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replaced R by mathbb{Z} and made clear(er) that the d_i are rationals
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Johannes Hahn
Johannes Hahn
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Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$$GL_M(\mathbb{Z})$ and $GL_N(R)$$GL_N(\mathbb{Z})$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$$d_1, d_2,\dotsc\in\mathbb{Q}$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}RQ^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R^T\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$ and $GL_N(R)$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}RQ^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R^T\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(\mathbb{Z})$ and $GL_N(\mathbb{Z})$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc\in\mathbb{Q}$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}RQ^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R^T\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

added 2 characters in body
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Amritanshu Prasad
Amritanshu Prasad
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Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$ and $GL_N(R)$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}QR^{-1}$$D = P^{-1}RQ^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R\mathbb Z^M/\mathbb Z^N$$R^T\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$ and $GL_N(R)$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}QR^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$ and $GL_N(R)$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}RQ^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R^T\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.

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Amritanshu Prasad
Amritanshu Prasad
  • 5.7k
  • 1
  • 38
  • 54

Put $R$ in Smith normal form. While this is usually defined for integer matrices, for a rational matrix $R$, we may write $R = P D Q$, where $P$ and $Q$ are in $GL_M(R)$ and $GL_N(R)$ respectively, and $D$ is a diagonal matrix with diagonal entries $d_1, d_2,\dotsc$ such that $d_{i+1}$ is an integer multiple of $d_i$ for $i = 1, 2, \dotsc$.

Now note that replacing $R$ by $D = P^{-1}QR^{-1}$ does not change the isomorphism class of $R\mathbb Z^N/\mathbb Z^M$ or $R\mathbb Z^M/\mathbb Z^N$ (since $Q^{-1}\mathbb Z^N = \mathbb Z^N$ and $P^{-1T}\mathbb Z^M = \mathbb Z^M$).

Since $D$ is diagonal, $D\mathbb Z^N/\mathbb Z^N$ and $D^T \mathbb Z^M/\mathbb Z^M$ are both $\bigoplus_i (d_i \mathbb Z/\mathbb Z)$, and therefore isomorphic.