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 $B$.


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