Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am a little bit confused about some basic terminology: What exactly do we mean by a morphism of the affine space $\mathbb{A}_k^n \rightarrow \mathbb{A}_k^n$, where e.g. $k$ is algebraically closed field?

How is this definition adapted if we assume that we have a structural sheaf on $\mathbb{A}_k^n$?

share|cite|improve this question
That depends on what formalism you're working with (in that there are several definitions which are equivalent but it takes a little work to prove that they are equivalent). When you say "morphism" you need to specify both a source and a target; if they're both affine space, then the simplest definition is a tuple of polynomials. – Qiaochu Yuan Oct 3 '12 at 18:18
@QiaochuYuan: Thanks. If we consider the Zariski topology, does it make any difference? A simple choice is a tuple of polynomials because they preserve the structural sheaf? – Manos Oct 3 '12 at 18:26
The affine category on its own doesn't have any notion of multiplication with which to define polynomials-of course this depends on the context, but an affine space morphism normally just means an affine linear function, i.e. an equivariant map for the action of $k^n$ on $\mathbb{A}^n$. – Kevin Carlson Oct 3 '12 at 18:28
up vote 4 down vote accepted

If you're thinking of $\mathbf{A}_k^n$ as the affine scheme $\mathrm{Spec}(k[X_1,\ldots,X_n])$, then a morphism $\mathbf{A}_k^n\rightarrow\mathbf{A}_k^n$ means a morphism of $k$-schemes. The $\mathrm{Spec}$ functor is fully faithful, so any morphism from $\mathbf{A}_k^n$ to itself is $\mathrm{Spec}(\varphi)$ for a unique $k$-algebra map $\varphi:k[X_1,\ldots,X_n]\rightarrow k[X_1,\ldots,X_n]$. The universal property of the polynomial algebra over $k$ says that such a morphism is equivalent to the data of an $n$-tuple of elements in the target, i.e., $n$ polynomials $f_1,\ldots,f_n$. So specifying a morphism is the same as giving an $n$-tuple of polynomials in $k[X_1,\ldots,X_n]$. This extends to $k$-morphisms $X\rightarrow\mathbf{A}_k^n$ for any $k$-scheme $X$: such a morphism is uniquely determined by the data of an $n$-tuple of global sections of $\mathscr{O}_X$, i.e., an element of $\mathscr{O}_X(X)^n$. In fact it works for $R$-morphisms $X\rightarrow\mathbf{A}_R^n$ for any $R$-scheme $X$ and any ring $R$...or even any base scheme $S$.

share|cite|improve this answer
Could you please give me a reference to the "universal property of the polynomial algebra"? – Manos Oct 3 '12 at 18:58
I would think that almost any abstract algebra textbook that develops polynomial rings from scratch will state this property (e.g. Lang, Hungerford): if $R$ is a ring and $A$ is an $R$-algebra, then given $a_1,\ldots,a_n\in A$, there is a unique $R$-algebra homomorphism $R[X_1,\ldots,X_n]\rightarrow A$ with $X_i\mapsto a_i$. – Keenan Kidwell Oct 3 '12 at 19:04
This looks like the evaluation homomorphism. – Manos Oct 3 '12 at 19:10
Yes. The universal property says that any set of $n$ elements gives rise to an ``evaluation" homomorphism. The point is that the multiplication law in $R[X_1,\ldots,X_n]$, i.e., multiplication of formal polynomials, is universal. It remains valid when you replace all the variables with elements of any $R$-algebra. – Keenan Kidwell Oct 3 '12 at 19:17
Got it, thanks! – Manos Oct 3 '12 at 19:25

Another answer has described morphisms of $\mathbb{A}^n$ within the category of affine schemes. Here's a discussion of maps in the category of affine spaces, since both are useful in many contexts.

$\mathbb{A}^n$ is a $k^n$-torsor, which means there's a regular group action of $k^n$ on $\mathbb{A}^n$, namely the one sending $x\in\mathbb{A}^n\mapsto x+v,v\in k^n$. This is usually intuitively described as $\mathbb{A}^n$ being $k^n$ without an origin. (Indeed there's a forgetful functor $U$ taking a vector space to its underlying affine space, as well as a functor $D$ taking $\mathbb{A}^n$ to $k^n$. These aren't quite adjoints, because $UD$ isn't naturally isomorphic to the identity functor on the affine category.)

So the only operation we get inside $\mathbb{A}^n$ is subtraction: $x-y$ is the unique $v\in k^n$ such that $x+v=y$. Then the endomorphisms of $\mathbb{A}^n$ are its endomorphisms as a $k^n$-torsor, which might be described as $f:\mathbb{A}^n\to\mathbb{A^n}$ such that $f(x)-f(y)=A(x-y)$ for $A$ a linear transformation $k^n\to k^n$, a map that preserves subtraction "up to a linear transformation."

If you were to (non-naturally) associate $\mathbb{A}^n$ with $k^n$, you would see that the affine linear maps $f: v\mapsto Av+u$ satisfy the above definition, while conversely you could get $u$ as the image of whatever affine point was sent to $0$ and $A$ as the $f-u$, but it's cleaner to avoid using the functor $D$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.