[proofplan] The strategy is to sandwich $\mathfrak{a}$ between two free $\mathbb{Z}$-modules of the same rank — namely $N(\mathfrak{a})\mathcal{O}_L \subseteq \mathfrak{a} \subseteq \mathcal{O}_L$ — and invoke the structure theorem for subgroups of $\mathbb{Z}^n$ of finite index, which provides an upper-triangular change-of-basis matrix. Choosing an integral basis of $\mathcal{O}_L$ converts the question into one about sublattices of $\mathbb{Z}^n$, where the lemma produces a $\mathbb{Z}$-basis $\alpha_1, \ldots, \alpha_n$ of $\mathfrak{a}$ whose change-of-basis determinant has absolute value $|\mathcal{O}_L/\mathfrak{a}| = N(\mathfrak{a})$. The discriminant identity $\Delta(\alpha_1, \ldots, \alpha_n) = \det(A)^2 \Delta(\alpha_1', \ldots, \alpha_n')$ under a change-of-basis matrix $A$, combined with $\Delta(\alpha_1', \ldots, \alpha_n') = D_L$ by definition, yields the formula $\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L$. [/proofplan]

[step:Sandwich $\mathfrak{a}$ between two free $\mathbb{Z}$-modules of rank $n$]By [Norm Lies in the Ideal](/theorems/1592), $N(\mathfrak{a}) \in \mathfrak{a} \cap \mathbb{Z}$. Since $\mathfrak{a}$ is an ideal of $\mathcal{O}_L$ and $N(\mathfrak{a}) \in \mathfrak{a}$, we have $N(\mathfrak{a}) \cdot \beta \in \mathfrak{a}$ for every $\beta \in \mathcal{O}_L$, so \begin{align*} N(\mathfrak{a})\mathcal{O}_L \subseteq \mathfrak{a} \subseteq \mathcal{O}_L. \end{align*} By the [Existence of an Integral Basis](/theorems/1581), there exist $\alpha_1', \ldots, \alpha_n' \in \mathcal{O}_L$ such that \begin{align*} \mathcal{O}_L = \alpha_1'\mathbb{Z} \oplus \cdots \oplus \alpha_n'\mathbb{Z}, \end{align*} making $\mathcal{O}_L$ a free $\mathbb{Z}$-module of rank $n = [L:\mathbb{Q}]$. Under the $\mathbb{Z}$-module isomorphism \begin{align*} \varphi: \mathcal{O}_L &\xrightarrow{\sim} \mathbb{Z}^n \\ \sum_{i=1}^n c_i \alpha_i' &\mapsto (c_1, \ldots, c_n), \end{align*} the chain becomes \begin{align*} N(\mathfrak{a}) \mathbb{Z}^n \subseteq \varphi(\mathfrak{a}) \subseteq \mathbb{Z}^n. \end{align*}[/step]

[guided]We need to show $\mathfrak{a}$ is a free $\mathbb{Z}$-module of rank exactly $n$. The natural ambient is $\mathcal{O}_L$, which is known to be free of rank $n$ over $\mathbb{Z}$ by the [Existence of an Integral Basis](/theorems/1581); we must show $\mathfrak{a}$ sits inside $\mathcal{O}_L$ with finite quotient, so that the rank does not drop. **Lower bound on $\mathfrak{a}$.** By the [Norm Lies in the Ideal](/theorems/1592), the ideal norm $N(\mathfrak{a}) = |\mathcal{O}_L/\mathfrak{a}|$ belongs to $\mathfrak{a} \cap \mathbb{Z}$. Ideals are closed under multiplication by ring elements, so for every $\beta \in \mathcal{O}_L$ the product $N(\mathfrak{a}) \cdot \beta$ lies in $\mathfrak{a}$. This gives $N(\mathfrak{a})\mathcal{O}_L \subseteq \mathfrak{a}$. **Upper bound on $\mathfrak{a}$.** By the definition of an ideal of $\mathcal{O}_L$, $\mathfrak{a} \subseteq \mathcal{O}_L$. Combining, \begin{align*} N(\mathfrak{a})\mathcal{O}_L \subseteq \mathfrak{a} \subseteq \mathcal{O}_L. \end{align*} **Translating to $\mathbb{Z}^n$.** Fix an integral basis $\alpha_1', \ldots, \alpha_n'$ of $\mathcal{O}_L$ and consider the $\mathbb{Z}$-linear isomorphism \begin{align*} \varphi: \mathcal{O}_L &\xrightarrow{\sim} \mathbb{Z}^n \\ \sum_{i=1}^n c_i \alpha_i' &\mapsto (c_1, \ldots, c_n). \end{align*} Because $\varphi$ is a $\mathbb{Z}$-module isomorphism, subgroups correspond bijectively under $\varphi$, and $\varphi(N(\mathfrak{a})\mathcal{O}_L) = N(\mathfrak{a})\mathbb{Z}^n$. Let $M := \varphi(\mathfrak{a}) \subseteq \mathbb{Z}^n$. Then \begin{align*} N(\mathfrak{a})\mathbb{Z}^n \subseteq M \subseteq \mathbb{Z}^n. \end{align*} The quotient $\mathbb{Z}^n/M \cong \mathcal{O}_L/\mathfrak{a}$ has order $N(\mathfrak{a})$, which is finite: that is the leverage we need for the next step.[/guided]

