I'm trying to give a motivational explanation for why morphisms between algebraic sets $X \subset \mathbb{A}^n$ and $Y \subseteq \mathbb{A}^m$ is defined the way that it is defined, that is, that it is the set of polynomial maps $\mathbb{A}^n \rightarrow \mathbb{A}^m$ restricted to the domain $X$ and codomain $Y$. Many texts simply say that it is natural to define morphisms between varieties to be polynomial maps between algebraic sets, because algebraic sets are zero loci of polynomials, but I just cannot seem to get to the point where I feel myself "getting it", so to speak. Therefore, I have been trying to come up with a motivational explanation of my own. But here to, I run into problems.

Now, let's say that $X \subset \mathbb{A}^n$ and $Y \subseteq \mathbb{A}^m$ are algebraic sets. Then that means that $X$ and $Y$ are the zero loci of some set of polynomials $f_i \in K[x_1, \dots, x_n]$ (call it $T$) and $g_i \in K[y_1 , \dots , y_m]$ (call it $S$). Let's say then that we have a map $\varphi : \mathbb{A}^n \rightarrow \mathbb{A}^m$, and let's ask what conditions we need to put on it to make it a morphism $X \rightarrow Y$.

$\varphi$ then induces a map from the set of functions $\mathbb{A}^m \rightarrow K$ (denote it $\mathcal{F}[\mathbb{A}^m]$) to the set of functions $\mathbb{A}^n \rightarrow K$ (denote it $\mathcal{F}[\mathbb{A}^n]$), $\varphi^{*} : \mathcal{F}[\mathbb{A}^n] \rightarrow \mathcal{F}[\mathbb{A}^m]$, by $g(y_1 , \dots , y_m) \mapsto f(z_1 , \dots , z_n) = g (y_1 (z_1 , \dots , z_n), \dots , y_m (z_1 , \dots , z_n))$.

We want our morphism to be such that points in $X$ are mapped to points in $Y$. Since $Y = \{ (y_1 , \dots , y_m) \in \mathbb{A}^m | g_i (y_1 , \dots , y_m) = 0 , \forall g_i \in S \}$, this then means that we will have $g_i (\varphi(x_1 , \dots , x_n)) = 0$ for all $(x_1 , \dots , x_n) \in X$, and so $X$ may be described as $X = \{ (x_1 , \dots , x_n) \in \mathbb{A}^n | g_i (\varphi (x_1 , \dots , x_n)) = 0 , \forall g_i \in S \}$.

Now this is the point at which I am struggling.

It is known that the zero locus of general functions $h_i : \mathbb{A}^n \rightarrow K$ is not necessarily an algebraic set. For example, the set $\{ (x,y) | y - \sin^2(x) = 0 \}$ can be shown not to be algebraic. If I could show that we need to have $\varphi$ be a polynomial map for the $\{ (x_1 , \dots , x_n) \in \mathbb{A}^n | g_i (\varphi (x_1 , \dots , x_n)) = 0 , \forall g_i \in S \}$ to actually define an algebraic set, I would basically have the whole definition justified.

The problem is, though, I don't really know how to go about doing that. Is that even possible? Can I find non-polynomial $\varphi: X \rightarrow Y$ such that $\{ (x_1 , \dots , x_n) \in \mathbb{A}^n | g_i (\varphi (x_1 , \dots , x_n)) = 0 , \forall g_i \in S \}$ still defines an algebraic set in $\mathbb{A}^n$? Do I need to add further requirements?

As always, any and all help will be greatly appreciated!

  • $\begingroup$ I think the explanation is much simpler than you're making it; we consider polynomial mappings because we are interested in studying polynomial mappings. $\endgroup$ – Hurkyl Sep 1 '17 at 1:25
  • $\begingroup$ Yes, well, you see, the problem is that I don't really get that explanation. $\endgroup$ – StormyTeacup Sep 1 '17 at 1:40
  • 1
    $\begingroup$ The answer to your question in your last paragraph is trivially yes: Just consider any map $X=\mathbb{C}^n\to\mathbb{C}^m=Y$ that's not algebraic. What you're missing is the requirement that the maps into the field are algebraic, in which case the motivation is the same as in the case of topological or smooth manifolds. $\endgroup$ – user347489 Sep 1 '17 at 2:03
  • 1
    $\begingroup$ @StormyTeacup: I'm saying the motivation here is, in large part, exactly the opposite direction -- it would be more like we thought group homomorphisms were interesting, so we invented groups to help study them! $\endgroup$ – Hurkyl Sep 1 '17 at 2:34
  • 1
    $\begingroup$ Linear mappings leading to the idea of vector spaces might be a better example, since one is likely to encounter various kinds of linear transformations before one conceive the idea of the abstract vector spaces to study them. $\endgroup$ – Hurkyl Sep 1 '17 at 2:39

Your Answer

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

Browse other questions tagged or ask your own question.