[proofplan] We present $M$ as a quotient of a free module $R^n$ by a submodule of relations $K$. Since $R$ is a PID, $K$ is itself free. The relation matrix $A$ representing the inclusion $K \hookrightarrow R^n$ can be diagonalised via the Smith normal form: there exist invertible matrices $P$ and $Q$ such that $PAQ$ is diagonal with entries $d_1, \ldots, d_s, 0, \ldots, 0$ satisfying $d_1 \mid d_2 \mid \cdots \mid d_s$. The coordinate change induced by $P$ and $Q$ converts the quotient $R^n / K$ into a direct sum of cyclic modules $R/(d_i)$ and free copies of $R$. Uniqueness follows from the fact that the invariant factors are determined by the sequence of gcd-ideals of minors of $A$, which are invariants of $M$ itself. [/proofplan]

[step:Present $M$ as a quotient of a free module by a free submodule of relations]Since $M$ is finitely generated, choose generators $m_1, \ldots, m_n \in M$ and define the surjective $R$-module homomorphism \begin{align*} \pi: R^n &\to M \\ (r_1, \ldots, r_n) &\mapsto r_1 m_1 + \cdots + r_n m_n. \end{align*} Let $K := \ker(\pi) \le R^n$. By the [First Isomorphism Theorem for Modules](/theorems/862), $M \cong R^n / K$. Since $R$ is a PID, every submodule of a free $R$-module of finite rank is itself free of rank at most $n$ (this is the [Submodules of Free Modules over a PID](/theorems/???) theorem). Therefore $K$ is a free $R$-module of rank $m$ for some $0 \leq m \leq n$. Choose a basis $k_1, \ldots, k_m$ of $K$. Each $k_j$ is a vector in $R^n$, so we may write $k_j = \sum_{i=1}^n a_{ij} e_i$ where $e_1, \ldots, e_n$ is the standard basis of $R^n$. Define the **relation matrix** $A \in R^{n \times m}$ by $(A)_{ij} = a_{ij}$, so that the columns of $A$ are the coordinate vectors of the basis elements of $K$.[/step]

[guided]Our goal is to understand the structure of $M$ by expressing it as a quotient of a known object. Since $M$ is finitely generated, say by elements $m_1, \ldots, m_n$, we can build a surjection from the free module $R^n$ onto $M$. Define \begin{align*} \pi: R^n &\to M \\ (r_1, \ldots, r_n) &\mapsto r_1 m_1 + \cdots + r_n m_n. \end{align*} This is an $R$-module homomorphism because $M$ is an $R$-module and scalar multiplication distributes over addition. It is surjective because $m_1, \ldots, m_n$ generate $M$: every element of $M$ is an $R$-linear combination of these generators. Let $K := \ker(\pi)$. By the [First Isomorphism Theorem for Modules](/theorems/862), $M \cong R^n / K$. The key question is: what does $K$ look like? Since $R$ is a PID, the [Submodules of Free Modules over a PID](/theorems/???) theorem guarantees that every submodule of a free $R$-module of finite rank is itself free of rank at most that of the ambient module. Therefore $K \cong R^m$ as an $R$-module for some $0 \leq m \leq n$. Why does this matter? Because the problem of classifying $M$ is now reduced to understanding the quotient $R^n / K$ where both $R^n$ and $K$ are free. The inclusion $K \hookrightarrow R^n$ is described by the relation matrix $A \in R^{n \times m}$: choosing a basis $k_1, \ldots, k_m$ of $K$, each $k_j = \sum_{i=1}^n a_{ij} e_i \in R^n$, the matrix $A$ has columns equal to the coordinate vectors of the $k_j$ with respect to the standard basis $e_1, \ldots, e_n$ of $R^n$.[/guided]

[step:Reduce the relation matrix to Smith normal form]Since $R$ is a PID, the [Smith Normal Form](/theorems/861) theorem applies to the matrix $A \in R^{n \times m}$: there exist an invertible matrix $P \in GL_n(R)$ and an invertible matrix $Q \in GL_m(R)$ such that \begin{align*} PAQ = D := \begin{pmatrix} d_1 & & & & 0 & \cdots & 0 \\ & d_2 & & & 0 & \cdots & 0 \\ & & \ddots & & \vdots & & \vdots \\ & & & d_s & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 0 & 0 & \cdots & 0 \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ 0 & \cdots & 0 & 0 & 0 & \cdots & 0 \end{pmatrix}, \end{align*} where $d_1, \ldots, d_s \in R$ are nonzero elements satisfying the divisibility chain $d_1 \mid d_2 \mid \cdots \mid d_s$, and $s = \operatorname{rank}(A)$. The matrices $P$ and $Q$ are products of elementary row and column operations over $R$ (adding a multiple of one row/column to another, multiplying a row/column by a unit, swapping rows/columns), each of which is invertible over $R$.[/step]

