In the course of reading a paper , I've encountered the following property of interest.

If $R$ is a ring, say it satisfies (*) if: For any smooth, irreducible $R$-algebra $B$ of finite type such that all the fibers of $Spec B$ over points of codimension one in $Spec R$ are irreducible, then $(B \otimes_R K)^* = B^* K^*$, where $K$ is the fraction field of $R$.

The author remarks that it is easy to verify property ( * ) for UFDs. However, I don't see how to do this. What's a proof UFDs satisfy ( * )?

The application I'm interested in is actually where $R$ is a DVR. I feel that in this case, one should be able to give an even simpler argument.

Here is the motivation: we have some rational function $f$ on the generic fiber of an abelian scheme, so it's invertible away from its divisor $D$ on the generic fiber. We want to multiply $f$ by an element of $K^*$ to extend it to the complement over $\overline{D}$ on the entire abelian scheme.

  • $\begingroup$ Sorry, the fraction field. $\endgroup$
    – Tony
    Jan 5, 2013 at 17:26

1 Answer 1


Here is a sketch. The hypothesis, when $R$ is a ufd says that any prime $p\in R$ remains a prime in $B$. So, if $a,b\in B\otimes K$ are units such that $ab=1$, then clearing denominators (from $R$), we get an equation $a'b'=f$ where $a',b'\in B, 0\neq f\in R$. Further we may assume that no prime in $R$ divides $a'$ or $b'$. If $f$ is not a unit in $R$, then pick a prime dividing it and then it will divide either $a'$ or $b'$, which is contrary to our assumption. Thus $f$ is a unit and so $a',b'$ are units in $B$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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