# Why is regular function?

Suppose that we have two varieties $V,W$ (affine or projective, arbitrary). They are two algebraic object and as usual, we can define the map $\varphi: V\longrightarrow W$. It's easy to think about it as $\varphi(P)=(\varphi_{1}(P),...,\varphi_{m}(P))$ for any point $P=(a_1,...,a_{n})$ of $V$.

Then if we want to study the map $\varphi$, we have to care about $\varphi_{i} : V\longrightarrow k$, where $k$ is the base field.

My question is, why does $\varphi_{i}$ need to be a polynomial map ? Why do people care about the regular property? What did lead the mathematician to the idea of regular function ?

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What do you mean by "arbitrary variety" if you're defining functions coordinate-wise? This is not quite clear to me. – Nils Matthes Sep 11 '12 at 12:01
I mean if $V$ is affine/projective then $W$ is affine/projective correspondingly. Then we can define the morphism. – Arsenaler Sep 11 '12 at 12:57
But varieties are objects that are defined by polynomial equations. Isn't it the obvious choice that the morphisms in the category are polynomial? Varieties were essentially invented to study solutions to polynomial equations. What book did you learn the definition of a variety from? – Matt Sep 11 '12 at 13:05
The case polynomial map is quite understandable. I emphasize on the problem of regular function. I got an idea: each morphism $\varphi: V\longrightarrow W$ gives us a $k$-algebra homomorphism $varphi^* : k[W]\longrightarrow k[V]$. From this we want to extend to a homomorphism from $k(W)$ to $k(V)$, then we have to consider the regularity of maps. But I do not think that is all. May be the picture is more complicated. – Arsenaler Sep 11 '12 at 13:51
Oh. I see. You should just think of "regular" as "locally polynomial." It is analogous to the case of a smooth manifold. You want the functions between smooth manifolds to be "smooth," but the only way to do this is to define locally smooth. It is completely equivalent to say a function $f:M\to N$ is smooth if and only if for every smooth $\phi: N\to \mathbb{R}$ the composition $\phi\circ f: M\to \mathbb{R}$ is smooth. If you approach smooth manifolds from this viewpoint it is a direct translation that the definition of regular function is what corresponds to locally polynomial. – Matt Sep 11 '12 at 17:00