[guided]We want to simplify the relation matrix $A \in R^{n \times m}$ as much as possible using operations that do not change the isomorphism type of the quotient $R^n / K$. The operations available are: (1) changing the basis of $R^n$ (row operations on $A$), and (2) changing the basis of $K \cong R^m$ (column operations on $A$). The question is: how simple can we make $A$ using these operations over a PID? The answer is the [Smith Normal Form](/theorems/861) theorem: for any matrix $A \in R^{n \times m}$ over a PID $R$, there exist invertible matrices $P \in GL_n(R)$ and $Q \in GL_m(R)$ such that \begin{align*} PAQ = D := \begin{pmatrix} d_1 & & & & 0 & \cdots & 0 \\ & d_2 & & & 0 & \cdots & 0 \\ & & \ddots & & \vdots & & \vdots \\ & & & d_s & 0 & \cdots & 0 \\ 0 & \cdots & 0 & 0 & 0 & \cdots & 0 \\ \vdots & & \vdots & \vdots & \vdots & & \vdots \\ 0 & \cdots & 0 & 0 & 0 & \cdots & 0 \end{pmatrix}, \end{align*} where $d_1, \ldots, d_s \in R$ are nonzero elements satisfying the divisibility chain $d_1 \mid d_2 \mid \cdots \mid d_s$, and $s = \operatorname{rank}(A)$. Why does this work over a PID but not over an arbitrary ring? The key is that in a PID, any two elements $a, b \in R$ have a gcd expressible as an $R$-linear combination: $\gcd(a, b) = \alpha a + \beta b$ for some $\alpha, \beta \in R$. This Bezout property allows us to perform row and column operations that replace a pair of entries by their gcd and zero — the analogue of Gaussian elimination, but using gcd reduction instead of division. By repeatedly applying this process, we can clear all off-diagonal entries, arriving at the diagonal form $D$. The invertible matrices $P$ and $Q$ are products of three types of elementary matrices over $R$: (i) swapping two rows or columns, (ii) multiplying a row or column by a unit $u \in R^\times$, and (iii) adding an $R$-multiple of one row or column to another. Each of these operations is invertible over $R$ (swaps are self-inverse, unit scaling is inverted by $u^{-1}$, and adding $\lambda$ times one row to another is inverted by subtracting $\lambda$ times that row). Therefore $P \in GL_n(R)$ and $Q \in GL_m(R)$. The divisibility chain $d_1 \mid d_2 \mid \cdots \mid d_s$ is not an accident but a structural feature: $d_t$ equals the gcd of all $t \times t$ minors of $A$ divided by the gcd of all $(t-1) \times (t-1)$ minors. This characterisation also guarantees the uniqueness of the $d_i$ up to units, a fact we will exploit in the uniqueness argument below.[/guided]

[step:Perform the change of basis to diagonalise the quotient $R^n / K$]The matrix $P$ defines an $R$-module automorphism $\varphi_P: R^n \to R^n$ via left multiplication by $P$, and $Q$ defines an automorphism $\varphi_Q: R^m \to R^m$. Since $K$ is the image of the map $R^m \to R^n$ defined by $v \mapsto Av$, the change of basis gives \begin{align*} K = A(R^m) = P^{-1} D Q^{-1} (R^m) = P^{-1} D(R^m), \end{align*} where the last equality uses the fact that $Q^{-1}$ is an automorphism of $R^m$ and hence $Q^{-1}(R^m) = R^m$. Define the new basis $f_1, \ldots, f_n$ of $R^n$ by $f_i := P^{-1} e_i$, i.e., the columns of $P^{-1}$. In this basis, $K$ is spanned by $d_1 f_1, d_2 f_2, \ldots, d_s f_s$. Since $\varphi_P$ is an automorphism of $R^n$, it induces an isomorphism \begin{align*} R^n / K \cong R^n / D(R^m). \end{align*} The submodule $D(R^m) \le R^n$ is generated by the columns of $D$, which are $d_1 e_1, d_2 e_2, \ldots, d_s e_s$. Since $R^n = R e_1 \oplus R e_2 \oplus \cdots \oplus R e_n$ and $D(R^m) = (d_1) e_1 \oplus (d_2) e_2 \oplus \cdots \oplus (d_s) e_s \oplus 0 \oplus \cdots \oplus 0$, the quotient decomposes as a direct sum of cyclic modules: \begin{align*} R^n / D(R^m) &\cong \frac{R e_1}{(d_1) e_1} \oplus \frac{R e_2}{(d_2) e_2} \oplus \cdots \oplus \frac{R e_s}{(d_s) e_s} \oplus R e_{s+1} \oplus \cdots \oplus R e_n \\ &\cong R/(d_1) \oplus R/(d_2) \oplus \cdots \oplus R/(d_s) \oplus R^{n-s}. \end{align*}[/step]

