This is an homework exercise of the Algebra lecture.

I need to evaluate the Smith normal form of the following matrix $$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} \in M(3\times 3;\Bbb{Z}[\xi]),$$ where $\xi := \frac{1+\sqrt{-19}}{2}$.

We have seen in the lecture, that similarly as for rings, there exists the Smith Normal Form also for PID's. To solve the problem (we don't need to find $S, T \in M(3 \times 3; \Bbb{Z}[\xi])$, such that $SAT$ is in Smith Normal Form) we applied row, and column operations, paying particular attention to not multiply rows and columns by nonunits, since this is not allowed. (Also according to this answer). The steps that we performed are:

\begin{align*}\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} &\overset{2\text{column}+\xi \text{ first column}}{\leadsto} \begin{pmatrix}1 & 0 & \xi-1\\2 \xi&8+2\xi^2&8\xi+7\\\xi& 4+\xi^2 & 3\xi +2 \end{pmatrix}\\ \overset{3\text{column}-(\xi-1) \text{ first column}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\2 \xi&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix} &\overset{2\text{row}-(2\xi) \text{ first row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\ \overset{3\text{row}-(\xi) \text{ first row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}& \overset{2\text{row}-2 \text{ third row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&0&2\xi+3\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\ \overset{\text{swap columns}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& -\xi^2+4\xi+2 & 4+\xi^2 \end{pmatrix} & \\ \overset{\text{second column $+$ third column}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 4 \xi +6 & 4+\xi^2 \end{pmatrix} & \overset{\text{third row } - 2 \text{ second row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 0 & 4+\xi^2 \end{pmatrix}=: B \end{align*}

Unfortunately the term $b_{33}$ is not divisible by the term $b_{22}$ as it should be. We have done the steps many many times, but we can't find the error (I'm doing this homework exercise with my classmates). Do you maybe find it? Thank you in advance for any help.

  • 3
    $\begingroup$ Although ths Smith normal form exists over any PID, computing it algorithmically requires having effective Bezout coefficients. This is easy in a Euclidean domain, but the PID they gave you is (one of the rare ones that is known to be) not a Euclidean domain. So you will need al alternative method to find Bezout coefficients (at least int the instances your encounter). Try to think of that first; then apply the (final step of) the description in the link. $\endgroup$ Commented Oct 3, 2014 at 7:41

1 Answer 1


$\xi^2=\xi-5$, and therefore $4+\xi^2=\xi-1$ (which is a prime element). On the other side, $2\xi+3=2(\xi-1)+5=2(\xi-1)+\xi-\xi^2=-(\xi-1)(\xi-2)$, so $b_{33}\mid b_{22}$.

The SNF of your matrix is $$\begin{pmatrix}1 & 0 & 0\\0&\xi-1&0\\0& 0 & (\xi-1)(\xi-2) \end{pmatrix}.$$


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