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!


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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Hideyuki Kabayakawa Dec 6 '09 at 2:35
  • $\begingroup$ 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. $\endgroup$ – David E Speyer Dec 6 '09 at 2:39
  • $\begingroup$ 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). $\endgroup$ – Hideyuki Kabayakawa Dec 6 '09 at 3:30
  • $\begingroup$ 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. $\endgroup$ – David E Speyer Dec 6 '09 at 4:16
  • $\begingroup$ I think last example is definitive. Thanks $\endgroup$ – Hideyuki Kabayakawa Dec 8 '09 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.