[step:Apply the structure theorem for finite-index subgroups of $\mathbb{Z}^n$ to produce an upper-triangular basis]We apply the following standard result from the theory of finitely generated abelian groups: [claim:Finite-index sublattices of $\mathbb{Z}^n$ admit upper-triangular bases] Let $M \leq \mathbb{Z}^n$ be a subgroup such that $\mathbb{Z}^n/M$ is finite. Then $M$ is a free $\mathbb{Z}$-module of rank $n$, and there exist a $\mathbb{Z}$-basis $v_1, \ldots, v_n$ of $M$ and an ordering of the standard basis $e_1, \ldots, e_n$ of $\mathbb{Z}^n$ such that the matrix $A \in \mathbb{Z}^{n \times n}$ expressing $(v_1, \ldots, v_n)$ in terms of $(e_1, \ldots, e_n)$ is upper triangular. Moreover, \begin{align*} |\mathbb{Z}^n/M| = |\det A|. \end{align*} [/claim] [proof] This is the Hermite Normal Form statement for $\mathbb{Z}$-lattices. Since $N(\mathfrak{a})\mathbb{Z}^n \subseteq M$, the subgroup $M$ has finite index and in particular is a free $\mathbb{Z}$-module of rank $n$ (the containment $N(\mathfrak{a})\mathbb{Z}^n \subseteq M$ forces rank $\geq n$; $M \subseteq \mathbb{Z}^n$ forces rank $\leq n$). The row-reduction procedure over $\mathbb{Z}$ — applying elementary column operations on the matrix whose columns generate $M$ — transforms any generating matrix into upper-triangular form with positive diagonal entries while preserving the $\mathbb{Z}$-span. The resulting columns $v_1, \ldots, v_n$ give the desired basis, and the matrix $A$ they produce is upper triangular. For the index formula, the quotient $\mathbb{Z}^n/M$ is the cokernel of multiplication by $A$; for an upper-triangular integer matrix this cokernel decomposes as \begin{align*} \mathbb{Z}^n/M \cong \bigoplus_{i=1}^n \mathbb{Z}/A_{ii}\mathbb{Z}, \end{align*} which has order $\prod_i |A_{ii}| = |\det A|$. [/proof] Applying the claim to $M = \varphi(\mathfrak{a})$, we obtain $\mathbb{Z}$-basis vectors $v_1, \ldots, v_n$ of $\varphi(\mathfrak{a})$ with \begin{align*} |\det A| = |\mathbb{Z}^n/\varphi(\mathfrak{a})| = |\mathcal{O}_L/\mathfrak{a}| = N(\mathfrak{a}). \end{align*} Pulling back through $\varphi^{-1}$, we obtain $\alpha_1, \ldots, \alpha_n \in \mathfrak{a}$ with $\alpha_i := \varphi^{-1}(v_i)$, and \begin{align*} \mathfrak{a} = \alpha_1\mathbb{Z} \oplus \cdots \oplus \alpha_n\mathbb{Z}. \end{align*} The relation between the two bases is \begin{align*} \alpha_i = \sum_{j=1}^n A_{ji}\, \alpha_j', \end{align*} where $A = (A_{ji})$ is the upper-triangular matrix produced by the claim.[/step]

