Affine $k$-domain, which is also a PID and whose group of units is same as that of $k$

Question

Let $k$ be an algebraically closed field of characteristic zero. Let $R$ be a finitely generated $k$-algebra such that $R$ is a non-field PID and $R^*=k^*$, where $R^*$ denotes the group of units of $R$. Then is it true that $R=k[f]$ for some $f \in k[X]$, i.e., $R$ is isomorphic to $k[X]$ as $k$-algebras ?

Answer 1

I recently found this old question, and it bugged me that there seems to be no elementary solution, so here is my attempt at a "minimally complicated" solution. I hope this is of use for someone :)

Let $X = \operatorname{Spec}(R)$ be the affine variety corresponding to $R$, this is a smooth affine curve since $R$ is a PID (and PID's are regular and of dimension 1). Let $\overline{X} \supset X$ be its smooth compactification, it is the projective curve obtained by embedding $X \hookrightarrow \mathbb{A}^n \subset \mathbb{P}^n$, closing up and taking the normalization.

Let $\overline{X} \setminus X = \{p_0,\dots,p_r\}$, then we have the following right-exact sequence of Weil divisor class groups [essentially Hartshorne, Prop. II.6.5]

$$\bigoplus_{i=0}^r \mathbb{Z}\cdot p_i \to \operatorname{Cl}(\overline{X}) \to \operatorname{Cl}(X) \to 0.$$

The kernel of the left map are exactly those linear combinations of the $p_i$ which are principal divisors in $\operatorname{Div}(\overline X)$. In other words, they correspond to rational functions $0 \neq f \in K(\overline{X})$ which have zeros and poles only in $p_0,\dots,p_r$, so elements in $\mathcal{O}_{\overline X}(X)^\times$. Since $X = \operatorname{Spec} R$ is affine, this is exactly the group of units $R^\times$. Two rational functions define the same divisor if and only if they are the same up to a scalar in $k^\times$ (this mildly uses $\overline{k} = k$). Summing up, we have an exact sequence

$$0 \to k^\times \to R^\times \to \bigoplus_{i=0}^r \mathbb{Z}\cdot p_i \to \operatorname{Cl}(\overline{X}) \to \operatorname{Cl}(X) \to 0.$$

This shows that the condition $R^\times = k^\times$ is equivalent to $\mathbb{Z}^{r+1} \to \operatorname{Cl}(\overline{X})$ being injective. On the other hand, for a smooth affine curve $X$ we have that $\operatorname{Cl}(X) = 0$ if and only if $X$ is a PID, see for example [Hartshorne, Prop. II.6.2]. So in this sense $X$ being a PID is equivalent to $\mathbb{Z}^{r+1} \to \operatorname{Cl}(\overline{X})$ being surjective. Under your assumptions we thus see that $\operatorname{Cl}(\overline X) \cong \mathbb{Z}^{r+1}$, and hence the subgroup of degree $0$ divisors satisfies $\operatorname{Cl}^0(\overline X) \cong \mathbb{Z}^r$.

This is where some non-trivial facts about smooth projective curves enter (and I'm not sure how to avoid this). The group $\operatorname{Cl}^0(\overline X)$ is in bijection to the (closed points of the) Jacobian $\operatorname{Jac}(\overline X)$, an abelian variety; an irreducible projective algebraic group of dimension $g = g(X)$. We would like to argue that $\mathbb{Z}^r$ is only an abelian variety if $r=0$. This can be done for example by noting that abelian varieties are divisible groups (as suggested by Mohan), or by the fact that for $\operatorname{char}(k) \nmid n$ the group of $n$-torsion points is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{2g}$. If $k$ is uncountable, then one can also argue that if $g = \dim \operatorname{Jac}(\overline X) \geq 1$, then it must be uncountable. This is best seen over $\mathbb{C}$ if one accepts the description of abelian groups $$\operatorname{Jac}(\overline{X}) \cong \operatorname{H}^1(\overline X,\mathcal{O}_{\overline X}) / \operatorname{H}^1_{\text{sing}}(\overline X, \mathbb{Z}) \cong \mathbb{C}^g/\Lambda, \qquad \Lambda \cong \mathbb{Z}^{2g}.$$

Okay, assume we could somehow show that $r=0$, i.e. $\overline X = X \cup p_0$, then we are almost done. Since $\operatorname{Cl}(\overline X) = \mathbb{Z}$, any two points $p,q$ are linearly equivalent, and the rational function having a simple zero at $p$ and pole at $q$ gives an isomorphism $\phi \colon \overline X \to \mathbb{P}^1$ [Hartshorne, Exa. II.6.10.1] (alternatively this also follows from $g=0$). So $X \cong \mathbb{P}^1 \setminus \phi(p_0) \cong \mathbb{A}^1$ (see for example here, this can fail if $k$ is not algebraically closed). This concludes $R = \mathcal{O}_{X}(X) \cong \mathcal{O}_{\mathbb{A}^1}(\mathbb{A^1}) = k[x]$.