[proofplan] We compare the discriminant of a natural $\mathbb{Z}$-basis of the principal ideal $\langle \alpha \rangle$ against the [Norm and Discriminant](/theorems/1594) identity. Starting from an integral basis $\alpha_1, \ldots, \alpha_n$ of $\mathcal{O}_L$, multiplication by $\alpha$ produces the $\mathbb{Z}$-basis $\alpha\alpha_1, \ldots, \alpha\alpha_n$ of $\langle \alpha \rangle$. The discriminant identity then gives $\Delta(\alpha\alpha_1, \ldots, \alpha\alpha_n) = N(\langle \alpha \rangle)^2\, D_L$, while a direct computation using the multiplicativity of $\sigma_i$ on products yields $\Delta(\alpha\alpha_1, \ldots, \alpha\alpha_n) = N_{L/\mathbb{Q}}(\alpha)^2\, D_L$. Comparison and extraction of square roots (with the correct sign on the positive integer $N(\langle \alpha \rangle)$) gives the result. [/proofplan] [step:Handle the case $\alpha = 0$ separately] If $\alpha = 0$, then $\langle \alpha \rangle = \{0\}$ is the zero ideal, for which $N(\langle 0 \rangle)$ is not defined (the ideal norm is defined only for nonzero ideals). Moreover, the theorem implicitly assumes $\alpha \neq 0$ for $\langle \alpha \rangle$ to have a well-defined nonzero norm; this is the standard convention when stating "$N(\langle \alpha \rangle) = |N_{L/\mathbb{Q}}(\alpha)|$" since $N_{L/\mathbb{Q}}(0) = 0$ and the identity $N(\langle 0 \rangle) = 0$ is consistent only under the convention $N(\langle 0 \rangle) := 0$. We may assume $\alpha \neq 0$ throughout. Then $\alpha$ is invertible in $L$ (as $L$ is a field), and $\langle \alpha \rangle$ is a nonzero ideal of $\mathcal{O}_L$. [guided] The theorem statement involves the ideal norm $N(\langle \alpha \rangle) = |\mathcal{O}_L/\langle \alpha \rangle|$, which requires the quotient to be finite. For $\alpha \neq 0$ this is automatic, since $\mathcal{O}_L/\langle \alpha \rangle$ is a finite ring by the [Nonzero Ideals Have Bounded Quotients](/theorems/1583) theorem. For $\alpha = 0$, $\langle \alpha \rangle = \{0\}$ and $\mathcal{O}_L/\{0\} = \mathcal{O}_L$ is infinite, so the ideal norm in its usual definition does not apply; and $N_{L/\mathbb{Q}}(0) = 0$, so the formula holds under the convention $N(\langle 0 \rangle) := 0$. We focus on the interesting case $\alpha \neq 0$. Then: - $\alpha$ is a nonzero element of $\mathcal{O}_L \subseteq L$; since $L$ is a field, $\alpha \in L^\times$. - $\langle \alpha \rangle = \alpha \mathcal{O}_L$ is a nonzero ideal of $\mathcal{O}_L$. - The quotient $\mathcal{O}_L/\langle \alpha \rangle$ is finite of size $N(\langle \alpha \rangle)$. [/guided] [/step] [step:Produce a $\mathbb{Z}$-basis of $\langle \alpha \rangle$ by multiplying an integral basis by $\alpha$] Let $\alpha_1, \ldots, \alpha_n$ be an integral basis of $\mathcal{O}_L$ (which exists by the [Existence of an Integral Basis](/theorems/1581)). Define \begin{align*} \beta_i := \alpha \alpha_i \in \mathcal{O}_L, \quad i = 1, \ldots, n. \end{align*} We claim $\beta_1, \ldots, \beta_n$ is a $\mathbb{Z}$-basis of $\langle \alpha \rangle$. Every element of $\langle \alpha \rangle = \alpha \mathcal{O}_L$ has the form $\alpha \gamma$ for some $\gamma \in \mathcal{O}_L$. Expanding $\gamma = \sum c_i \alpha_i$ with $c_i \in \mathbb{Z}$, \begin{align*} \alpha \gamma = \alpha \sum_{i=1}^n c_i \alpha_i = \sum_{i=1}^n c_i\, \beta_i. \end{align*} This shows $\beta_1, \ldots, \beta_n$ span $\langle \alpha \rangle$ over $\mathbb{Z}$. For $\mathbb{Z}$-linear independence: a relation $\sum c_i \beta_i = 0$ expands to $\alpha \sum c_i \alpha_i = 0$, and since $\alpha \neq 0$ in the field $L$, we get $\sum c_i \alpha_i = 0$, whence $c_i = 0$ by linear independence of $\alpha_1, \ldots, \alpha_n$. Therefore $\langle \alpha \rangle = \beta_1\mathbb{Z} \oplus \cdots \oplus \beta_n\mathbb{Z}$. [guided] **Why this basis?** The natural way to describe $\langle \alpha \rangle = \alpha \mathcal{O}_L$ is: every element factors as $\alpha$ times an element of $\mathcal{O}_L$. Expanding the second factor in the integral basis expresses every element of $\langle \alpha \rangle$ as a $\mathbb{Z}$-linear combination of $\alpha \alpha_1, \ldots, \alpha \alpha_n$. So $\beta_i := \alpha \alpha_i$ is the natural candidate for a $\mathbb{Z}$-basis. **Spanning.** Let $x \in \langle \alpha \rangle$. Then $x = \alpha\gamma$ for some $\gamma \in \mathcal{O}_L$. Writing $\gamma = \sum_i c_i \alpha_i$ with $c_i \in \mathbb{Z}$ (unique such decomposition, since $\alpha_i$ is an integral basis), \begin{align*} x = \alpha\gamma = \alpha \sum_i c_i \alpha_i = \sum_i c_i (\alpha \alpha_i) = \sum_i c_i \beta_i. \end{align*} **$\mathbb{Z}$-linear independence.** Suppose $\sum c_i \beta_i = 0$ with $c_i \in \mathbb{Z}$. Factoring, \begin{align*} \alpha \left( \sum_i c_i \alpha_i \right) = 0 \quad \text{in } L. \end{align*} Since $L$ is a field and $\alpha \neq 0$, we can divide by $\alpha$: $\sum_i c_i \alpha_i = 0$. But $\alpha_1, \ldots, \alpha_n$ is an integral basis, in particular $\mathbb{Z}$-linearly independent, so all $c_i = 0$. **Conclusion.** $\beta_1, \ldots, \beta_n$ spans $\langle \alpha \rangle$ over $\mathbb{Z}$ and is $\mathbb{Z}$-linearly independent, hence \begin{align*} \langle \alpha \rangle = \beta_1 \mathbb{Z} \oplus \cdots \oplus \beta_n \mathbb{Z}. \end{align*} [/guided] [/step] [step:Compute $\Delta(\beta_1, \ldots, \beta_n)$ via the structure theorem for ideal bases] By [Norm and Discriminant](/theorems/1594), for any $\mathbb{Z}$-basis of a nonzero ideal $\mathfrak{a}$, the discriminant satisfies $\Delta = N(\mathfrak{a})^2 D_L$. Applying this to the basis $\beta_1, \ldots, \beta_n$ of $\langle \alpha \rangle$, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = N(\langle \alpha \rangle)^2\, D_L. \tag{$*$} \end{align*} [guided] Step 2 established that $\beta_1, \ldots, \beta_n$ is a $\mathbb{Z}$-basis of $\langle \alpha \rangle$. The [Norm and Discriminant](/theorems/1594) theorem applies directly: for any nonzero ideal $\mathfrak{a} \unlhd \mathcal{O}_L$ and any $\mathbb{Z}$-basis of $\mathfrak{a}$, the discriminant of that basis equals $N(\mathfrak{a})^2 D_L$. **Hypothesis verification.