[guided]**Applying the structure theorem.** We have placed $\varphi(\mathfrak{a})$ as a subgroup of $\mathbb{Z}^n$ with finite index $|\mathcal{O}_L/\mathfrak{a}| = N(\mathfrak{a})$. The claim above then applies directly: it gives a $\mathbb{Z}$-basis $v_1, \ldots, v_n$ of $\varphi(\mathfrak{a})$ such that the matrix $A$ expressing $(v_i)$ in terms of the standard basis $(e_i)$ is upper triangular, with \begin{align*} |\det A| = |\mathbb{Z}^n/\varphi(\mathfrak{a})| = N(\mathfrak{a}). \end{align*} **Why upper triangular is important.** The upper-triangularity is not itself used downstream — what matters for the discriminant computation is only that $A$ is an integer matrix with $|\det A| = N(\mathfrak{a})$. We record the upper-triangular form because it is part of the standard HNF statement, but the determinant identity is the load-bearing conclusion. **Pulling back to $\mathcal{O}_L$.** Set $\alpha_i := \varphi^{-1}(v_i) \in \mathcal{O}_L$. Since $\varphi$ is a $\mathbb{Z}$-linear isomorphism of $\mathcal{O}_L$ onto $\mathbb{Z}^n$ sending $\mathfrak{a}$ bijectively to $\varphi(\mathfrak{a})$, the elements $\alpha_1, \ldots, \alpha_n$ form a $\mathbb{Z}$-basis of $\mathfrak{a}$. That is, \begin{align*} \mathfrak{a} = \alpha_1\mathbb{Z} \oplus \cdots \oplus \alpha_n\mathbb{Z}. \end{align*} In particular $\mathfrak{a}$ is a free $\mathbb{Z}$-module of rank $n$, which is half of statement 1 of the theorem. **Recording the change-of-basis matrix.** The relation $v_i = \sum_j A_{ji} e_j$ in $\mathbb{Z}^n$ translates under $\varphi^{-1}$ to \begin{align*} \alpha_i = \sum_{j=1}^n A_{ji}\, \alpha_j', \end{align*} with the same matrix $A$. This is the bridge to the discriminant computation in Step 4.[/guided]

[step:Verify that $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Q}$-basis of $L$]The elements $\alpha_1, \ldots, \alpha_n \in L$, being $\mathbb{Z}$-linear combinations of $\alpha_1', \ldots, \alpha_n'$ via the matrix $A$, span the $\mathbb{Q}$-subspace $A(\alpha_1' \mathbb{Q} + \cdots + \alpha_n' \mathbb{Q}) \subseteq L$. Since $\alpha_1', \ldots, \alpha_n'$ is a $\mathbb{Q}$-basis of $L$ (it is a $\mathbb{Z}$-basis of $\mathcal{O}_L$ and $L = \operatorname{Frac}(\mathcal{O}_L)$ by the [Fraction Field of the Ring of Integers](/theorems/1571)), and $\det A \neq 0$ (since $|\det A| = N(\mathfrak{a}) \geq 1$), the linear map $A$ is an isomorphism of $\mathbb{Q}^n$ onto itself, and therefore $\alpha_1, \ldots, \alpha_n$ is also a $\mathbb{Q}$-basis of $L$.[/step]

[guided]We have two bases in play: $\alpha_1', \ldots, \alpha_n'$ is a $\mathbb{Z}$-basis of $\mathcal{O}_L$, and $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Z}$-basis of $\mathfrak{a}$. We want to show the latter is also a $\mathbb{Q}$-basis of $L$. **$\alpha_1', \ldots, \alpha_n'$ is a $\mathbb{Q}$-basis of $L$.** By the [Fraction Field of the Ring of Integers](/theorems/1571), $L$ is the fraction field of $\mathcal{O}_L$. Since $\mathcal{O}_L = \bigoplus \alpha_i'\mathbb{Z}$ and localising commutes with direct sums, \begin{align*} L = \alpha_1'\mathbb{Q} \oplus \cdots \oplus \alpha_n'\mathbb{Q}, \end{align*} so $\alpha_1', \ldots, \alpha_n'$ is a $\mathbb{Q}$-basis of $L$. **Transporting to $\alpha_i$.** The $\mathbb{Q}$-linear map \begin{align*} T_A: \mathbb{Q}^n &\to L \\ (c_1, \ldots, c_n) &\mapsto \sum_{i=1}^n c_i \alpha_i \end{align*} factors as $T_A = T_{A'} \circ A$, where $A: \mathbb{Q}^n \to \mathbb{Q}^n$ is the matrix from Step 2 (now viewed over $\mathbb{Q}$), and $T_{A'}: \mathbb{Q}^n \to L$, $(c) \mapsto \sum c_i \alpha_i'$ is an isomorphism by the previous paragraph. **Invertibility of $A$ over $\mathbb{Q}$.** $A$ is invertible over $\mathbb{Q}$ iff $\det A \neq 0$. From Step 2, $|\det A| = N(\mathfrak{a})$. Since $\mathfrak{a} \neq 0$ is a nonzero ideal, $\mathcal{O}_L/\mathfrak{a}$ is a nontrivial finite ring, hence $N(\mathfrak{a}) \geq 1$ and $\det A \neq 0$. **Conclusion.** $T_A$ is a composition of two $\mathbb{Q}$-isomorphisms, so $T_A$ is itself a $\mathbb{Q}$-isomorphism. Equivalently, $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Q}$-basis of $L$, completing statement 1.[/guided]

