I recently asked a question about the regular functions in $\mathbf{GL}(n,\mathbb{C})$ but now that I have read the appendix I am again confused; here is my past question: Understanding regular functions of $\mathbf{GL}(n,\mathbb{C})$.
Again, I want to know formally how the matrix entry functions $x_{i,j}$ and the function $\text{det}(x)^{-1}$ generate the algebra of regular functions $\mathcal{O}[\mathbf{GL}(n,\mathbb{C})]$. I am following the book Symmetry, Representations and Invariants.
First some context. The definition of a polynomial on a vector space:
If $V$ is a finite dimensional $\mathbb{C}$-vector space, a function $f\colon V\to \mathbb{C}$ is a polynomial of degree $\leq k$ if for some basis $\{e_{1},\dots,e_{n}\}$ of $V$ one has $$f\left (\sum_{i=1}^{n}x_{i}e_{i}\right ) = \sum_{|I|\leq k}a_{I}x^{I}.$$ If there is a multi-index $I$ with $|I|=k$ and $a_{I}\neq 0$, then we say $f$ has degree $k$. If $a_{I}=0$ when $|I|\neq k$, we say $f$ is homogeneous of degree $k$. Let $\mathcal{P}(V)$ be the set of all polynomials on $V$.
$\mathcal{P}(V)$ is an algebra, and it is generated by the linear coordinate functions $x_{1},\dots,x_{n}$; thus we have $\mathcal{P}(V)\cong \mathbb{C}[x_{1},\dots,x_{n}]$. Moreover, a subset $X\subseteq V$ is an affine algebraic set if there exist functions $f_{j}\in \mathcal{P}(V)$ such that $$X=\{v\in V\colon\: f_{j}(v) = 0 \text{ for } j=1,\dots,m\}$$
I assume that when a group $G\subseteq V$ is an affine algebraic set, we call it an algebraic group. By this definition I understand why $\mathbf{GL}(n,\mathbb{C})$ is an algebraic group (we take de zero polynomial). Now, the affine ring $\mathcal{O}[X]$ of regular functions on $X$ is define as $$\mathcal{O}[X] = \{f|_{X}\colon \: f\in \mathcal{P}(V)\}.$$ Also, if $J_{X} = \{f\in \mathcal{P}(V)\colon \: f|_{X} = 0\}$, then we have $\mathcal{O}[X]\cong \mathcal{P}(V)/J_{X}$.
So, if the assertion is that $\mathcal{O}[\mathbf{GL}(n,\mathbb{C})]$ is generated by the matrix entry functions $x_{i,j}$ and $\text{det}(x)^{-1}$, as I see it $$ \mathcal{O}[\mathbf{GL}(n,\mathbb{C})] = \{ f|_{\mathbf{GL}(n,\mathbb{C})}\colon \: f\in \mathcal{P}(M_{n}(\mathbb{C})) \} $$
I want to understand why $\text{det}(x)^{-1}$ is necessary and see how it is that these functions generate the algebra.
Another question, $\text{det}(x)^{-1}$ makes sense for invertible matrices, but for $M_{n}(\mathbb{C})$ it does not; is $\text{det}(x)^{-1}$ a polynomial so that it can generate this algebra? As far as I can see, it is not the restriction of a polynomial of $V$.
There's one big problem here: if you want to apply the formula $\mathcal{O}[X] = \{f|_X: f\in \mathcal{P}(V)\}$, you need $X$ to be a closed subvariety of $V$. But $Gl(n,\Bbb C)$ is not closed in $M_{n\times n}(\Bbb C)$! What you have here is instead the complement of the vanishing locus of the function $\det$, which is open. The correct way to see what you're after here is to note that $GL(n,\Bbb C)$ embeds as a closed subvariety in $\Bbb A^{n^2+1}$ with $n^2$ coordinates $x_{ij}$ and one coordinate $t$ as the vanishing locus of $t\cdot\det -1$, and now viewing $t$ as $\det^{-1}$ you have exactly the claim you're looking to prove by the method you mention in your post.