Let $K$ denote an algebraically closed field. Define that an algebraic variety over $K$ is a ringed space that can be covered by open sets, each of which is isomorphic to an affine algebraic variety, possibly subject to some further conditions if it helps.

Question. Let $(f,\alpha) : A \rightarrow B$ and $(f,\beta) : A \rightarrow B$ be two morphisms of locally ringed spaces. Are $\alpha$ and $\beta$ necessarily equal as natural transformations? If not, is this at least true for "nice" spaces and/or "nice" fields? If not, how is it possible to define a function between algebraic varieties and then "prove" that this function is a morphism, when the corresponding natural transformation isn't unique?


2 Answers 2


No, as the following example proves.

Example. Let $\mathbb{K}=\overline{\mathbb{Z}_p}$ be the algebraic closure of the field of integer numbers modulo $p\in\mathbb{P}$, and let $X=Y=\mathbb{A}^1_{\mathbb{K}}$ as locally ringed space; considering the following maps of $X$ in itself: \begin{gather} O:x\in X\to 0\in X,\\ F:x\in X\to x^p-x\in X; \end{gather} one has that as maps $O=F$, but as regular maps $O\neq F$ because $O^{*}$ is the zero endomorphism of $\mathbb{K}[t]$ and $F^{*}$ is the Frobenius endomorphism of $\mathbb{K}[t]$. $\triangle$

And if one fixes a continuous map $f:(X,\mathcal{O}_X)\to(Y,\mathcal{O}_Y)$ then the pull-back $f^{*}:\mathcal{O}_Y\to f_{*}\mathcal{O}_X$ is well-defined ever; and if $f^{*}$ is a morphism of sheaves then $(f,f^{*})$ is a morphism of locally ringed spaces.

  • $\begingroup$ Consider $\mathbb{F}_p$ to refer to the field of $p$ elements, since its unambiguous; $\mathbb{Z}_p$ is also used to to the ring of $p$-adic integers. $\endgroup$
    – user14972
    Sep 15, 2018 at 11:01
  • $\begingroup$ Yes, I know; indeed I specify what is $\mathbb{Z}_p$. ;) $\endgroup$ Sep 15, 2018 at 11:03
  • $\begingroup$ So is it correct to say that each continuous map $f$ between ringed spaces induces a canonical morphism of ringed spaces, namely $(f,f^*)$? $\endgroup$ Sep 15, 2018 at 11:40
  • $\begingroup$ No, it is not! For example: any permutation of $\mathbb{A}^1_{\mathbb{K}}$ with Zariski topology is a continuous map, but not any permutation is a morphism of locally ringed spaces. $\endgroup$ Sep 15, 2018 at 11:49
  • 1
    $\begingroup$ Wait why is $O=F$? We don't have $$X^p=X$$ for all $$X \in \overline{\mathbb{F}_P}$$ right.. $\endgroup$
    – M. Van
    Mar 1, 2022 at 14:08

Here's another example (I have problems with the other answer given, see my comments below that): take the algebraic closure $k$ of $\mathbb{F}_p$ as your ground field and consider $X= \text{Spec}(k)$. Then any two distinct automorphisms of $k$ give the same map on the level of sets, since $X$ consists of only one point. You can take the frobenius map and the identity map for example.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .