t - 2\rho)\cdot \rho$ times $2\pi$ because the integrand does not depend on $\theta$. [/proof] With the base cases and the two recurrences, we verify the closed form $V(r, s, t) = 2^r (\pi/2)^s t^n / n!$ by induction on $n = r + 2s$, stepping in $r$ or $s$ as needed. **Induction on $r$ (fixing $s$).** Assume $V(r, s, t) = 2^r (\pi/2)^s t^n/n!$ with $n = r + 2s$. By the real recursion, \begin{align*} V(r+1, s, t) = 2 \int_0^t V(r, s, t-u)\, d\mathcal{L}^1(u) = 2 \cdot 2^r \left(\frac{\pi}{2}\right)^s \frac{1}{n!} \int_0^t (t-u)^n\, d\mathcal{L}^1(u). \end{align*} The substitution $v = t - u$ (with $d\mathcal{L}^1(v) = -d\mathcal{L}^1(u)$ and domain $[0, t] \to [0, t]$ reversed) gives $\int_0^t (t-u)^n\, d\mathcal{L}^1(u) = \int_0^t v^n\, d\mathcal{L}^1(v) = t^{n+1}/(n+1)$. Therefore \begin{align*} V(r+1, s, t) = 2^{r+1} \left(\frac{\pi}{2}\right)^s \frac{t^{n+1}}{(n+1)!}, \end{align*} which is the formula for $(r+1, s)$ with $n+1 = (r+1) + 2s$. **Induction on $s$ (fixing $r$).** Assume $V(r, s, t) = 2^r (\pi/2)^s t^n/n!$. By the complex recursion, \begin{align*} V(r, s+1, t) = 2\pi \cdot 2^r \left(\frac{\pi}{2}\right)^s \frac{1}{n!} \int_0^{t/2} (t - 2\rho)^n\, \rho\, d\mathcal{L}^1(\rho). \end{align*} Substitute $u = t - 2\rho$ so that $\rho = (t - u)/2$ and $d\mathcal{L}^1(u) = -2\, d\mathcal{L}^1(\rho)$, with $\rho \in [0, t/2]$ mapped to $u \in [0, t]$ (reversed): \begin{align*} \int_0^{t/2} (t - 2\rho)^n\, \rho\, d\mathcal{L}^1(\rho) = \int_0^t u^n \cdot \frac{t-u}{2} \cdot \frac{1}{2}\, d\mathcal{L}^1(u) = \frac{1}{4} \int_0^t u^n (t - u)\, d\mathcal{L}^1(u). \end{align*} The last integral is a Beta function evaluation: \begin{align*} \int_0^t u^n (t - u)\, d\mathcal{L}^1(u) = \int_0^t t u^n\, d\mathcal{L}^1(u) - \int_0^t u^{n+1}\, d\mathcal{L}^1(u) = \frac{t^{n+2}}{n+1} - \frac{t^{n+2}}{n+2} = \frac{t^{n+2}}{(n+1)(n+2)}. \end{align*} Therefore \begin{align*} V(r, s+1, t) = 2\pi \cdot 2^r \left(\frac{\pi}{2}\right)^s \frac{1}{n!} \cdot \frac{1}{4} \cdot \frac{t^{n+2}}{(n+1)(n+2)} = 2^r \left(\frac{\pi}{2}\right)^{s+1} \frac{t^{n+2}}{(n+2)!}, \end{align*} which is the formula for $(r, s+1)$ with $(n+2) = r + 2(s+1)$. (We used $2\pi / 4 = \pi/2$, multiplying the $(\pi/2)^s$ factor.) Starting from the base cases and iterating the two steps, every $(r, s)$ is reached, establishing the closed form.[/step]

[guided]The volume formula $V(r, s, t) = 2^r (\pi/2)^s t^n / n!$ generalizes two familiar facts: - For $s = 0$ (totally real field): $V(r, 0, t) = 2^r t^r / r!$ is the volume of the $\ell^1$-ball of radius $t$ in $\mathbb{R}^r$, equivalently the cross-polytope. - For $r = 0, s = 1$ (imaginary quadratic): $V(0, 1, t) = \pi t^2 / 4$ is the area of the disc $\{2|z| \leq t\}$ in $\mathbb{C}$. **Strategy for computing $V(r, s, t)$.