The definition of a rational representation of a linear algebraic group
Mia Lopez 
I'm reading Concini-Procesi's "the invariant theory of matrices". My question regards their setup in the introduction, and its relation/translation into the language of modern algebraic geometry.
The setup
Here is the setup they use in $\S$2.3:
Let $F$ be an algebraically closed field. We view $GL_n(F)$ as the affine subvariety of $F^{n^2+1} = M_n(F)\times k$ defined by the pairs $(A,u)$ satisfying $\det(A)u = 1$. Thus its coordinate ring is $F[x_{i,j}][\det^{-1}]$ (polynomial ring in $n^2$ coordinates with the determinant inverted).
They define a linear algebraic group $G$ over $F$ to be a subgroup of $GL_n(F)$ cut out by polynomial equations. It's coordinate ring is defined to be the restriction of the coordinate ring of $GL_n(F)$ to the subvariety $G$, denoted $A[G]$. Thus $A[G]$ is the quotient of $F[x_{i,j}][\det^{-1}]$ by some ideal.
They define an $N$-dimensional rational representation of $G$ to be a homomorphism$$\rho : G\longrightarrow GL_N(F)$$such that the matrix elements $\rho(g)_{h,k}$ for $h,k = 1,\ldots,N$ belong to $A[G]$.
My question
Here, I assume they mean to take $\rho(g)_{h,k}$ as functions on $G$? Is this the same as saying that $\rho$ defines a morphism of affine group schemes?
Wikipedia defines a representation to be rational if $\rho$ defines a rational map of algebraic groups $G\rightarrow GL_N$, so e.g. the morphism needs only be defined on a dense open.
Are these two definitions equivalent?
$\endgroup$1 Answer
$\begingroup$No, a “rational representation” is an honest morphism.
The terminology is in contrast to “polynomial representation” which would mean the morphism extends to the closure of $G$ in the space of matrices $M_n(F)$, i.e. the representation doesn’t involve negative powers of the determinant.
For example the action of $GL_n$ on a one-dimensional vector space by the $n$-th power of the determinant is rational for all $n$ and polynomial for $n\geq 0$.
$\endgroup$
