$\phi \ : \ \mathbb{P}^n \rightarrow \mathbb{P}^m$ continuous $\ \Longrightarrow \ \phi$ morphism of algebraic varieties?

I am confused about regular functions on $\mathbb{P}^n(K)$, (K algebraicly closed field.) It is about the following result:

Regular functions on $\mathbb{P}^n$ are locally constant.

Does that mean that all continuous maps $\mathbb{P}^n \rightarrow \mathbb{P}^m$ are immediately morphisms of algebraic varieties? That is what I thought I proved below:

Claim: All continuous maps $\mathbb{P}^n \rightarrow \mathbb{P}^m$ are morphisms of algebraic varieties.

"Proof": Let $\phi \ : \ \mathbb{P}^n \rightarrow \mathbb{P}^m$ be continuous. We will show $\phi$ is compatible with regular functions. Let $f \ : \ \mathbb{P}^m \rightarrow K$ be regular, i.e locally constant. That means that $$\forall Q \in \mathbb{P}^m, \ \exists U_Q \stackrel{\text{open}}{\subseteq} \mathbb{P}^m, \ \text{ such that } \# \phi(U_Q) = 1$$ Where $U_Q$ denotes a neighbourhood of $Q$. Now we consider $f \circ \phi \ : \ \mathbb{P}^n \rightarrow K$. Let $P \in \mathbb{P}^n$. For $\phi(P) \in \mathbb{P}^m$ we obtain an open neighbourhood $U_{\phi(P)}$ as described above. Hence $V_P \ := \ \phi^{-1}(U_{\phi(P)})$ is an open neighbourhood for $P$ on which $f \circ \phi$ is constant, as desired.

It seems so strange because I would say being a morphism of algebraic varieties is a much stronger property then being a continuous map. This is what you could help me with:

• Confirm that my reasoning is wrong.
• Point out where it went wrong.
• Provide a counterexemple to the statement I thought I proved.

• Someone phrased this really terribly to you. The way this is stated makes one think that local constancy is equivalent to being algebraic. In fact, the reason that the regular functions on $\mathbb{P}^n$ are locally constant is because they're constant. This is then related to another issue you have which is that you think that an algebraic map is precisely one that sends globally regular functions to globally regular functions--this is not true. It is one that sends regular functions on any open to regular functions on the preimage. In particular, if one looks at the open $\mathbb{A}^n$ Nov 24 '16 at 10:13
• inside of $\mathbb{P}^n$, then this has lots, and lots of regular functiosn (they're precisely polynomials in $n$-variables!) and your argument certainly fails there. PS, the maps $\mathbb{P}^n\to\mathbb{P}^m$ are ,essentially, maps of the form $[x_0:\ldots:x_n]\mapsto [\varphi_1(x_0,\ldots,x_n):\ldots:\varphi_m(x_0,\ldots,x_n)]$ where $\varphi_i$ are homogenous polynomials of the same degree which vanish nowhere simultaneously on $\mathbb{P}^n$. Nov 24 '16 at 10:13
• Dear @Alex Youcis, could you make an answer of this and could you extend your explanation according to points that I would like to be clarified? Nov 24 '16 at 10:20
• I am currently doing that--although I thought I did answer your bullets above. Nov 24 '16 at 10:22

Confirm that my reasoning is wrong.

Dear friend, your reason is, in fact, wrong. Or, more to the point it's based off of fallacious assumptions that I will address below.

Point out where it went wrong.

First, you should totally ignore the statement that regular functions on $\mathbb{P}^n$ are locally constant. While true, that's just...silly to say, and really misleading. In fact, recall that a locally constant function on a connected space is automatically constant. So, since most varieties we consider are connected, local constancy is the same thing as constancy. That, there, is the more instructive statement: every regular function $\mathbb{P}^n\to K$ is constant.

The second issue that you have is the following. To show that a map of varieties $f:X\to Y$ is regular, you need to show that for every open $V\subseteq Y$ and every regular function $\varphi:V\to K$ the composition $\varphi\circ f:f^{-1}(V)\to K$ is regular. You have only shown in your proof when $X=\mathbb{P}^n$ and $Y=\mathbb{P}^m$ that this is true for $V=Y$--this is certainly not enough. For example, what do you do when you take $V=\mathbb{A}^m$ whose regular functions are just polynomials in $n$-variables?

Give me a counterexample to what I thought I proved.

As I mentioned above, the regular maps $\mathbb{P}^n\to\mathbb{P}^m$ are of the form

$$[x_0:\cdots:x_n]\mapsto [\varphi_1(x_0,\ldots,x_n):\cdots:\varphi_m(x_0,\ldots,x_n)]$$

where $\varphi_i$ are homogeneous polynomials of the same degree which have no common zeros on $\mathbb{P}^n$.

So, any function not of that form, but continuous would work.

As a nudge towards a more concrete example, note that $\mathbb{P}^1$ has the cofinite topology. So any set-theoretic automorphism of $\mathbb{P}^1$ is continuous. I'm sure you can write one down that is not of the above form.

• Thank you! first one small question. Why do you call the map $\mathbb{P}^n \rightarrow \mathbb{P}^m$ you described regular? Do you mean a morphism? So far I only heard about regular maps from an algebraic variety to a field. Nov 24 '16 at 10:38
• @KoenraadvanDuin Uh, I don't know. To be frank, it's been a while since I thought about things in 'classical language'. Do you not call them regular maps? Maybe 'locally regular' would be better? Anyways, yes, it means morphism. Nov 24 '16 at 10:40
• Ok. Let me just point out in how far I understand your answer. You stated several things that are new for me: connectedness of projective spaces and the characterisation of maps $\mathbb{P}^n \rightarrow \mathbb{P}^m$ you gave me. I won't ask a prove for these statements, I will just assume that they are true or look up some proofs. Nov 24 '16 at 10:41
• @KoenraadvanDuin The connectedness of projective space is easy--$\mathbb{P}^n$ is covered by $\mathbb{A}^n$'s which all intersect. So, it suffices to show that $\mathbb{A}^n$ is connected. But, if $\mathbb{A}^n=U\sqcup V$ for opens $U$ and $V$ then the regular functions on $\mathbb{A}^n$ would be a product of regular functions on $U$ and regular functions on $V$. That means that the regular functions on $\mathbb{A}^n$, which are just $K[x_1,\ldots,X_n]$, as a ring would be isomorphic to the product of two non-trivial rings $R\times S$ (the rings of regular functions on $U$ and $V$). But, Nov 24 '16 at 10:43
• Another thing, do we need that $K$ is algebraicly closed here? Nov 24 '16 at 10:44