** Use Tonelli to strip off one coordinate at a time, producing a one-dimensional integral over the new coordinate against $V(r', s', t - \text{used budget})$ where the budget $t$ is reduced by the $\ell^1$-cost of the stripped coordinate. Each type (real, complex) gives a distinct recursion. **Real coordinate recursion.** Tonelli's theorem applies to the non-negative measurable indicator $\mathbb{1}_{B_{r+1, s}(t)}$ (non-negativity is the condition, which is automatic; measurability holds because $B_{r+1,s}(t)$ is closed). We slice by $y_1$: for each $y_1 \in [-t, t]$, the $(y_2, \ldots, z_s)$-section is exactly $B_{r, s}(t - |y_1|)$ (empty if $|y_1| > t$). Hence \begin{align*} V(r+1, s, t) = \int_{-t}^t V(r, s, t - |y_1|)\, d\mathcal{L}^1(y_1). \end{align*} By evenness of $|y_1|$ under $y_1 \mapsto -y_1$, we fold into $2\int_0^t$ with $u = y_1$. **Complex coordinate recursion.** Similarly, the $z_1$-slice of $B_{r, s+1}(t)$ has volume $V(r, s, t - 2|z_1|)$, so \begin{align*} V(r, s+1, t) = \int_{\{z_1 \in \mathbb{C} : 2|z_1| \leq t\}} V(r, s, t - 2|z_1|)\, d\mathcal{L}^2(z_1). \end{align*} We apply the polar coordinates formula for the two-dimensional Lebesgue measure, which states that under $z_1 = \rho e^{i\theta}$, $d\mathcal{L}^2(z_1) = \rho\, d\mathcal{L}^1(\theta)\, d\mathcal{L}^1(\rho)$, with the domain $\{|z_1| \leq t/2\}$ becoming $[0, t/2] \times [0, 2\pi)$. Since the integrand is $\theta$-independent, the $\theta$-integral contributes a factor of $2\pi$. **Induction on $n$.** We establish $V(r, s, t) = 2^r (\pi/2)^s t^n/n!$ by induction, stepping $r \to r+1$ (for fixed $s$) and $s \to s+1$ (for fixed $r$). Starting at $(r, s) = (0, 0)$ where $V(0, 0, t) = 1$ (a single point with "volume" 1 in the degenerate zero-dimensional case), or equivalently starting at the base cases $(1, 0)$ and $(0, 1)$, we reach every $(r, s)$. The $r$-induction uses $\int_0^t (t-u)^n\, d\mathcal{L}^1(u) = t^{n+1}/(n+1)$ (substitute $v = t - u$). The $s$-induction uses $\int_0^{t/2}(t - 2\rho)^n \rho\, d\mathcal{L}^1(\rho) = t^{n+2}/[4(n+1)(n+2)]$ (substitute $u = t - 2\rho$, then expand $u^n(t-u)/4$). In both cases, the combinatorial factor $1/n!$ transforms as $1/(n+1)!$ and $1/(n+2)!$ respectively, and the leading constants pick up exactly $2$ (from the real case) or $\pi/2$ (from the complex case). This matches the predicted formula.[/guided]

[step:Choose $t$ so that $\operatorname{vol}(B_{r,s}(t)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$]Set $t = t^*$ where \begin{align*} (t^*)^n = \left(\frac{4}{\pi}\right)^s n! \cdot |D_L|^{1/2} N(\mathfrak{a}). \end{align*} We verify that this choice gives $\operatorname{vol}(B_{r,s}(t^*)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$. By Step 3, \begin{align*} \operatorname{vol}(B_{r,s}(t^*)) = 2^r \left(\frac{\pi}{2}\right)^s \frac{(t^*)^n}{n!}. \end{align*} Substituting the choice of $(t^*)^n$: \begin{align*} \operatorname{vol}(B_{r,s}(t^*)) = 2^r \left(\frac{\pi}{2}\right)^s \frac{1}{n!