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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.


  • 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!

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up vote 2 down vote accepted

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.

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Sorry, my English is very rough. I understand your contraexemple but I need to cofirm the condition (R:(R:J))=J with J generates by a, b, c and d. – Francisco Perdomo Dec 6 '09 at 2:35
In this case, (R:J) is R, so you are fine. Using your description, you need to show that there is no height one prime containing a, b, c and d, which is also easy. – David Speyer Dec 6 '09 at 2:39
I made a mistake and it must say (R:(R:J))=R. Certainly it seems(R:J)=R and (R:(R:J))=R. If the ring is almost factorial, do you think a similar contraexample go on? (especially with the condition of J in the Gabriel filter). – Francisco Perdomo Dec 6 '09 at 3:30
Based on a quick google search, it looks like "almost factorial" is the same as "Q-factorial"? If so, I think I can still give counter-examples. Look at my update above. – David Speyer Dec 6 '09 at 4:16
I think last example is definitive. Thanks – Francisco Perdomo Dec 8 '09 at 0:52

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