Points of \(\mathbb{G}_m\) Are the Units of \(R\)
Let \(k\) be a field and \(R\) a commutative \(k\)-algebra. There is a natural isomorphism of abelian groups
\[ \mathbb{G}_{m,\operatorname{Spec} k}(R) \;\cong\; R^\times, \]
where \(R^\times\) denotes the group of multiplicative units of \(R\).
Proof
The \(R\)-points of \(\mathbb{G}_m\) over \(\operatorname{Spec} k\) are \(k\)-algebra homomorphisms \(k[T, T^{-1}] \to R\). Such a homomorphism is determined by the image of \(T\), which must be a unit (since \(T \cdot T^{-1} = 1\) forces the image of \(T^{-1}\) to be the inverse of the image of \(T\)). Conversely, any unit \(u \in R^\times\) gives such a homomorphism by sending \(T \mapsto u\) and \(T^{-1} \mapsto u^{-1}\). Hence
\[ \mathbb{G}_m(R) \;\cong\; \operatorname{Hom}_{k\text{-Alg}}(k[T, T^{-1}], R) \;\cong\; R^\times. \]
To verify the bijection is a group isomorphism, evaluate the
multiplication on pairs \((u, v) \in R^\times \times R^\times\). The
comultiplication \(\Delta(T) = T \otimes T\) sends \(T\) in the composed
hom to \(uv\), recovering the multiplication of \(R^\times\). The
identity corresponds to \(\varepsilon(T) = 1\), and the inverse to
\(S(T) = T^{-1}\). All three match the multiplicative group structure on
\(R^\times\).
\(\square\)