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Well, to quote from Wolfram MathWorld directly,

Given an affine variety $V$ in the $n$-dimensional affine space $K^n$, where $K$ is an algebraically closed field, the coordinate ring of $V$ is the quotient ring $K[V] = K[x_1 , \dots , x_n] / I(V)$.

My question is simply this, why is it called the "coordinate ring"? In what sense does this ring give or can be said to define coordinates on the affine variety?

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    $\begingroup$ $\mathbb{C}[X,Y]/(Y^2-X^3+aX+b)$ is the coordinate ring of the elliptic curve $E : y^2 = x^3-ax-b$. It is also the ring of polynomial functions $E \to \mathbb{C}$. Its fraction field is the field of rational functions $E \to \mathbb{C}$. Writing them as $\mathbb{C}[x,y]$ and $\mathbb{C}(x,y)$ with $x,y$ satisfying the algebraic relation $E$ helps seeing what it means. $\endgroup$
    – reuns
    Commented Aug 17, 2017 at 18:58
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    $\begingroup$ It's the ring generated by the coordinate functions $x_i$. $\endgroup$ Commented Aug 17, 2017 at 19:14
  • $\begingroup$ Sorry, I'm not quite getting what you're trying to tell me. You wouldn't mind elaborating a little further? The coordinates of the affine variety, or of the space the affine variety exist in, or...? $\endgroup$ Commented Aug 17, 2017 at 19:16
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    $\begingroup$ I gave the motivating example : the ring of polynomial functions $V \to K$, generated by the $x_1,\ldots,x_n$ variables satisfying the algebraic relation of $V$. For a projective curve or variety the definitions are a little bit more complicated, the function field contains only rationals functions invariant under $(x,y,z) \mapsto (\lambda x,\lambda y,\lambda z)$. $\endgroup$
    – reuns
    Commented Aug 17, 2017 at 19:24
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    $\begingroup$ Aaaaaaaaaah. Yeah, thanks! Took a little while for the puzzle pieces to fall into place in my head. But, yes. Just as we define a variety in terms of polynomials, if we want to look at features of a variety, we have to define those in terms of the polynomials that exist specifically on that variety. Hence the motivation for finding the coordinate ring in the first place, and by coordinates, you mean "those particular coordinates which are in the variety". It suddenly makes sense! :) $\endgroup$ Commented Aug 17, 2017 at 20:03

2 Answers 2

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A lot of algebraic geometry is developed by analogy with differential geometry.

In differential geometry, we speak of manifolds rather than varieties. The main feature of the definition of a manifold is that it is covered by coordinate charts: open sets $U$ together with a (smooth) homeomorphism $x : U \to \mathbb{R}^n$ (or alternatively, to an open subset of $\mathbb{R}^n$). The individual components of $x$ are called coordinates.

As an example, consider the pair of functions $(x,y)$ defined on the Euclidean plane after choosing coordinate axes.


The situation here is not too different. On $\mathbb{A}^n_K$, affine $n$-space over a field $K$, the standard 'coordinate chart' has coordinates given by the $n$ variables. Since we're doing algebra, the 'functions' we can make out of these are polynomials.

The term "coordinate function" here refers to a function we can build out of coordinates — much like the ring of smooth continuous real-valued functions on a manifold. The coordinate ring is the ring of all such functions. (although, I don't think the phrase "coordinate function" tends to be used this way in differential geometry)

For subvarieties $V \subseteq \mathbb{A}^n_K$, we understand coordinate functions to be restrictions of coordinate functions on $\mathbb{A}^n_K$, thus we use the quotient ring of the coordinate ring of $\mathbb{A}^n_K$.

Furthermore, since every element of $K[V]$ can be viewed as the image of $t$ under a unique $K$-homomorphism $K[t] \to K[V]$, we can actually identify coordinate functions on $V$ with algebraic maps $V \to \mathbb{A}^1_K$.

(if we're doing absolute geometry, the target would instead be $\mathbb{A}^1_\mathbb{Z} = \operatorname{Spec} \mathbb{Z}[t]$)

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The $i$th "coordinate function" on $K^n$ is the function $K^n\to K$ taking $(x_1,\dots,x_n)$ to $x_i$, its $i$th coordinate. Thinking of elements of $K[x_1,\dots,x_n]$ as functions on $K^n$ in the usual way, $x_i$ is the $i$th coordinate function. The ring $K[V]$ is then the $K$-algebra of functions on $V$ generated by the restrictions of all of these coordinate functions to $V$. So, as the algebra generated by the coordinate functions, it is the "coordinate ring".

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