I'm studying basic algebraic geometry, and I confused with the topology in the product of varieties. In my definitions, an affine algebraic variety is a ringed space $(X,\mathscr{O} _X)$ which is isomorphic to Zariski closed of some affine space $\mathbb{A}^n$ (zero locus of polynomials) with his sheaf of regular functions, and more generally, an algebraic variety is a ringed space $(X,\mathscr{O}_X)$ that is locally an affine variety, i.e., for all $x\in X$ exists an open subset $x\in U\subset X$ such that $(U,\mathscr{O}|_U)$ is affine (isomorphic to Zariski closed).

Now, it can be proved that every product of Zariski closed sets is Zariski closed, and then use this result to prove that finite products in the category of algebraic varieties (in the sense defined above) exists. For this, let $X,Y$ two algebraic varieties. By definition, exists $\{U_i\},\{V_j\}$ open covers of $X,Y$ respectively such that $U_i,V_j$ are affine for every $i,j$. Clearly, $\{U_i\times V_j\}$ collection covers $Z:=X\times Y$ and you want to use this collection to give $Z$ algebraic variety structure. This is done as follows: We say that an open set of $X\times Y$ is such that his intersection with every $U_i\times V_j$ is open in $U_i\times V_j$. But my question is how to visualize this topology, because $U_i,V_j$ aren't Zariski closed, they are only isomorphic to Zariski closed, and then i don't know what is the topology and structure sheaf in the product $U_i\times V_j$. I wish someone could explain this to me.

  • $\begingroup$ That definition of algebraic variety is a bit strange. There is no separatedness condition. That would be called a prevariety in, for example, JS Milne's notes $\endgroup$
    – David Lui
    Oct 2, 2022 at 22:20

1 Answer 1


Note that the topology on $X \times Y$ is not the product topology on $X \times Y$. However, we know that if $U \subseteq X$ and $V \subseteq Y$ are affine, then $U \times V$ is the variety whose coordinate ring is the tensor product of the coordinate rings of $U$ and $V$.

For general varieties $X$ and $Y$, we can cover $X$ and $Y$ with affine open sets $(U_i)_{i \in I}$ and $(V_j)_{j \in J}$. We know what $U_i \times V_j$ looks like by above. We can then glue them together to get $X \times Y$ (see any textbook on algebraic geometry on how to do this). Note that $\pi_1^{-1}(U_i) \cap \pi_2^{-1}(V_j)$ is an open subset of $X \times Y$ isomorphic to $U_i \times V_j$. (If you're familiar with schemes then see Stacks project Lemma 26.17.3 )

Therefore, the topology on $X \times Y$ is such that the natural morphisms $U_i \times V_j \rightarrow X \times Y$ are open immersions. That is : inclusions of open subsets.

In general topology there is a theorem that if $T$ is a topological space with open cover $U_i$, and $S \subseteq T$, then $S$ is closed iff $S \cap U_i$ is closed in each $U_i$ with its subspace topology.

Apply that theorem to $X \times Y$ and the open cover $U_i \times V_j$ to get the result.


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