In my recent reading of various books and notes on algebraic geometry and scheme theory, I have come across three definitions of affine $n$-space over a field $k$:

  1. $\mathbb{A}_k^n$ is $k^n$ 'without an origin';

  2. $\mathbb{A}_k^n$ is simply $k^n$ with the Zariski topology (i.e. zero loci of sets of polynomials are closed);

  3. $\mathbb{A}_k^n=\operatorname{Spec}k[x_1,\ldots,x_n]$.

I know that things like the Nullstellensatz give us a nice way of going from one point of view to another when it comes to polynomial rings and affine varieties, but here I'm having some confusion. For $k=\mathbb{C}$ (which is algebraically closed, so things should all act reasonably nicely) we have that the points of $\operatorname{Spec}\mathbb{C}[x]$ are in bijective correspondence with points of $\mathbb{C}$, apart from the general point $[(0)]$. So it seems to me that these definitions can't all be the same, but searching for affine space on Google or in books leads to any one of the above definitions, depending on the level of the text.

Am I just making a silly mistake, or is there a difference in these definitions? If so, which one is the 'right' one?

  • 3,009
  • 1
  • 17
  • 43
  • 1
    (1) is a good thought but doesn't seem super precise (no topology is specified, for example). (3) is a scheme and certainly has more points. What is shown in, for example, Hartshorne's book is that if $k=\bar{k}$ then there is a fully faithful functor from classical varieties over $k$ to schemes over $k$. (3) is "the" scheme corresponding to (2) under this functor. – Hoot Jul 06 '15 at 00:36
  • 4
    If you find this disturbing, wait till you discover all the different things that are called $X$. – WillO Jul 06 '15 at 02:42
  • 1
    "Same" is more useful in the sense of "can I do the same things with them?" than in the sense of "are these the same mathematical object?" For a more more dramatic example, any commutative ring $R$ is the *same* as the collection of all $R$-modules and homomorphisms between them. Some things are easier to do with $R$, and some things are easier to do with $R$-mod, but (up to isomorphism) anything you can do you can do with the other. (e.g. starting with $R$ you can always then construct $R$-mod and continue from there. Similarly, from $R$-mod you can recover $R$) –  Jul 06 '15 at 04:39
  • @user14972, can you give an example for things that are easier to be done with the category of modules over a ring than with the ring itself? and how it translates to ring-theoretic world? – Fawzy Hegab May 31 '20 at 09:05
  • Does the notation $\mathbb{A}^{n}_{k}$ include an 'Affine frame' while the notation $k^{n}$ does not include an Affine frame? – crecycreak Feb 07 '21 at 19:43

2 Answers2


So it seems to me that these definitions can't all be the same, but searching for affine space on Google or in books leads to any one of the above definitions, depending on the level of the text.

Right, they're not all the same. They define objects in three different categories, which are respectively

  1. The category of affine spaces over $k$. This requires a bit of elaboration. One way to define a vector space is as a set equipped with an $n$-ary operation for every $n$-tuple $(r_1, r_2, \dots r_n)$ of elements of $k$, corresponding to the linear combination operation $(x_1, x_2, \dots x_n) \mapsto \sum r_i x_i$, together with various axioms relating these. An affine space is a slightly restricted version of this where you only allow operations where $\sum r_i = 1$ ("affine linear combinations"). In particular you can't multiply by zero, so you don't know where your origin is. In this category, endomorphisms $k^n \to k^n$ correspond to maps of the form $v \mapsto Av + w$ where $A$ is a matrix and $w$ is a vector.

  2. The category of what I'll call "naive affine varieties" over $k$. This is the category whose objects are Zariski-closed subsets of $k^n$ for some $n$ and whose morphisms are polynomial functions over $k$. The big difference between this option and #1 is that polynomial functions of degree greater than $1$ are allowed. I don't want to call these things varieties because if $k$ is not algebraically closed this is the wrong category to work in. In this category, endomorphisms $k^n \to k^n$ correspond to $n$-tuples of polynomials of $n$ variables over $k$.

  3. The category of varieties over $k$, which itself has several definitions depending on your level of sophistication. The big difference between this option and #2 is that given a variety $X$ over a field $k$ it makes sense to talk about the points $X(L)$ of that variety over any field extension $L$ of $k$ (and in fact much more generally than this, but let's stick to field extensions for simplicity). For example, if $k = \mathbb{R}$ then $\{ x^2 + y^2 = -1 \}$ defines the empty set in option #2 but it defines an interesting variety (with interesting points over $\mathbb{C}$) in this option. In this category, endomorphisms $k^n \to k^n$ look the same as above, but by looking at field extensions you can get a more interesting endomorphism object involving polynomials over field extensions of $k$.

The relationship between these categories is that there are functors from #1 to #3 to #2. The tendency to give an object the same name in different categories where it appears when they're related by a functor is pretty common in mathematics (for example, think of how many categories have an object called $\mathbb{R}$) and it's good to get used to it sooner rather than later.

