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I'm reading this set of notes and I'm trying to understand this passage where they explain how to put a topology on $GL_n(R)$ when $R$ is a topological ring, which I am not completely following. The following is the passage (slightly paraphrased):

Let $GL_n$ be defined over $\mathbb{Q}$. Then there is a closed immersion $GL_n \rightarrow gl_n \times \mathbb{G}_a$ given on points in a $\mathbb{Z}$-algebra $R$ by $GL_n(R) \rightarrow gl_n(R) \times \mathbb{G}_a(R)$ where $g \rightarrow (g, \det g^{-1})$. Then identifying $gl_n$ with $\mathbb{G}_a^{n^2}$ we can put a subspace topology on $GL_n(R)$ via this way.

In the notes affine schemes are defined by functor of points so if $X = Spec A$ then $X : Alg_k \rightarrow \ Set$ where $R \rightarrow Hom_{Alg_k}(A, R)$, and $X(R)$ the $R$-valued points of $X$.

What I'm not understanding is:

1) $GL_n$ is defined over $\mathbb{Q}$. So it takes in $\mathbb{Q}$-algabra R; I understand the definition of $GL_n(R)$ when $R$ is a $\mathbb{Q}$-algebra. But how does one make sense of this for $\mathbb{Z}$-algebra $R$ as above?

2) Is the fact that $GL_n \rightarrow gl_n \times \mathbb{G}_a$ is a closed immersion important here? or are we just using the fact that $GL_n(R) \rightarrow gl_n(R) \times \mathbb{G}_a(R)$ where $g \rightarrow (g, \det g^{-1})$ is injective here? (to view it as a subspace of the right hand side)

Any explanation would be appreciated... Thank you.

PS here $\mathbb{G}^n_a:=Spec(\mathbb{Q}[t_1, ..., t_n])$ and $GL_n = Spec(\mathbb{Q}[x_{i,j}][y]/ (det(x_{i,j}) y - 1) )$

PPS Here is the link to the notes I am reading: Chapter 1 https://services.math.duke.edu/~hahn/Chapter1.pdf (Page 9 is where algebraic groups are defined, and in the first sentence of Section 1.5 $k$ is set to be a field.) So I thought algebraic groups were defined only over fields (at least in this notes...)

Chapter 2 https://services.math.duke.edu/~hahn/Chapter2.pdf The passage that I paraphrased is the first paragraph of Section 2.3 page 38. Then on page 40 line -8 they say "Now assume that $G$ is an algebraic group over the global field $F$" so I had guessed that $GL_n$ mentioned on page 38 was defined over a global field $F$ (for simplicity I said $\mathbb{Q}$ above) since it didn't mention where it was defined over.

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    $\begingroup$ Why would $GL_n$ be defined over $\mathbb{Q}$? It seems to me that it is defined over $\mathbb{Z}$ ($GL_n=\mathrm{Spec}\mathbb{Z}[a_{ij},\det^{-1}]$) $\endgroup$
    – Denis Nardin
    Commented Apr 10, 2019 at 16:36
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    $\begingroup$ You refer twice to "these notes" but you don't say which notes, so it's hard to point out the issue. Certainly defining schemes over subrings (not restricted to be subfields) is crucial to simultaneously view them as schemes over fields of different characteristics. $\endgroup$
    – YCor
    Commented Apr 10, 2019 at 20:20
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    $\begingroup$ If it is defined over $\mathbb{Z}$, it is defined over any field by base change. Of course this is the case for $GL_n$. $\endgroup$
    – abx
    Commented Apr 11, 2019 at 11:55
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    $\begingroup$ Note that a $\mathbb{Z}$-algebra is just a ring. The definition you give for $GL_n$ is explicitly the spectrum of a ring that is the base change from the corresponding ring using integer coefficients, as the polynomial $det(x_{ij})y-1$ is in $\mathbb{Z}[x_{ij}][y]$. The notes may just have been written in a slightly sloppy way. $\endgroup$
    – David Roberts ♦
    Commented Apr 11, 2019 at 14:46
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    $\begingroup$ For (2), yes it's necessary that it be a closed immersion. 1. Forget group schemes. 2. Set $n=1$, which covers everything interesting. 3. Directly prove that for a topological ring $R$ that the units in the antidiagonal subspace topology $R^\times\subset R^2$ forms a topological group. 4. For $R$ the adeles, show that the units in the obvious subspace topology $R^\times\subset R=gl_1(R)$ have a different topology, not a topological group. 5. Use this to learn about schemes, not vice versa. $\endgroup$
    – Ben Wieland
    Commented Apr 11, 2019 at 21:51

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