Let $R$ be a Discrete Valutation Ring and let $A$ be a $n \times n$ matrix taking values in $R$. Suppose $A$ has invariant factors $(\alpha\_1, \alpha\_2, \cdots, \alpha\_n)$ (i.e. performing elementary rows and columns moves we get $diag(\alpha\_1, \cdots, \alpha\_n$)). Suppose $U$ is invertible:



> 
> 
> > 
> > Does $AU$ have the same invariant factors $(\alpha\_1, \cdots, \alpha\_n)$?
> > 
> > 
> > 
> 
> 
> 



Invariant Factors of a product $AU$ with $U$ an invertible matrix
I'll assume that elementary row and column transformations have determinant one. Otherwise the question is trivial, as darij grinberg pointed out in his comment above. My answer expands on Mohan's comment in this case.


The answer is **yes** and follows from



> 
> 
> > 
> > **Claim**. Let $R$ be a local ring and let $A$ be an $n$-by-$n$ matrix over $R$. Then the following are equivalent:  
> > 
> >  1. There exist $B, C \in GL\_n(R)$ such that $BAC$ is a diagonal matrix.  
> > 
> >  2. There exist $B, C \in E\_n(R)$ such that $BAC$ is a diagonal matrix. 
> > 
> > 
> > **Corollary.** If $R$ is a Discrete Valuation Ring (DVR), then every matrix $A$ over $R$ can be reduced, by means of elementary row and column transformations, to a diagonal matrix $\text{diag}(\alpha\_1, \dots,\alpha\_d)$ with $\alpha\_1 \notin R^{\times}$ and $\alpha\_i \vert \alpha\_{i + 1}$. The elements $\alpha\_i$ are, up to multiplication by a unit, uniquely determined by the latter conditions; they are called *the invariant factors of $A$*. 
> > 
> > 
> > 
> 
> 
> 


Here we denote by $E\_n(R)$ the subgroup of $GL\_n(R)$ generated by matrices that differ from the identity by a single off-diagonal element.



> 
> 
> > 
> > **Proof of the claim.** Assume that assertion 1 holds, so that $\alpha := BAC$ is diagonal for some $B, C \in GL\_n(R)$. Since $R$ is Generalized Euclidean ring in the sense of P. M. Cohn (see [1, §4], the result is due to Klingenberg, 1962, valid for non-commutative local ring as well), and since $E\_n(R)$ is normalized by invertible matrices in $GL\_n(R)$, we can write $B = DE$ and $C = E'D'$ with $E, E' \in E\_n(R)$ and $D, D'$ are invertible diagonal matrices. Therefore $\beta = EAE'$ is diagonal, which establishes assertion 2. Moreover, $\beta$ is obtained from $\alpha$ by multiplying invertible diagonal matrices. The converse is obvious.
> > 
> > 
> > **Proof of the Corollary.** Since $R$ is a principal ideal domain (PID), assertion 1 of the above claim holds. Indeed, the Smith Normal Form Theorem for PIDs guarantees the existence of $B, C \in GL\_n(R)$ and of the diagonal matrix $\alpha = \text{diag}(\alpha\_1, \dots, \alpha\_n)$ such that $\alpha = BAC$. It guarantees also the the additional divisibility constraints on the invariant factors $\alpha\_i$ of $A$ and their uniqueness, up to multiplication by units.
> > 
> > 
> > 
> 
> 
> 




---


[1] P. Cohn, "On the structure of the $GL\_2$ of a ring", 1966.


