On the definition of the affine subspace and complex lines

Question

I'm reading "Algebraic Geometry: A problem solving approach" and on page 197 the author mentions:

For a field $k$, the affine n-space over k is the set $$\mathbb{A}^n(k) = \{(a_1,a_2,...,a_n) : a_i \in k \text{ for } i = 1,...,n\} $$

Then it is mentioned that $\mathbb{A}^1(\mathbb{C})$ is referred to as the complex line. But it isn't this just the complex plane? In other words, isn't $\mathbb{A}^1 = \{(a) : a \in \mathbb{C}\} = \mathbb{C}$?

What is the definition of affine n-space saying and what is the difference between $\mathbb{A}^1(k)$ and $k$?

Answer 1

In a first approach, there seems to be very little difference between $k$ and $\mathbb{A}^1(k)$ – or even between $\mathbb{A}^n(k)$ and $k^n$. And in many situations you can indeed think of them as similar (as a guide for intuition, if nothing else). The interest of the notation $\mathbb{A}^n$ is that it turns a set of “scalars” ($k$) into a “variety”, ie some form of “shape”. In this way, $\mathbb{A}^n$ has the same “type” as, say, $\mathbb{P}^n$, or any set of zeroes of any family of polynomials.

Very soon in algebraic geometry, it is understood that the notion of zero set of a family of polynomials is insufficient of itself, or troublesome to manipulate. And so such a set is enriched with a ring (and, in more modern versions, a sheaf of rings because some phenomena need to be studied locally to be understood) of regular functions. Again, the notation $\mathbb{A}^n$ makes the distinction between the scalar values of functions and the topological space made up with the elements of $k^n$, and polynomial functions as regular functions.

When $k=\mathbb{C}$, we then realize, through the Nullstellensatz, that the points of $\mathbb{A}^n$ can be paired up with an invariant directly defined in terms of its ring of regular functions, instead of conflating “points” (geometric) and “values” (algebraic) – every maximal ideal of $k[x_1,\ldots,x_n]$ is some $(x_1-a_1,\ldots,x_n-a_n)$ for some scalars $a_i$. So we can just forget the “point characterization” of $\mathbb{A}^n$ to just remember that it is the space of maximal ideals (this has been outdated since 1960-ish and schemes but let’s ignore that) with the ring of regular functions $k[x_1,\ldots,x_n]$ – and recall that shorthand for “the maximal ideal $(x_1-a_1,\ldots,x_n-a_n)$” is “the point $(a_1,\ldots,a_n)$”.

There’s an issue with this definition, though. A huge issue: this creates “phantom” points when the base field isn’t algebraically closed. For instance, consider $\mathbb{A}^1$ on $k=\mathbb{R}$ (and try and think of the same on $\mathbb{Q}$ if you’re brave enough). With our new definitions, the points are the maximal ideals of $\mathbb{R}[x]$, so they’re essentially the irreducible polynomials of $\mathbb{R}$. They are of two kinds: the first kind is that of polynomials $x-a$ – the points $a$ – and then the second-degree irreducible polynomials, for instance $x^2+1$... but what kind of point is that?

Wait a second... we said that $a$ was the scalar corresponding to the ideal $(x-a)$ – because (more or less) the polynomial $x-a$ vanishes at $a$... so the “scalar” corresponding to $x^2+1$ would be $i$? Or perhaps $-i$? It’s both at once, actually...

So $\mathbb{A}^1$ knows. $\mathbb{A}^1$ remembers. Even if you consider it on a non-algebraically closed field, $\mathbb{A}^1$ has some points for algebraic numbers.

And $\mathbb{A}^n(k)$? Well, sometimes we really don’t want to consider these “appearing points” for algebraic numbers that aren’t even numbers. So $\mathbb{A}^n(k)$ is the set of points of $\mathbb{A}^n$ which correspond to actual scalars, that is, the maximal ideals of the form $(x_1-a_1,\ldots,x_n-a_n)$ where the $a_i$ are in $k$.

TL; DR: actually $\mathbb{A}^n$ is a very big, very rich object of nature different from $k^n$, and $\mathbb{A}^n(k)$ simply is the subset of points of $\mathbb{A}^n$ that are nice enough to have “coordinates” in $k$.

As for your first question, it’s one basic thing in algebraic geometry to make the concept of dimension field-invariant: for instance, if the real line had dimension one but $\mathbb{A}^1(\mathbb{C})$ had dimension two, what would be the dimension of $\mathbb{A}^1(\mathbb{Q})$? Or $\mathbb{A}^1(\mathbb{F}_p)$? Intuitively, these objects are “similar” over different fields (not unlike linear algebra)... so it’s normal that they be given the same algebraic dimension.

Answer 2

Mindlack already gave a complete answer, so I just want to underscore something in the complex case. If we work over $\Bbb{C}$, we can choose an underlying real structure to see $\Bbb{C}$ as a $2-$dimensional vector space over $\Bbb{R}$. Topologically, we think of this as a real plane. However, when we work purely over $\Bbb{C}$, it is actually more correct to think of $\Bbb{C}$ (by extension $\Bbb{A}^1_{\Bbb{C}})$ as being $1-$dimensional, i.e. as a "line".

At first, this does not really make sense. However, when we think of the intersection theoretic properties of copies of $\Bbb{C}$ in $\Bbb{C}^2$ it becomes clear why this is more correct. Even though a $1-$dimensional subspace $V(\cong \Bbb{C})$ of $\Bbb{C}^2$ is like a "real plane", any two such $V$ and $V'$ which intersect in a point besides the origin are identical. So, as far as complex/algebraic geometry is concerned, copies of $\Bbb{C}$ really behave more like lines than planes.

Actually, this phenomenon persists when one thinks about algebraic curves. A nonsingular algebraic curve over $\Bbb{C}$ can be thought of as a Riemann surface also, which we think of as topologically a real surface. However, if these are "plane" curves (in the sense that they embed in $\Bbb{P}^2)$, they satisfy Bézout's Theorem:

If $X$ and $Y$ are plane curves in $\Bbb{P}^2$ the number of points at which they intersect (counted with multiplicity) is $\deg(X)\cdot \deg(Y)$.

Here, degree is the degree of the polynomial cutting out the curve. So, in spite of being (classically) topologically $2-$dimensional as $\Bbb{R}-$manifolds, these curves intersect in discrete sets of points unless they are equal. This is quite confusing at first, but coming to terms with this makes us realize that $1-$dimensional complex varieties look more like "curves" than they do like "surfaces."