[step:Compute the discriminant via the change-of-basis formula]Fix any field embeddings $\sigma_1, \ldots, \sigma_n: L \hookrightarrow \mathbb{C}$, given by the [Embeddings of a Number Field](/theorems/1576). Recall that for any $\mathbb{Q}$-basis $\beta_1, \ldots, \beta_n$ of $L$, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) := \det(\sigma_i(\beta_j))^2. \end{align*} Writing $\alpha_i = \sum_{j=1}^n A_{ji}\, \alpha_j'$ (the relation from Step 2), linearity of $\sigma_i$ gives \begin{align*} \sigma_i(\alpha_k) = \sum_{j=1}^n A_{jk}\, \sigma_i(\alpha_j'). \end{align*} In matrix form, if $M = (\sigma_i(\alpha_j')) \in \mathbb{C}^{n \times n}$ and $M' = (\sigma_i(\alpha_j)) \in \mathbb{C}^{n \times n}$, then $M' = M \cdot A$. Taking determinants, \begin{align*} \det M' = \det M \cdot \det A, \end{align*} and squaring both sides, \begin{align*} \Delta(\alpha_1, \ldots, \alpha_n) = (\det M')^2 = (\det A)^2 (\det M)^2 = (\det A)^2 \Delta(\alpha_1', \ldots, \alpha_n'). \end{align*} By Step 2, $(\det A)^2 = N(\mathfrak{a})^2$. By the definition of the [field discriminant](/page/Field%20Discriminant), $\Delta(\alpha_1', \ldots, \alpha_n') = D_L$. Substituting: \begin{align*} \Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L. \end{align*}[/step]

[guided]**The discriminant is a quadratic form in the basis matrix.** For any $\mathbb{Q}$-basis $\beta_1, \ldots, \beta_n$ of $L$, and any tuple of embeddings $\sigma_1, \ldots, \sigma_n: L \hookrightarrow \mathbb{C}$, \begin{align*} \Delta(\beta_1, \ldots, \beta_n) = \det(\sigma_i(\beta_j))^2. \end{align*} The matrix on the right squares nicely under a change of basis, as we now verify. **Change of basis under $A$.** From Step 2, $\alpha_k = \sum_j A_{jk}\, \alpha_j'$. Applying the $\mathbb{Q}$-linear embedding $\sigma_i$ (embeddings are automorphisms over $\mathbb{Q}$, hence $\mathbb{Q}$-linear), \begin{align*} \sigma_i(\alpha_k) = \sum_{j=1}^n A_{jk}\, \sigma_i(\alpha_j'). \end{align*} Introduce the $n \times n$ complex matrices $M_{ij} := \sigma_i(\alpha_j')$ and $M'_{ik} := \sigma_i(\alpha_k)$. The sum above reads \begin{align*} M'_{ik} = \sum_{j=1}^n M_{ij}\, A_{jk}, \end{align*} i.e., $M' = M \cdot A$ as matrix products. **Multiplicativity of $\det$ and squaring.** The determinant is multiplicative: $\det(M') = \det(M)\, \det(A)$. Squaring both sides and using the definitions, \begin{align*} \Delta(\alpha_1, \ldots, \alpha_n) = (\det M')^2 = (\det M)^2 (\det A)^2 = \Delta(\alpha_1', \ldots, \alpha_n') \cdot (\det A)^2. \end{align*} **Identifying both factors.** - $\Delta(\alpha_1', \ldots, \alpha_n') = D_L$: by the definition of the field discriminant, $D_L$ is the discriminant of any integral basis of $\mathcal{O}_L$. That this value is independent of the chosen integral basis follows from the same change-of-basis formula applied to two integral bases: the matrix relating them has determinant $\pm 1$ (since both directions are integer matrices), so its square is $1$. - $(\det A)^2 = N(\mathfrak{a})^2$: from Step 2, $|\det A| = N(\mathfrak{a})$, so $(\det A)^2 = N(\mathfrak{a})^2$. **Substituting,** \begin{align*} \Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2\, D_L. \end{align*} This completes statement 2. **Why the identity matters.** The identity is a bridge between two kinds of discriminants: the intrinsic $D_L$, which measures the arithmetic of $\mathcal{O}_L$, and the discriminant $\Delta(\alpha_1, \ldots, \alpha_n)$ of an arbitrary chosen basis of $\mathfrak{a}$. The multiplicative gap is exactly $N(\mathfrak{a})^2$ — the square of the index. This identity is the key technical tool behind [Square-Free Discriminant Criterion](/theorems/1595).[/guided]