# Smith Normal Forms over Fields and PIDs

I need to reduce the following matrices into the Smith Normal form over the field $(\mathbb{Z}/2\mathbb{Z})[x]$:

$$M_{1} = \left ( \begin{array}{ccc} x & 1 & 0 \\ 0 & x & 1 \\ 0 & 0 & x \end{array} \right)$$ and $$M_{2}= \left ( \begin{array}{ccc} x & 0 & 0 \\ 0 & x & 1 \\ 0 & 0 & x \end{array} \right) .$$

Do I need to calculate it in the same way as with principal ideal domains, I am kind of lost since I have found that according to Dummit, the SNF is the matrix $M -xI$ when we have a field $F$, whereas in other sources it is performed by column, row operations in the original matrix when we have a PID?

When I calculate the matrix $B_{i} = M_{i}-xI$ for $i=1,2$ I get that the characteristic polynomial is equal to the minimal polynomial and are $C_{B_{i}}(x) = 0 = m_{B_{i}}(x)$. So this means that $M_{1}, M_{2}$ are similar?

Thanks

• No, Dummit and Foote define it the same way. Theorem 21, p. 479: "the $n\times n$ matrix $xI - A$ with entries from $F[x]$ can be put into the diagonal form (called the Smith Normal Form for $A$) with monic nonzero elements $a_1(x), a_2(x), \ldots , a_m(x)$ of $F[x]$ with degrees at least one and satisfying $a_1(x) \mid a_2(x) \mid \cdots \mid a_m(x)$." – André 3000 Feb 10 '15 at 22:28
• Also $\mathbb{Z}/2\mathbb{Z}[x]$ is not a field, but it is a PID. – André 3000 Feb 10 '15 at 22:37

$R=(\mathbb{Z}/2\mathbb{Z})[x]$ is a PID (not a field!) and $M_1,M_2\in M_3(R)$, so you have to compute the SNF of these matrices. (It's not hard to see that $SNF(M_1)=\mathrm{diag}(1,1,x^3)$, and $SNF(M_2)=\mathrm{diag}(1,x,x^2)$, so they are not equivalent.)

For some reasons which I don't understand Dummit and Foote call the SNF of a matrix $A\in M_n(F)$, where $F$ is a field, the SNF of its characteristic matrix $xI-A$ which belongs to $M_n(F[x])$. (Note that $F[x]$ is a PID and one can talk about the SNF of a matrix whose entries are in $F[x]$.)
I have to mention that usually we call the invariant factors of a matrix $A\in M_n(F)$ as being the (non-constant) invariant factors of $xI-A$, but this is another story. (A matrix over a field has also a SNF, but this has only $1$'s and $0$'s on the main diagonal.)