Qiaochu Yuan
  • 359,788
  • 42
  • 777
  • 1,145

You aren't being silly and the reason for the various definitions is in part due to just being a recent topic and in part due to the lack of coverage of geometry in a modern setting.

So let's go over all of the definitions and see what they share and what they don't. The goal of our definition at the end of the day should be to define a "space" which we can study the geometry of in both an algebraic and geometric way.

Because of this, there is no really great way to define a geometric space. In reality, we formally define a set of objects and then show that we can think of this as being a geometric space which we are familiar with. Our model can be the plane, for example. Since $\mathbb{R}$ has a linear ordering, we can view this as a line. It is a miracle of Descartes to introduce the orthogonal description of $\mathbb{R}^2$ and to assign to each point of the plane an element of $\mathbb{R}^2$ in a systematic manner. Namely, take $(a,b)$ and send it to the coordinate we are familiar with. Of course, we could do the same thing with $\mathbb{C}$ which was shown by Gauss.

Now, this is great except there is no really true way to choose which point represents what. Our choices were completely arbitrary just so we could work with geometry in an algebraic setting. This is explicitly seen in linear algebra where we could, for example, replace the orthogonal lines of the $x,y$-axis with the lines $y=2x$ and the $x$-axis. This gives another totally valid representation of the plane where there is a bijection between "points" and elements of $\mathbb{R}^2$. Alright, well we should probably avoid any ambiguity if we want to just define what a plane is then in this algebro-geometric setting.

Now here's something even more confusing. At first we were only looking at the plane. But what about $3$-space? There are tons of planes in $3$-space. So what difference does the geometry of the plane $\mathbb{R}^2\times \{0\}$ have (in whichever way you decide to describe it) with $\mathbb{R}^2\times \{1\}$? The geometry in these two planes is completely the same. One is simply a translation of the other. But the first we would consider as a subspace of the vector space $\mathbb{R}^3$ and the second we wouldn't consider if we wanted to work in a completely algebraic setting. How could we work algebraically in the latter? Even the way we add points would be wrong (as the third component clearly doesn't stay at $1$ if we add two points component-wise).

Alright, here's where we get into the definitions. Definition 1: Affine $n$-space is $k^n$ without the origin. No; this is wrong. Affine $n$-space is our geometric idea of what an arbitrary $k^n$ should look like. Say we are looking at a plane before we have assigned a coordinate system $\mathbb{R}^2$ to it. Then there is no difference between a plane, and a plane lying above the other. These are both affine planes. Affine $n$-space should generalize this idea to the $n$-dimensional setting where, if we were to view a geometric object and it "looked" like $n$-space and we could assign to it the points of $k^n$ in a one-to-one correspondence, then we will call it $\mathbb{A}_k^n$. In this way both the plane, and every other plane in $3$-space represent $\mathbb{A}_k^n$ before we have given names to its points.

People say topology and geometry are related. In reality, a set $A$ is the same set with or without a topology on it. Definition 2 is where people try to associate the algebraic nature with affine space but it really doesn't make sense. This is just as "general" as the above definition but we don't need to impose a topology on the space. A typical reason why this definition is used in, say, Hartshorne's text is to get people away from the idea that we first need to assign names to the points of a geometric space before we can work with them. Why not just work with the set of $n$-tuples of elements of $k$, and then if we want to apply them to a geometric space we can do so whenever we choose? So to avoid the familiarity with $k^n$ we will give it a new topology, the Zariski topology, and then you won't think of it as being the old object you are used to making more confusing than need be.

Finally, definition 3. Arguably the most rigorous definition and in the aim of where we want to go but it almost isn't even a definition. This is the most algebraic way to describe the geometric space but this is only because we have a one to one correspondence between maximal ideals and points in $k^n$. Usually there are even more points if $k$ is not algebraically closed, and these correspond to varieties in the Zariski topology. Still, this needs to be proved and is not obvious. That is why it really should not be taken as the definition of affine space.

But the reasons why we want to work with affine space in the first place are philosophical. When we introduce coordinate systems into a geometric space we lose some of the geometry. That's why most textbooks don't talk about this; it's philosophy not mathematics. If you want to think of affine $n$-space you can choose any of the above definitions. If you are planning to do commutative algebra I would choose the last definition. But in reality, make sure to separate the geometric space and the coordinate system and realize that the intrinsic properties of the space are not lost when we decide to choose a different way to label things.

  • 5,560
  • 1
  • 16
  • 34
  • This is a really great answer with lots of background information in it, and I'm sorry that I can't accept two answers! – Tim Jul 06 '15 at 10:22
  • 5
    @Tim As long as you get an answer! I'm sure anyone who's ever tried to learn algebra geometry in the last twenty years has asked this same question. – Eoin Jul 06 '15 at 13:05