[guided]We now translate the Smith normal form of the relation matrix into a decomposition of $M$. The key idea is that invertible changes of basis in $R^n$ and $R^m$ do not alter the isomorphism type of the quotient $R^n / K$. The inclusion $K \hookrightarrow R^n$ factors through the map $\iota_A: R^m \to R^n$ defined by $\iota_A(v) = Av$, whose image is $K = A(R^m)$. The Smith normal form $PAQ = D$ can be rewritten as $A = P^{-1} D Q^{-1}$, so \begin{align*} K = A(R^m) = P^{-1} D Q^{-1}(R^m) = P^{-1} D(R^m), \end{align*} where we used the fact that $Q^{-1}: R^m \to R^m$ is an automorphism, hence $Q^{-1}(R^m) = R^m$. The automorphism $\varphi_P: R^n \to R^n$ given by $v \mapsto Pv$ sends $K = P^{-1} D(R^m)$ to $D(R^m)$. Since an automorphism of $R^n$ sends quotients to isomorphic quotients, we obtain \begin{align*} M \cong R^n / K \cong R^n / D(R^m). \end{align*} Now we read off the structure. The submodule $D(R^m)$ is the image of $R^m$ under the diagonal map $D$, hence it is generated by the columns of $D$: $d_1 e_1, d_2 e_2, \ldots, d_s e_s$ (the remaining $n - s$ columns are zero). Since these generators lie in distinct direct summands of $R^n = Re_1 \oplus \cdots \oplus Re_n$, the quotient splits: \begin{align*} R^n / D(R^m) \cong \bigoplus_{i=1}^{s} Re_i / (d_i) e_i \oplus \bigoplus_{j=s+1}^{n} Re_j \cong \bigoplus_{i=1}^{s} R/(d_i) \oplus R^{n-s}. \end{align*} Each summand $Re_i / (d_i) e_i \cong R/(d_i)$ via the isomorphism $r e_i + (d_i) e_i \mapsto r + (d_i)$. The summands $Re_j$ for $j > s$ have no relations imposed and contribute free copies of $R$.[/guided]

[step:Discard unit invariant factors and identify the free rank and torsion part]Among $d_1, \ldots, d_s$, some may be units of $R$. If $d_i \in R^\times$, then $(d_i) = R$ and $R/(d_i) = 0$, so the corresponding summand vanishes. Discard all unit invariant factors. Relabel the remaining nonunit elements as $d_1, \ldots, d_k$ (preserving the divisibility chain $d_1 \mid d_2 \mid \cdots \mid d_k$). The number of discarded unit factors is $s - k$, and the free rank becomes $r := n - s + (s - k) = n - k$. We conclude \begin{align*} M \cong R^r \oplus R/(d_1) \oplus R/(d_2) \oplus \cdots \oplus R/(d_k), \end{align*} where $r = n - k \geq 0$ and $d_1 \mid d_2 \mid \cdots \mid d_k$ with each $d_i$ nonzero and nonunit. The summand $R^r$ is the **free part** and $\bigoplus_{i=1}^k R/(d_i)$ is the **torsion part** of $M$ (each element of $R/(d_i)$ is annihilated by $d_i$, hence by $d_k$).[/step]