** - $\langle \alpha \rangle$ is a nonzero ideal: verified in Step 1. - $\beta_1, \ldots, \beta_n$ is a $\mathbb{Z}$-basis of $\langle \alpha \rangle$: verified in Step 2. The theorem gives \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = N(\langle \alpha \rangle)^2\, D_L. \tag{$*$} \end{align*} [/guided] [/step] [step:Compute $\Delta(\beta_1, \ldots, \beta_n)$ directly via multiplicativity of embeddings] Let $\sigma_1, \ldots, \sigma_n: L \hookrightarrow \mathbb{C}$ be the field embeddings of $L$ (given by [Embeddings of a Number Field](/theorems/1576)). By definition of the discriminant, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = \det(\sigma_i(\beta_j))^2 = \det(\sigma_i(\alpha \alpha_j))^2. \end{align*} Each $\sigma_i$ is a ring homomorphism, so $\sigma_i(\alpha \alpha_j) = \sigma_i(\alpha) \sigma_i(\alpha_j)$. Therefore the matrix $(\sigma_i(\beta_j))$ factors as \begin{align*} \sigma_i(\alpha \alpha_j) = \sigma_i(\alpha) \cdot \sigma_i(\alpha_j), \end{align*} which is the row-scaled matrix $D \cdot M$ where $D = \operatorname{diag}(\sigma_1(\alpha), \ldots, \sigma_n(\alpha))$ and $M_{ij} = \sigma_i(\alpha_j)$. Taking determinants, \begin{align*} \det(\sigma_i(\alpha \alpha_j)) = \det(D) \det(M) = \left( \prod_{i=1}^n \sigma_i(\alpha) \right) \det(\sigma_i(\alpha_j)). \end{align*} By the [Norm and Trace via Embeddings](/theorems/1577) theorem, $N_{L/\mathbb{Q}}(\alpha) = \prod_{i=1}^n \sigma_i(\alpha)$. Squaring, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = N_{L/\mathbb{Q}}(\alpha)^2 \cdot \det(\sigma_i(\alpha_j))^2 = N_{L/\mathbb{Q}}(\alpha)^2 \cdot \Delta(\alpha_1, \ldots, \alpha_n) = N_{L/\mathbb{Q}}(\alpha)^2\, D_L, \tag{$**$} \end{align*} where the last equality uses that $\alpha_1, \ldots, \alpha_n$ is an integral basis and so $\Delta(\alpha_1, \ldots, \alpha_n) = D_L$ by definition. [guided] We compute $\Delta(\beta_1, \ldots, \beta_n)$ a second time, now directly from its defining formula, and factor out the $\alpha$ contribution. **Discriminant formula.** For a $\mathbb{Q}$-basis $\gamma_1, \ldots, \gamma_n$ of $L$ and any enumeration of the embeddings $\sigma_1, \ldots, \sigma_n: L \hookrightarrow \mathbb{C}$, \begin{align*} \Delta(\gamma_1, \ldots, \gamma_n) := \det(\sigma_i(\gamma_j))^2. \end{align*} **Factoring the embedding matrix.** Each $\sigma_i$ is a ring homomorphism $L \to \mathbb{C}$, so it respects multiplication: $\sigma_i(\alpha \alpha_j) = \sigma_i(\alpha) \sigma_i(\alpha_j)$. Assembling into a matrix, \begin{align*} (\sigma_i(\beta_j))_{ij} = (\sigma_i(\alpha) \sigma_i(\alpha_j))_{ij} = D \cdot M, \end{align*} where \begin{align*} D = \operatorname{diag}(\sigma_1(\alpha), \ldots, \sigma_n(\alpha)), \qquad M_{ij} = \sigma_i(\alpha_j). \end{align*} (The $i$-th row of $D \cdot M$ equals $\sigma_i(\alpha)$ times the $i$-th row of $M$, matching $\sigma_i(\alpha)\, \sigma_i(\alpha_j)$.) **Determinant.** Since $D$ is diagonal with entries $\sigma_i(\alpha)$, \begin{align*} \det(D) = \prod_{i=1}^n \sigma_i(\alpha), \end{align*} and by multiplicativity of $\det$, \begin{align*} \det(\sigma_i(\beta_j)) = \det(D) \det(M) = \left( \prod_{i=1}^n \sigma_i(\alpha) \right) \det(\sigma_i(\alpha_j)). \end{align*} **Identifying the product of embeddings as the norm.