Points of \(\operatorname{GL}_n\) Are Invertible \(n \times n\) Matrices over \(R\)
Let \(k\) be a field, \(n\) a positive integer, and \(R\) a commutative \(k\)-algebra. There is a natural isomorphism of groups
\[ \operatorname{GL}_{n, \operatorname{Spec} k}(R) \;\cong\; \operatorname{GL}_n(R), \]
where the right-hand side is the group of \(n \times n\) matrices over \(R\) with invertible determinant under multiplication.
Proof
The \(R\)-points of \(\operatorname{GL}_n\) over \(\operatorname{Spec} k\) are \(k\)-algebra homomorphisms \(k[x_{ij}][\det^{-1}] \to R\). Such a homomorphism is determined by the images \(r_{ij} \in R\) of the indeterminates \(x_{ij}\), subject to the condition that the image of \(\det^{-1}\) is the multiplicative inverse of \(\det(r_{ij})\); equivalently, the matrix \((r_{ij}) \in M_n(R)\) must have an invertible determinant in \(R\). Hence
\[ \operatorname{GL}_n(R) \;\cong\; \{(r_{ij}) \in M_n(R) \,:\, \det(r_{ij}) \in R^\times\} \;=\; \operatorname{GL}_n(R). \]
To verify the bijection is a group isomorphism, compute the product of
two matrix points \((r_{ij}), (s_{ij}) \in \operatorname{GL}_n(R)\)
under the comultiplication
\(\Delta(x_{ij}) = \sum_k x_{ik} \otimes x_{kj}\). The image of
\(x_{ij}\) under the composed hom is \(\sum_k r_{ik} \cdot s_{kj}\),
which is the \((i, j)\)-entry of the matrix product \((r_{ij})(s_{ij})\).
The identity corresponds to \(\varepsilon(x_{ij}) = \delta_{ij}\),
giving the identity matrix; the inverse to \(S(x_{ij}) = \det^{-1} \cdot
C_{ji}\), giving the matrix inverse. All three match the standard group
structure on \(\operatorname{GL}_n(R)\).
\(\square\)