[guided]Not all diagonal entries $d_1, \ldots, d_s$ need be nonunits. If $d_i$ is a unit in $R$, then the ideal $(d_i)$ equals $R$, so $R/(d_i) \cong R/R = 0$: that summand contributes nothing. The divisibility chain ensures that if $d_i$ is a unit, then $d_j$ is also a unit for all $j \leq i$ (since $d_j \mid d_i$ and units divide only units up to units, but more precisely: $d_j \mid d_i$ and $d_i \in R^\times$ means $(d_j) \supseteq (d_i) = R$, so $(d_j) = R$ and $d_j \in R^\times$). Therefore the unit factors form an initial segment $d_1, \ldots, d_{s-k}$ and the nonunit factors are $d_{s-k+1}, \ldots, d_s$. After discarding the zero summands and relabeling the remaining nonunit invariant factors as $d_1, \ldots, d_k$, the decomposition becomes \begin{align*} M \cong R^{n-s} \oplus R^{s-k} \oplus R/(d_1) \oplus \cdots \oplus R/(d_k) = R^{n-k} \oplus \bigoplus_{i=1}^k R/(d_i). \end{align*} Setting $r := n - k$, this is the decomposition claimed in the theorem. The torsion submodule $\operatorname{Tor}(M) := \{m \in M : rm = 0 \text{ for some nonzero } r \in R\}$ is precisely $\bigoplus_{i=1}^k R/(d_i)$, since each element of $R/(d_i)$ is annihilated by $d_i \neq 0$, and the free part $R^r$ has no torsion: if $r \cdot (a_1, \ldots, a_r) = 0$ with $r \neq 0$ in the integral domain $R$, then each $a_j = 0$.[/guided]

[step:Prove uniqueness of the free rank $r$]Let $F := \operatorname{Frac}(R)$ denote the fraction field of the integral domain $R$. The functor $- \otimes_R F$ is exact (since $F$ is a flat $R$-module, being a localisation of $R$). Applying it to the decomposition: \begin{align*} M \otimes_R F \cong F^r \oplus \bigoplus_{i=1}^k R/(d_i) \otimes_R F. \end{align*} For each $i$, $R/(d_i) \otimes_R F = 0$: the element $\overline{1} \otimes (a/b)$ in $R/(d_i) \otimes_R F$ equals $\overline{1} \otimes (d_i \cdot a / (d_i b)) = \overline{d_i} \otimes (a/(d_i b)) = \overline{0} \otimes (a/(d_i b)) = 0$, since $\overline{d_i} = 0$ in $R/(d_i)$. Therefore \begin{align*} M \otimes_R F \cong F^r. \end{align*} The dimension $\dim_F(M \otimes_R F) = r$ is an invariant of $M$ alone (independent of the chosen decomposition), so $r$ is uniquely determined.[/step]

[guided]To show that the rank $r$ is independent of the decomposition, we use a standard technique: **extend scalars** to the fraction field to kill all torsion. Let $F := \operatorname{Frac}(R)$ denote the field of fractions of $R$. Since $R$ is an integral domain, $F$ exists and is a flat $R$-module (localisation is always flat). The functor $- \otimes_R F$ therefore preserves exact sequences and distributes over direct sums. Applying it to the decomposition $M \cong R^r \oplus \bigoplus_{i=1}^k R/(d_i)$: \begin{align*} M \otimes_R F \cong (R \otimes_R F)^r \oplus \bigoplus_{i=1}^k (R/(d_i) \otimes_R F) \cong F^r \oplus \bigoplus_{i=1}^k (R/(d_i) \otimes_R F). \end{align*} We claim each torsion summand vanishes: $R/(d_i) \otimes_R F = 0$. To verify this, take any pure tensor $\overline{a} \otimes (p/q) \in R/(d_i) \otimes_R F$ where $\overline{a} \in R/(d_i)$ and $p/q \in F$. Then \begin{align*} \overline{a} \otimes \frac{p}{q} = \overline{a} \otimes \frac{d_i p}{d_i q} = \overline{a} \cdot d_i \otimes \frac{p}{d_i q} = \overline{a d_i} \otimes \frac{p}{d_i q} = \overline{0} \otimes \frac{p}{d_i q} = 0. \end{align*} The key step is pulling the scalar $d_i$ across the tensor: $\overline{a} \otimes (d_i \cdot x) = \overline{a} \cdot d_i \otimes x = \overline{ad_i} \otimes x$, and $\overline{ad_i} = 0$ in $R/(d_i)$ because $ad_i \in (d_i)$. Therefore $M \otimes_R F \cong F^r$, an $F$-vector space of dimension $r$. Since $M \otimes_R F$ depends only on $M$ (not on the decomposition), and the dimension of a vector space over a field is unique, the rank $r$ is an invariant of $M$.[/guided]