** The [Norm and Trace via Embeddings](/theorems/1577) theorem gives \begin{align*} N_{L/\mathbb{Q}}(\alpha) = \prod_{i=1}^n \sigma_i(\alpha). \end{align*} **Squaring.** Squaring the determinant identity, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = \det(\sigma_i(\beta_j))^2 = N_{L/\mathbb{Q}}(\alpha)^2 \cdot \det(\sigma_i(\alpha_j))^2 = N_{L/\mathbb{Q}}(\alpha)^2\, \Delta(\alpha_1, \ldots, \alpha_n). \end{align*} By definition of $D_L$, $\Delta(\alpha_1, \ldots, \alpha_n) = D_L$ since $\alpha_i$ is an integral basis. So \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = N_{L/\mathbb{Q}}(\alpha)^2\, D_L. \tag{$**$} \end{align*} [/guided] [/step] [step:Equate the two expressions and extract the norm] From ($*$) and ($**$), \begin{align*} N(\langle \alpha \rangle)^2\, D_L = N_{L/\mathbb{Q}}(\alpha)^2\, D_L. \end{align*} The field discriminant $D_L$ is nonzero — for instance by the [Non-Degeneracy of the Trace Form](/theorems/1580), which implies that the Gram matrix of the trace form on any integral basis has nonzero determinant, and $D_L$ equals $\pm$ this determinant (up to sign). We may therefore cancel $D_L$: \begin{align*} N(\langle \alpha \rangle)^2 = N_{L/\mathbb{Q}}(\alpha)^2. \end{align*} Both sides are integers. Taking the positive square root of both sides and using $N(\langle \alpha \rangle) > 0$, \begin{align*} N(\langle \alpha \rangle) = |N_{L/\mathbb{Q}}(\alpha)|. \end{align*} [guided] **Equating the two computations.** Steps 3 and 4 both compute $\Delta(\beta_1, \ldots, \beta_n)$, giving the two expressions \begin{align*} N(\langle \alpha \rangle)^2\, D_L = \Delta(\beta_1, \ldots, \beta_n) = N_{L/\mathbb{Q}}(\alpha)^2\, D_L. \end{align*} **Cancelling $D_L$.** Is $D_L \neq 0$? Yes: by the [Non-Degeneracy of the Trace Form](/theorems/1580), the bilinear form $(x, y) \mapsto \operatorname{Tr}_{L/\mathbb{Q}}(xy)$ on $L \times L$ is non-degenerate, and the discriminant $D_L = \det(\operatorname{Tr}_{L/\mathbb{Q}}(\alpha_i' \alpha_j'))$ is exactly the determinant of its Gram matrix on an integral basis. Non-degeneracy forces this determinant to be nonzero. Dividing both sides by $D_L$: \begin{align*} N(\langle \alpha \rangle)^2 = N_{L/\mathbb{Q}}(\alpha)^2. \end{align*} **Extracting square roots.** Both sides are integers. An equation $a^2 = b^2$ with $a, b \in \mathbb{Z}$ gives $a = \pm b$. Which sign? We know $N(\langle \alpha \rangle) = |\mathcal{O}_L/\langle \alpha \rangle| > 0$ (a positive integer, being the cardinality of a nontrivial finite group), while $N_{L/\mathbb{Q}}(\alpha) \in \mathbb{Z}$ can be negative. So the positive square root of each side yields \begin{align*} N(\langle \alpha \rangle) = |N_{L/\mathbb{Q}}(\alpha)|. \end{align*} **The absolute value is necessary.** Example: $L = \mathbb{Q}(i)$, $\alpha = 2 + i$. Then $N_{L/\mathbb{Q}}(2+i) = (2+i)(2-i) = 5 > 0$ and $N(\langle 2 + i \rangle) = 5$ agree directly. But taking $\alpha' = -(2+i)$ gives $N_{L/\mathbb{Q}}(\alpha') = (-(2+i))(-(2-i)) = 5$ and $\langle \alpha' \rangle = \langle 2+i \rangle$, also $5$. The absolute value reflects that the ideal norm forgets the sign of the generator. [/guided] [/step]