Let $K$ be an algebraic closed field and $G=GL(n, K)$. There is a Hopf algebra structure on $K[G]=K[T_{11}, T_{12}, \ldots, T_{nn}, d^{-1}]$, $d=\det (T_{ij})$ given by $e^*(T_{ij})=\delta_{ij}$, $\mu^*(T_{ij})=\sum_{h} T_{ih} \otimes T_{hj}$, $\iota^*(T_{ij})=(-1)^{i+j}d^{-1}\det(T_{rs})_{r\neq j, s\neq i}$. My question is:
(1) Why $K[G]=K[T_{11}, T_{12}, \ldots, T_{nn}, d^{-1}]$?
(2) Why $e^*(T_{ij})=\delta_{ij}$, $\mu^*(T_{ij})=\sum_{h} T_{ih} \otimes T_{hj}$, $\iota^*(T_{ij})=(-1)^{i+j}d^{-1}\det(T_{rs})_{r\neq j, s\neq i}$?
I know that if we have an algebraic group $G$, with multiplication $\mu: G\times G \to G$, inverse $\iota: G \to G$, and identity map $e: \{e\} \to G$, then we have $\mu^*: K[G] \to K[G] \otimes K[G]$, $f \mapsto f\circ \mu$, $\iota^*: K[G] \to K[G]$, $f \mapsto f\circ \iota$, $e^{*}: K[G] \to K$, $f \mapsto f(e)$.
We have $e^*(T_{ij})=T_{ij}(e)$. Here $e$ is the identity matrix in $GL(n, K)$. Why $T_{ij}(e)=\delta_{ij}$?
Thank you very much.
I think your notation already raises some issues. I'd prefer
$$K[G] = K[ t_{11}, \dots, t_{nn}, d] / < d \cdot \det(t_{ij}) -1 >$$
$K$ can be a unital commutative ring, need not be algebraically closed.
I recommend to look up any book on linear algebraic groups. The equivalence of the categories of $K$-Hopf algebras and linear algebraic groups is general and explicit.
Here are some references:
http://en.wikipedia.org/wiki/Group_scheme