Let $X \subseteq \mathbb{C^m}, Y\subseteq \mathbb{C}^n$ be algebraic (not necessarily irreducible), and let $\phi\colon X \to Y$ be a map such that for each $p \in X$, there are there is an open neighborhood $\mathcal{O} \subseteq X$ containing $p$ and polynomials $f_1,\ldots,f_n,g_1,\ldots,g_n \in \mathbb{C}[X_1,\ldots,X_m]$ such that the $g_i$'s don't vanish on $\mathcal{O}$ and for each $x\in \mathcal{O}$ we have

$$\phi(x) = \left(\frac{f_1(x)}{g_1(x)},\ldots,\frac{f_n(x)}{g_n(x)} \right).$$

Is it true that $\phi$ must be a continuous map with respect to the Zariski topologies? I have read a few answers on this site, but I find them unsatisfactory:

Continuity of rational functions between affine algebraic sets : The answer seems to assume that if polynomials are continuous in the Zariski topology, then maps whose values are quotients of polynomials are also continuous. I am not confident that this is true in the Zariski topology. In the comments of the answer the answerer shows why polynomials are continuous in the Zariski topology, but this is already clear to me.

The continuity of the rational maps in the zariski topology : The answerer assumes that to prove that a rational map is continuous, it suffices to prove that a morphism of algebraic varieties is continuous. I am new to algebraic geometry and don't see why this is true. Moreover, it seems to me that the coordinate ring homomorphism $\phi$ he mentions may not be surjective, and so the answerer incorrectly assumes that the inverse image of a maximal ideal is maximal.


1 Answer 1


There's a lot going on here, but the first thing to note is that the description you provide is actually of a regular function, not a rational function (a rational function is a regular function defined on a dense open subset up to an equivalence relation). The next thing to know is that any such function on a closed subset $X\subset\Bbb C^m$ may be described globally as a polynomial (see here for instance, though there are many other sources). This is great news, because it means we can replace your map by a single map $$\phi(x)=(f_1(x_1,\cdots,x_m),\cdots,f_n(x_1,\cdots,x_m)).$$ Now it is easy to check the condition that this map is continuous with respect to the Zariski topology by showing that the preimage of a closed set is closed.

Suppose $Z\subset Y$ is a closed subset cut out by $h_1,\cdots,h_l$. Then the preimage of $Z$ under $\phi$ is the subset of $X$ cut out by $h_1(f_1,\cdots,f_n)=0$, $\cdots$, $h_l(f_1,\cdots,f_n)=0$. But these are just polynomials in the $x_i$, so $\phi^{-1}(Z)$ is Zariski closed in $X$, and we're done.

An attentive observer will note that we actually don't need the full strength of this condition if we know a little topology.

Topological fact: if $S$ is a subset of a topological space $X$ and $\{U_i\}_{i\in I}$ is an open cover of $X$, then $S$ is open/closed in $X$ iff $S\cap U_i$ is open/closed in each $U_i$.

For each point, apply your definition to find an open neighborhood where we can write $\phi(x)=(\frac{f_1}{g_1},\cdots,\frac{f_n}{g_n})$. If $Z\subset Y$ is a closed subset cut out by $h_1,\cdots,h_l$, then the preimage of $Z$ on $\mathcal{O}$ is the locus where the $h_i(\frac{f_1}{g_1},\cdots,\frac{f_n}{g_n})$ vanish. But expanding out and clearing denominators, we find that this condition is described by the vanishing of polynomials. This gives us that $\phi^{-1}(Z)\cap \mathcal{O}$ is closed, and we apply our topological fact to finish off the problem.

Let me also take a moment to address your issues with the problems you searched (which is a great idea, by the way). Link 1 has been dealt with above. For link 2, the big idea is that for any point in an open subset $U$ in a closed algebraic set $X\subset\Bbb C^n$, we can find a set of the form $D(f)$ which lies entirely inside it $U$ and contains $p$, and if $X$ is cut out by the ideal $I\subset \Bbb C[x_1,\cdots,x_n]$, then $D(f)$ is isomorphic to the subset of $\Bbb C^{n+1}$ cut out by $(I,zf-1)\subset \Bbb C[x_1,\cdots,x_n,z]$ by the map $(x_1,\cdots,x_n)\mapsto (x_1,\cdots,x_n,1/f)$. So we can deal with $D(f)$, where $1/f$ turns in to a polynomial, and since continuity is local, we're done.

Finally, about maximal ideals: if $f:R\to S$ is a $k$-map of finitely-generated $k$-algebras for some field $k$, it is always true that the preimage of a maximal ideal is a maximal ideal: if $I\subset S$ is maximal, then $k\hookrightarrow R/f^{-1}(I)\hookrightarrow S/I$. But this means that $R/f^{-1}(I)$ is a domain which is trapped between a finite field extension (use Zariski's lemma to show that $S/I$ is a finite extension of $k$), which must be a field.

  • $\begingroup$ Many thanks for the complete answer $\endgroup$
    – epsilon
    Commented Jun 15, 2021 at 9:55

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