Suppose we're working over an algebraically closed field $k$. If $V\subseteq\mathbb{A}^n$ and $W\subseteq\mathbb{A}^m$ are affine algebraic sets, there is a well known bijective correspondence $$ \operatorname{Mor}_\mathrm{reg}(V,W)\leftrightarrow\operatorname{Hom}_{k-\mathrm{alg}}(k[W],k[V]). $$

I assume this fails in general if $V$ and $W$ are not affine algebraic sets. Consider the case where $V=\{*\}$ is a point, and $W=\mathbb{A}^2\setminus\{(0,0)\}$. So $V$ is affine, but it's also a well known fact that $W$ is not affine. My question is, why does the natural map

$$ \operatorname{Mor}_{\mathrm{reg}}(V,W)\to\operatorname{Hom}_{k-\mathrm{alg}}(k[W],k[V]) $$

fail to be surjective, so that the bijection does not hold in this case?

  • 1
    $\begingroup$ The problem is that for $V$ and $W$ not affine, $k[V]$ (many times) does not give you much information about $V$. For example, for any projective variety $V$, $k[V]=k$. $\endgroup$ – rfauffar Nov 7 '14 at 12:53

This is similar to Hoot's answer, but since you wanted more detail, I thought I'd add it.

I tried to translate this result into classical language the best I could. Hopefully it makes some sense to you!

There is a very nice theorem, called algebraic Hartog's lemma, which says the following (in classical language)[you can find the proof in Vakil, or Qing Liu]

Theorem(Algebraic Hartog's Lemma): Let $X$ be a normal irreducible variety of dimension $n$, and let $Z\subseteq X$ be a variety of dimension less than or equal to $n-2$. Then, the restriction map $k[X]\to k[U]$ is an isomorphism, where $U=X-Z$.

The algebraic version of this theorem says that if $A$ is a normal domain, then


where the intersection takes place in $\text{Frac}(A)$. Intuitively, it says that since the zero set of a function has codimension 1, any open subset, which is the complement of a set of codimension at least $2$, can't invert anything. More symbolically, if $\displaystyle \frac{f}{g}$ is to be a function on $X-Z$, you need that the vanishing set $V(g)$ of $g$ is contained in $Z$. But, $V(g)$ has dimension $n-1$, so if $Z$ has dimension less than or equal to $n-2$, you can't have $V(g)\subseteq Z$. So, you can't invert anything.

In particular, this says that if we consider the inclusion $\mathbb{A}^2-\{(0,0)\}\to \mathbb{A}^2$, then the induced map of rings of functions $k[\mathbb{A}^2]\to k[\mathbb{A}^2-\{(0,0)\}]$ is an isomorphism of $k$-algebras. So, now, assume that the map

$$\text{Mor}(\ast,\mathbb{A}^2-\{(0,0)\})\to \text{Mor}(k[\mathbb{A}^2-\{(0,0)\}],k)$$

were a bijection. Then, consider the following commutative square

$$\begin{matrix}\text{Mor}(\ast,\mathbb{A}^2-\{(0,0)\}) & \to & \text{Mor}(k[\mathbb{A}^2-\{(0,0)\}],k)\\ \downarrow & & \uparrow\\ \text{Mor}(\ast,\mathbb{A}^2) & \to & \text{Mor}(k[\mathbb{A}^2],k)\end{matrix}$$

where the left hand vertical map comes from the inclusion $\mathbb{A}^2-\{(0,0)\}\to\mathbb{A}^2$, and the right hand side comes from the restriction map $k[\mathbb{A}^2-\{(0,0)\}]\to k[\mathbb{A}^2]$.

Now, since $\mathbb{A}^2$ is affine, the bottom horizontal arrow is a bijection. By algebraic Hartog's lemma, the right vertical map is a bijection. So, if we assume that the top horizontal arrow is a bijection, then we must conclude that the left vertical map is a bijection--but it's not!

The key to all of the above was not really just that $k[\mathbb{A}^2]\cong k[\mathbb{A}^2-\{(0,0)\}]$, but that (by algebraic Hartog's lemma) it was the restriction map that induced the isomorphism of $k$-algebras.

  • $\begingroup$ Dear Alex, thanks for the answer! This is more clear to me. $\endgroup$ – Jacqueline Pauwels Nov 24 '14 at 3:11
  • $\begingroup$ @JacquelinePauwels No problem! Let me know if there is anything I can clarify. :) $\endgroup$ – Alex Youcis Nov 24 '14 at 3:13

Here's one way to think about the problem, which will probably lead you to an answer even if the details aren't immediately clear. $$\begin{array}{lll} \mathbb{A}^2 - 0 & \to & \mathbb{A}^2 \\ & & \uparrow \\ & & * \end{array}$$ The horizontal morphism is just the inclusion. If you give me a $k$-homomorphism $\phi\colon k[x, y] \to k$ then that yields a vertical morphism because $\mathbb{A}^2$ is affine. I claim that if $\phi$ corresponded to a morphism $\pi\colon * \to \mathbb{A}^2 - 0$ then $\pi$ would make the above diagram into a commutative triangle.

[Note that in the bijection you give we only need that $W$ is affine.]

  • $\begingroup$ I apologize for giving practically the same answer you did! Hopefully it added something. $\endgroup$ – Alex Youcis Nov 24 '14 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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