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.

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.

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 $(B \otimes_R K)^* = B^* K^*$.

The author remarks that it is easy to verify property (*) for UFDs. I think I can sort of see the reasoning for PIDs: If $f \in (B \otimes_R K)^* $, we want to show that we may modify $f$ by some $c$ in $K^*$ such that for all primes $\mathfrak{p}$ of $(B \otimes_R K)$, $cf \notin \mathfrak{p}$. The image of the point $\mathfrak{p}$ under the map to $Spec R$ is some principal prime $(\alpha)$, so we modify by the appropriate power of $\alpha$.

What's a proof that UFDs satisfy (*)?

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.