} \cdot \left(\frac{4}{\pi}\right)^s n! \cdot |D_L|^{1/2} N(\mathfrak{a}) = 2^r \cdot 2^s \cdot |D_L|^{1/2} N(\mathfrak{a}), \end{align*} where we used $(\pi/2)^s (4/\pi)^s = (4/2)^s = 2^s$. Since $n = r + 2s$, we have $2^r \cdot 2^s \cdot 2^s = 2^{r + 2s} = 2^n$, so \begin{align*} \operatorname{vol}(B_{r,s}(t^*)) = 2^n \cdot 2^{-s} |D_L|^{1/2} N(\mathfrak{a}) = 2^n \operatorname{covol}(\sigma(\mathfrak{a})), \end{align*} using $\operatorname{covol}(\sigma(\mathfrak{a})) = 2^{-s} |D_L|^{1/2} N(\mathfrak{a})$ from Step 1.[/step]

[guided]We need to match the volume of $B_{r,s}(t)$ with $2^n \operatorname{covol}(\sigma(\mathfrak{a}))$, because [Minkowski's Theorem](/theorems/1608) part (2b) requires exactly this inequality (non-strictly, with $\geq$) for a closed convex symmetric body; the weakest version allowing us to find a non-zero lattice point is $\operatorname{vol} = 2^n \operatorname{covol}$. **Solving for $t$.** Equating $V(r, s, t) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$ and substituting both closed forms: \begin{align*} 2^r \left(\frac{\pi}{2}\right)^s \frac{t^n}{n!} = 2^n \cdot 2^{-s} |D_L|^{1/2} N(\mathfrak{a}) = 2^{n - s} |D_L|^{1/2} N(\mathfrak{a}). \end{align*} Rearranging: \begin{align*} t^n = \frac{2^{n - s} \cdot n!}{2^r (\pi/2)^s} |D_L|^{1/2} N(\mathfrak{a}) = \frac{2^{n - s - r}}{(\pi/2)^s} \cdot n! \cdot |D_L|^{1/2} N(\mathfrak{a}). \end{align*} Since $n - s - r = 2s - s = s$, the $2$-exponent simplifies to $2^s$: \begin{align*} t^n = \frac{2^s}{(\pi/2)^s} n! \cdot |D_L|^{1/2} N(\mathfrak{a}) = \left(\frac{2 \cdot 2}{\pi}\right)^s n! \cdot |D_L|^{1/2} N(\mathfrak{a}) = \left(\frac{4}{\pi}\right)^s n! \cdot |D_L|^{1/2} N(\mathfrak{a}). \end{align*} This is the value $(t^*)^n$. **Sanity check.** The $2^r \cdot 2^s \cdot 2^s = 2^{r+2s} = 2^n$ cancellation is the geometric content of "volume of $B_{r,s}(t)$ equals $2^n$ times the covolume." The $(4/\pi)^s$ factor absorbs the discrepancy between the $\mathbb{R}^1$-cross-polytope basic unit (a segment) and the $\mathbb{C}^1$-basic unit (a disc), which differ by a factor of $\pi/4$ per complex dimension.[/guided]

[step:Apply Minkowski's Theorem and extract an algebraic integer in $\mathfrak{a}$]The body $B_{r,s}(t^*)$ is closed (Step 2), convex (Step 2), symmetric about $\mathbf{0}$ (Step 2), and satisfies $\operatorname{vol}(B_{r,s}(t^*)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$ (Step 4). The lattice $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$ is full-rank (Step 1 via [Covolume of Ideal Lattices](/theorems/1609)). All hypotheses of [Minkowski's Theorem](/theorems/1608) part (2b) — closed convex symmetric body with $\operatorname{vol}(S) \geq 2^n \operatorname{covol}(\Lambda)$ — are satisfied. Hence there is a non-zero $\gamma \in \sigma(\mathfrak{a}) \cap B_{r,s}(t^*)$. Since $\sigma : \mathfrak{a} \to \sigma(\mathfrak{a})$ is injective (an embedding), $\gamma = \sigma(\alpha)$ for a unique non-zero $\alpha \in \mathfrak{a}$.[/step]

