Let R a normal domain, that is an integrally closed noetherian domain, like Dedekind domains, UFD, etc


Let A=(a i j ) a matrix with elements in R and dimension n x m. 


Suppose 


* rank A=1 ↔ all 2 x 2 minors are =0.
* J:= ideal generated by a i j  verify (R:(R:J))=R ↔ J is not included in any prime ideal with height 1.


If R is an UFD, with the preview conditions, we can write A like product of a n x 1 vector column C=(c i ) and a 1 x m vector file F=(f j ), that is a i j =c i ·f j .


I conjecture that is true in the general case, but I cannot make any progress. 


Have you contraexemples with normal rings?


I´m grateful for your advices!



Splitting matrix of rank one
I am having trouble understanding your English. But, if I understand you correctly, the following is a counter-example:


Let $k$ be a field and let $R$ be the ring $k[a,b,c,d]/(ab-cd)$. Then $R$ is normal and $\left( \begin{smallmatrix} a & c \\\\ d & b \end{smallmatrix} \right)$ has rank 1. However, we can not write this matrix as $\left( \begin{smallmatrix} w \\\\ x \end{smallmatrix} \right) \left( \begin{smallmatrix} y & z \end{smallmatrix} \right)$ for any $w$, $x$, $y$, $z \in R$.


I think your condition should almost imply that the ring is a UFD. If I have any non-unique factorization $ab=cd$, I can use it to build a counter-example like this one.


**UPDATE** Here are two more examples: $R=k[a,b,c]/(ac-b^2)$ and $\left( \begin{smallmatrix} a & b \\\\ b & c \end{smallmatrix} \right)$.


$R=\mathbb{Z}[\sqrt{-5}]$ and $\left( \begin{smallmatrix} 2 & 1+\sqrt{-5} \\\\ 1-\sqrt{-5} & 3 \end{smallmatrix} \right)$. 


These examples rule out most attempts I could think of to find a class of rings larger than UFDs for which the result holds.