[step:Prove uniqueness of the invariant factors via the gcd-ideal characterisation]For each $1 \leq t \leq \min(n, m)$, define the **$t$-th determinantal ideal** $\Delta_t(A) \trianglelefteq R$ to be the ideal generated by all $t \times t$ minors of the relation matrix $A$. Since elementary row and column operations do not change the ideal generated by minors of a given size (each operation multiplies the set of minors by a unit or takes $R$-linear combinations), $\Delta_t(A) = \Delta_t(D)$ where $D = PAQ$ is the Smith normal form. The $t \times t$ minors of the diagonal matrix $D$ are the products $d_{i_1} d_{i_2} \cdots d_{i_t}$ for $1 \leq i_1 < i_2 < \cdots < i_t \leq s$ (all other $t \times t$ minors are zero). Since $d_1 \mid d_2 \mid \cdots \mid d_s$, we have $d_j \mid d_{i_j}$ for each $1 \leq j \leq t$ whenever $j \leq i_j$ (because $i_1 < \cdots < i_t$ forces $i_j \geq j$, and the divisibility chain gives $d_j \mid d_{i_j}$). Therefore $d_1 d_2 \cdots d_t \mid d_{i_1} d_{i_2} \cdots d_{i_t}$ for every choice of $1 \leq i_1 < \cdots < i_t \leq s$, so the product $d_1 d_2 \cdots d_t$ generates $\Delta_t(D)$. Therefore \begin{align*} \Delta_t(A) = (d_1 d_2 \cdots d_t) \quad \text{for } 1 \leq t \leq s, \qquad \Delta_t(A) = (0) \quad \text{for } t > s. \end{align*} The individual invariant factors are recovered by the formula \begin{align*} (d_t) = \Delta_t(A) / \Delta_{t-1}(A) \quad \text{for } 1 \leq t \leq s, \end{align*} where $\Delta_0(A) := (1) = R$, and the quotient of ideals means: $d_t$ is the element such that $\Delta_t(A) = d_t \cdot \Delta_{t-1}(A)$, which is well-defined up to units because $R$ is a PID and $\Delta_{t-1}(A) \mid \Delta_t(A)$. Now, $\Delta_t(A)$ depends only on the submodule $K \le R^n$, not on the choice of basis for $K$ (a different basis $k_1', \ldots, k_m'$ produces a matrix $A' = A Q'$ for some $Q' \in GL_m(R)$, and $\Delta_t(AQ') = \Delta_t(A)$ by the Cauchy-Binet formula and the invertibility of $Q'$). Since $K = \ker(\pi)$ is determined by $M$ and the surjection $\pi$, it remains to verify that $\Delta_t(A)$ is independent of the presentation. Any two presentations of $M$ are related by elementary row operations (automorphisms of $R^n$), elementary column operations (automorphisms of $K$), and stabilisation (bordering $A$ with a new row and column whose intersection is $1$ and other new entries are $0$, corresponding to adding a redundant generator). The first two preserve $\Delta_t$ by the Cauchy-Binet argument above. For stabilisation, expanding a $t \times t$ minor of the bordered matrix along the new row shows it either equals a $t \times t$ minor of $A$ or a $(t-1) \times (t-1)$ minor of $A$; the latter lies in $\Delta_{t-1}(A) \subseteq \Delta_t(A)$, so $\Delta_t$ is unchanged. Therefore the sequence $(d_1), (d_2), \ldots, (d_k)$ is an invariant of $M$. Combining with the uniqueness of $r$ from the preceding step, this completes the proof: both the free rank $r$ and the invariant factor ideals $(d_1), \ldots, (d_k)$ are determined by $M$ alone, and since $R$ is a PID each $(d_i)$ is principal, so $d_i$ is unique up to multiplication by a unit of $R$.[/step]