[guided]We have set everything up so that [Minkowski's Theorem](/theorems/1608) part (2b) applies directly: **Hypothesis check.** - *Lattice:* $\sigma(\mathfrak{a}) \subset \mathbb{R}^n$ is a full-rank lattice, by [Covolume of Ideal Lattices](/theorems/1609). - *Closed:* $B_{r,s}(t^*)$ is closed by Step 2. - *Convex:* $B_{r,s}(t^*)$ is convex by Step 2. - *Symmetric:* $B_{r,s}(t^*)$ is symmetric about $\mathbf{0}$ by Step 2. - *Volume inequality:* $\operatorname{vol}(B_{r,s}(t^*)) = 2^n \operatorname{covol}(\sigma(\mathfrak{a}))$ by Step 4; this is the non-strict bound allowed for closed bodies in part (2b). **Conclusion.** Minkowski's theorem gives a non-zero lattice point $\gamma \in \sigma(\mathfrak{a}) \cap B_{r,s}(t^*)$. **Lifting to $\mathfrak{a}$.** The Minkowski embedding $\sigma : L \to \mathbb{R}^n$ is a $\mathbb{Q}$-algebra homomorphism and, when restricted to $\mathfrak{a}$, is an injective $\mathbb{Z}$-module homomorphism with image $\sigma(\mathfrak{a})$. Injectivity holds because $\sigma$ is $\mathbb{Q}$-linear with trivial kernel: if $\sigma(\alpha) = 0$, then $\sigma_1(\alpha) = 0$ (among others), but $\sigma_1$ is an injective $\mathbb{Q}$-algebra homomorphism (field embeddings are injective), so $\alpha = 0$. Therefore there is a unique non-zero $\alpha \in \mathfrak{a}$ with $\sigma(\alpha) = \gamma$.[/guided]

[step:Bound $|N(\alpha)|$ using AM-GM on the $\ell^1$-constraint]Write $\sigma(\alpha) = (y_1, \ldots, y_r, z_1, \ldots, z_s) \in B_{r, s}(t^*)$, with $y_i = \sigma_i(\alpha) \in \mathbb{R}$ and $z_j = \tau_j(\alpha) \in \mathbb{C}$. By [Norm and Trace via Embeddings](/theorems/1577), using that the complex embeddings come in conjugate pairs $\tau_j, \bar\tau_j$ with $\tau_j(\alpha) \bar\tau_j(\alpha) = |\tau_j(\alpha)|^2 = |z_j|^2$: \begin{align*} N_{L/\mathbb{Q}}(\alpha) = \prod_{i=1}^r \sigma_i(\alpha) \cdot \prod_{j=1}^s \tau_j(\alpha) \bar\tau_j(\alpha) = \prod_{i=1}^r y_i \cdot \prod_{j=1}^s |z_j|^2. \end{align*} Taking absolute values: \begin{align*} |N(\alpha)| = \prod_{i=1}^r |y_i| \cdot \prod_{j=1}^s |z_j|^2. \end{align*} We now apply the AM-GM inequality to the $n$ non-negative numbers $|y_1|, \ldots, |y_r|, |z_1|, |z_1|, \ldots, |z_s|, |z_s|$ (each $|z_j|$ counted twice), whose product is $\prod_i |y_i| \cdot \prod_j |z_j|^2 = |N(\alpha)|$. AM-GM states $\bigl(\prod_{k=1}^n a_k\bigr)^{1/n} \leq \tfrac{1}{n} \sum_k a_k$ for non-negative $a_k$. Applied here: \begin{align*} |N(\alpha)|^{1/n} = \left(\prod_i |y_i| \cdot \prod_j |z_j|^2\right)^{1/n} \leq \frac{1}{n}\left(\sum_i |y_i| + 2\sum_j |z_j|\right). \end{align*} Since $\sigma(\alpha) \in B_{r,s}(t^*)$, $\sum_i |y_i| + 2\sum_j |z_j| \leq t^*$, so \begin{align*} |N(\alpha)|^{1/n} \leq \frac{t^*}{n}, \qquad |N(\alpha)| \leq \frac{(t^*)^n}{n^n}. \end{align*}[/step]

