# Is there a $0$ in the Smith Normal Form?

Let $R$ be a PID.

Let $M$ be a $R$-module with generating set $\{m_1, m_2, ..., m_n\}$.

Let $R^{(n)}=\{f\mid f:\{1, 2, ..., n\}\to R\}$ be the free module with basis $$\{e_i\in R^{(n)}\mid e_i=(0, 0, ..., 0, 1_{i\text{th position}}, 0, ..., 0, 0), ~1\leq i\leq n\}.$$

Define $\phi:R^{(n)}\to M$ by $\phi(e_i)=m_i$. That is, $\phi\left(\sum_{i=1}^{n}r_i e_i\right)=\sum_{i=1}^{n}r_i m_i$. Then $\phi$ is an onto $R$-module homomorphism.

Since $K=\ker{\phi}$ is a submodule of the free module $R^{(n)}$, we know that $K$ has a finite linearly independent generating set $\{k_1, k_2, ..., k_m\}$, where $m\leq n$.

Suppose that $$(k_1, k_2, ..., k_m)=(e_1, e_2, ..., e_n)A.$$ for some $A\in M_{n\times m}(R)$.

The textbook (Jacobson's Algebra I and Goodman's Algebra) state that there exist invertible matrices $P\in M_n(R)$ and $Q\in M_m(R)$ such that $$PAQ=\text{diag}(\underbrace{d_1, d_2, ..., d_s, 0, ..., 0}_{\min{(n, m)}\text{ terms}}), ~r\leq \min{(n, m)}=m.$$

I have tried some concrete examples, for example, $M=\Bbb{Z}_2\oplus \Bbb{Z}_3\oplus \Bbb{Z}$ over $\Bbb{Z}$. But I can't find an example which has a $0$ in its smith normal form $\text{diag}(d_1, d_2, ..., d_s, 0, 0, ..., 0)$.

Could someone give an example $\phi$ which has a $0$ in the Smith Normal Form?

Edit: If $A$ is a $m$-by-$n$ matrix and $a_{ij}=0$ for every $i\neq j$, then we write $A=\text{diag}(a_{11}, a_{22}, ..., a_{kk})$, where $k=\min{(m, n)}$. For example, $$\begin{pmatrix} a_{11} & 0\\ 0 & 0\\ \hline 0 & 0\\ \end{pmatrix}=\text{diag}(a_{11},0).$$ $$\begin{pmatrix} a_{11} & 0 & 0\\ 0 & a_{22} & 0\\ 0 & 0 & 0\\ \end{pmatrix}=\text{diag}(a_{11},a_{22},0).$$ $$\begin{pmatrix} a_{11} & 0 & 0 & 0 & 0 & 0\\ 0 & a_{22} & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{pmatrix}=\text{diag}(a_{11},a_{22},0,0).$$

• The matrix $A$ has rank $m$ (for its columns are linearly independent), so clearly $PAQ$ is also a $n \times m$ matrix of rank $m$. I'm not sure what you mean by $\diag(\ldots)$ for a non-square matrix. – arkeet Oct 26 '16 at 4:33
• @arkeet Thank you for your hint. I got it! If $\{k_1, k_2, ..., k_m\}$ is linearly independent, then there has no $0$ in $\text{diag}(...)$. That why I can't find a proper example. Because I alway choose a linearly independent generating set of $K$. – bfhaha Oct 26 '16 at 9:13

1. $V=\Bbb{Z}_2\oplus \Bbb{Z}_3\oplus \Bbb{Z}\oplus \Bbb{Z}$
2. Define $\phi:\Bbb{Z}^{(4)}\to V$ by naturally.
3. $K=\ker{\phi}=\{r(2, 0, 0, 0)+s(0, 3, 0, 0)\mid r, s\in \Bbb{Z}\}$
4. $K=\langle (2, 0, 0, 0), (0, 3, 0, 0), (2, 3, 0, 0)\rangle$. (We choose a Non-linearly independent generating set of $K$ artificially.)
5. $A=\begin{pmatrix} 2 & 0 & 2\\ 0 & 3 & 3\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$
6. $PAQ=\begin{pmatrix} -1 & 1 & 0 & 0\\ 3 & -2 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\cdot \begin{pmatrix} 2 & 0 & 2\\ 0 & 3 & 3\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\cdot \begin{pmatrix} 1 & 3 & -1\\ 1 & 2 & -1\\ 0 & 0 & 1 \end{pmatrix}= \begin{pmatrix} 1 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 0\\ \hline 0 & 0 & 0 \end{pmatrix}= \text{diag}(1, 6, 0)$.