[guided]The uniqueness of the invariant factors is deeper than the uniqueness of the rank. We need to show that the ideals $(d_1), \ldots, (d_k)$ are determined by $M$ alone, not by the particular presentation $M \cong R^n / K$. The key tool is the **determinantal ideal** (also called the **Fitting ideal**). For each integer $t \geq 1$, define $\Delta_t(A)$ to be the ideal of $R$ generated by all $t \times t$ minors of $A$. We claim these ideals are invariants of the presentation $R^n \xrightarrow{A} R^n$ and ultimately of $M$ itself. **Step 1: Determinantal ideals are unchanged by elementary operations.** The Smith normal form gives $D = PAQ$ where $P \in GL_n(R)$ and $Q \in GL_m(R)$. We need $\Delta_t(A) = \Delta_t(D)$. By the Cauchy-Binet formula, a $t \times t$ minor of $PAQ$ is an $R$-linear combination of $t \times t$ minors of $A$ (with coefficients coming from minors of $P$ and $Q$), and vice versa (since $A = P^{-1} D Q^{-1}$). Since $P$ and $Q$ are invertible over $R$, the ideals generated by the $t \times t$ minors of $A$ and $D$ coincide: $\Delta_t(A) = \Delta_t(D)$. **Step 2: Read off the determinantal ideals of $D$.** The diagonal matrix $D$ has diagonal entries $d_1, \ldots, d_s, 0, \ldots, 0$. A $t \times t$ submatrix of $D$ has nonzero determinant only if it is a diagonal submatrix, in which case its determinant is a product $d_{i_1} \cdots d_{i_t}$ for some $1 \leq i_1 < \cdots < i_t \leq s$. The divisibility chain $d_1 \mid \cdots \mid d_s$ ensures that $d_1 \cdots d_t$ divides every other such product: for any $1 \leq i_1 < \cdots < i_t \leq s$, we have $i_j \geq j$ for each $j$, so the divisibility chain gives $d_j \mid d_{i_j}$, and therefore $d_1 \cdots d_t \mid d_{i_1} \cdots d_{i_t}$. Conversely, $d_1 \cdots d_t$ is itself a $t \times t$ minor of $D$ (the upper-left $t \times t$ block). Therefore $\Delta_t(D) = (d_1 \cdots d_t)$ for $t \leq s$ and $\Delta_t(D) = (0)$ for $t > s$. **Step 3: Recover individual invariant factors.** Setting $\Delta_0(A) := R$, the $t$-th invariant factor satisfies \begin{align*} (d_t) = \Delta_t(A) / \Delta_{t-1}(A), \end{align*} meaning $d_t$ is the generator of the principal ideal $\Delta_t(A) \cdot \Delta_{t-1}(A)^{-1}$ (which makes sense in a PID: $\Delta_{t-1}(A) \mid \Delta_t(A)$ since every $(t-1) \times (t-1)$ minor divides a cofactor expansion of a $t \times t$ minor). **Step 4: Independence of presentation.** We must show that $\Delta_t(A)$ depends only on $M$, not on the choice of generators or relations. Suppose $M \cong R^{n'} / K'$ is a different presentation with relation matrix $A' \in R^{n' \times m'}$. The two presentations are related by a sequence of three types of operations on the relation matrix: (1) elementary row operations (corresponding to automorphisms of $R^n$, i.e., change of generators of the ambient free module), (2) elementary column operations (corresponding to automorphisms of $R^m$, i.e., change of basis of the relation submodule $K$), and (3) stabilisation — adding a zero row and zero column bordered by a $1$ (corresponding to adding a redundant generator $m_{n+1} = 0$ and the relation $m_{n+1} = 0$, which enlarges the presentation without changing $M$). Operations (1) and (2) preserve $\Delta_t$ by the Cauchy-Binet argument from Step 1: left- or right-multiplication by an invertible matrix replaces each $t \times t$ minor by an $R$-linear combination of $t \times t$ minors, and the invertibility ensures the ideals coincide. Operation (3) also preserves $\Delta_t$: if $B$ is obtained from $A$ by bordering with a row and column whose intersection entry is $1$ (and all other new entries are $0$), then the $t \times t$ minors of $B$ include all $t \times t$ minors of $A$ plus additional minors involving the new row and column, but expanding along the new row shows each such minor equals a $(t-1) \times (t-1)$ minor of $A$, which already lies in $\Delta_{t-1}(A) \subseteq \Delta_t(A)$. Therefore $\Delta_t(B) = \Delta_t(A)$. Since any two presentations of $M$ are connected by a finite sequence of these operations and their inverses, $\Delta_t$ is an invariant of $M$. The invariant factors, recovered as $(d_t) = \Delta_t / \Delta_{t-1}$, are therefore uniquely determined up to units.[/guided]