[guided]We have an algebraic integer $\alpha \in \mathfrak{a}$ whose Minkowski embedding $\sigma(\alpha)$ satisfies the $\ell^1$-type constraint $\sum_i |y_i| + 2\sum_j |z_j| \leq t^*$. We want to convert this to a bound on the product $|N(\alpha)|$, which is where AM-GM (arithmetic mean $\geq$ geometric mean) enters. **Norm via embeddings.** [Theorem 1577](/theorems/1577) states that for a number field $L$ of degree $n = [L : \mathbb{Q}]$, the norm $N_{L/\mathbb{Q}}(\alpha)$ equals the product over all $n$ embeddings $L \hookrightarrow \mathbb{C}$: \begin{align*} N(\alpha) = \prod_{i=1}^r \sigma_i(\alpha) \cdot \prod_{j=1}^s \tau_j(\alpha) \bar\tau_j(\alpha). \end{align*} The hypotheses of Theorem 1577 (that $L$ is a number field) are satisfied by our setup. We have grouped the $2s$ complex embeddings into $s$ conjugate pairs; each pair contributes $\tau_j(\alpha) \bar\tau_j(\alpha) = |\tau_j(\alpha)|^2$ because $\bar\tau_j$ is the complex conjugate of $\tau_j$. **Taking absolute values.** With $y_i = \sigma_i(\alpha)$ and $z_j = \tau_j(\alpha)$: \begin{align*} |N(\alpha)| = \prod_i |y_i| \cdot \prod_j |z_j|^2. \end{align*} **Applying AM-GM.** The AM-GM inequality states: for non-negative $a_1, \ldots, a_n \geq 0$, $(a_1 \cdots a_n)^{1/n} \leq (a_1 + \cdots + a_n)/n$. We apply it to the list \begin{align*} \underbrace{|y_1|, \ldots, |y_r|}_{r \text{ terms}}, \underbrace{|z_1|, |z_1|, \ldots, |z_s|, |z_s|}_{2s \text{ terms}}, \end{align*} totalling $r + 2s = n$ non-negative reals. Their product is $\prod_i |y_i| \cdot \prod_j |z_j|^2 = |N(\alpha)|$, and their sum is $\sum_i |y_i| + 2\sum_j |z_j|$. The AM-GM conclusion: \begin{align*} |N(\alpha)|^{1/n} \leq \frac{1}{n}\left(\sum_i |y_i| + 2\sum_j |z_j|\right). \end{align*} **Using the $\ell^1$-constraint.** Because $\sigma(\alpha) \in B_{r,s}(t^*)$, by definition of this body, $\sum_i |y_i| + 2\sum_j |z_j| \leq t^*$. Hence \begin{align*} |N(\alpha)|^{1/n} \leq \frac{t^*}{n}. \end{align*} Raising to the $n$-th power: \begin{align*} |N(\alpha)| \leq \frac{(t^*)^n}{n^n}. \end{align*} **Why the factor 2 for the complex terms.** The $|z_j|$'s appear once as a product factor but we include them twice in the AM-GM list because $|z_j|^2 = |z_j| \cdot |z_j|$ is a product of two identical factors. The sum $|z_j| + |z_j| = 2|z_j|$ is then what appears on the AM side, matching the $2\sum_j |z_j|$ in the $\ell^1$-constraint.[/guided]

[step:Substitute the value of $(t^*)^n$ to obtain the Minkowski bound] Using the choice $(t^*)^n = (4/\pi)^s n! \cdot |D_L|^{1/2} N(\mathfrak{a})$ from Step 4 and the bound $|N(\alpha)| \leq (t^*)^n / n^n$ from Step 5: \begin{align*} |N(\alpha)| \leq \frac{(t^*)^n}{n^n} = \left(\frac{4}{\pi}\right)^s \frac{n!}{n^n} |D_L|^{1/2} N(\mathfrak{a}) = c_L N(\mathfrak{a}), \end{align*} where \begin{align*} c_L = \left(\frac{4}{\pi}\right)^s \frac{n!}{n^n} |D_L|^{1/2}. \end{align*} This gives the Minkowski bound $|N(\alpha)| \leq c_L N(\mathfrak{a})$, completing the proof. [/step]