I'm pretty sure this is true, but I have a feeling I've made a mistake in my reasoning somewhere - in most of the literature I've come across, I've seen the statement in the question proven for affine algebraic sets and projective algebraic sets separately, but never for quasi-projective algebraic sets.

I'm defining a projective algebraic set in $\mathbb{P}^n := \mathbb{P}^n(k)$ ($k$ algebraically closed) as a set of the form $\{[x_0: \dots : x_n] \in \mathbb{P}^n : f_1(x_0, \dots, x_n) = \dots = f_r(x_0, \dots, x_n) = 0\}$ for some homogeneous polynomials $f_i \in k[X_0, \dots X_n]$. Give $\mathbb{P}^n$ a topology by stipulating that the closed sets are precisely the projective algebraic sets. A quasi-projective algebraic set in $\mathbb{P}^n$ is then defined as the intersection of an open set $V \subseteq \mathbb{P}^n$ and a closed set $W \subseteq \mathbb{P}^n$.

Say a function $\phi: V \to W$ of quasi-projective algebraic sets $V \subseteq \mathbb{P}^n$, $W \subseteq \mathbb{P}^m$ is a regular map if, for every $x \in V$, there is an open neighbourhood $U$ of $x$ and polynomials $f_0, \dots f_m$, all homogeneous of the same degree, such that for all $y$ in $U$, we have $f_i(y) \neq 0$ for some $i$, and

$$\phi(y) = [f_0(y) : \dots : f_m(y)]$$

(where, if $y = [y_0 : \dots : y_n]$, $f_i(y)$ is shorthand for $f_i(y_0, \dots, y_n)$).

I would be grateful if anyone could check if the argument for the following claim is okay, or if I've gone wrong somewhere:

Claim: If $V \subseteq \mathbb{P}^n$, $W \subseteq \mathbb{P}^m$ are quasi-projective algebraic sets, and $\phi, \Phi: V \to W$ are regular maps that agree on some dense subset $X$ of $V$, then $\phi = \Phi$.

"Proof" Consider the subset $V' = \{x \in V : \phi(x) = \Phi(x) \}$ of $V$. We want to show that $V' = V$. For this, it suffices to show that $V'$ is closed in $V$ (as $V'$ contains $X$ and $X$ is dense in $V$).

Choose any point $x$ of $V$. There are open neighbourhoods $U^{\phi}_x$ and $U^{\Phi}_x$ of $x$ and homogeneous polynomials $f_0, \dots, f_m$ defining $\phi$ on $U^{\phi}_x$ (as in the definition of regular map above) and $g_0, \dots, g_m$ defining $\Phi$ on $U^{\Phi}_x$. Thus $U_x = U^{\phi}_x \cap U^{\Phi}_x$ is an open neighbourhood of $x$ such that $f_0, \dots, f_m$ define $\phi$ on $U_x$ and $g_0, \dots, g_m$ define $\Phi$ on $U_x$.


\begin{align*} V' \cap U_x &= \{y \in U_x : [f_0(y) : \dots : f_m(y)] = [g_0(y) : \dots : g_m(y)\} \\ &= \{y \in U_x : f_i(y) g_j(y) - f_j(y) g_i(y) = 0 \quad \forall \: i, j\} \\ &= U_x \cap \{y \in \mathbb{P}^n : f_i(y) g_j(y) - f_j(y) g_i(y) = 0 \quad \forall \: i, j\} \end{align*}

is closed in $U_x$ (as $\{y \in \mathbb{P}^n : f_i(y) g_j(y) - f_j(y) g_i(y) = 0 \quad \forall \: i, j\}$ is closed in $\mathbb{P}^n$).

So the $U_x$, $x \in V$, form an open cover of $V$ such that $V' \cap U_x$ is closed in $U_x$ for all $x \in V$. By basic topology, this means $V'$ is closed in $V$, which is what we wanted to show.

  • $\begingroup$ In regards to your worries about affine versus projective versus quasiprojective, this statement can be proven scheme-theoretically for all of these cases all at once, see math.stanford.edu/~vakil/0708-216/216class17.pdf theorem 4.1 for example. $\endgroup$ – KReiser May 1 '18 at 2:26
  • 1
    $\begingroup$ For me the argument is OK. $\endgroup$ – danneks May 1 '18 at 4:55

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.