An exercise about structure sheaf of product of two algebraic varieties

Question

I am interested in the Exercise 5.5.8 of this lecture notes. I have my own solution for this exercise but I need someone here to verify if there are any flaws in my arguments.

To recall, this exercise requires you to show that

(i) Given two algebraic varieties $X,Y$ and a fixed point $P_0 \in X$, then the map $\varphi: Y \to X \times Y, Q \mapsto (P_0,Q)$ is a morphism,

(ii) Furthermore, if $\mathcal{O}_X(X)=k$, then $\mathcal{O}_{X \times Y}(X \times Y) \cong \mathcal{O}_Y(Y)$.

My ideas are

(i) Each algebraic variety can be written as a union of (not necessarily finite) open affines, so we write $X=\bigcup_{i \in I} U_i$ and $Y=\bigcup_{j \in J} V_j$, where $U_i$ and $V_j$ are open affines. Then $P_0$ must live in some $U_{i_0}$ and similarly, a random element $Q \in X$ must live in some $V_{j_0}$. Now $\varphi(V_{j_0})=\{P_0\} \times V_{j_0} \subseteq U_{i_0} \times V_{j_0}$ and the restriction $\varphi_{|V_{j_0}}: V_{j_0} \to U_{i_0} \times V_{j_0}$ determined by $Q \mapsto (P_0,Q)$ is a morphism by Prop 4.3.9 in this lecture notes.

(ii) I use the map in part (i) to prove this part. Let $\pi_Y$ is the projection from $X \times Y \to Y$, it is not hard to see that $\pi_Y \circ \varphi$ is the identity over $Y$. Then $(\pi_Y \circ \varphi)^*=\varphi^* \circ \pi_Y^*$ is also the identity over $\mathcal{O}_Y(Y)$, which tells that $\varphi^*$ is an isomorphism having the inverse $\pi_Y^*$, and thus an isomorphism between the two $k$-algebras mentioned in the statement.

What I am concerned about is that I did not use the assumption $\mathcal{O}_X(X)=k$ at all, so I think my arguments for part (ii) are wrong. However I cannot figure out where I made the mistakes. Also, we knew that $\mathcal{O}_{X \times Y}(X \times Y) \cong \mathcal{O}_X(X) \otimes_k \mathcal{O}_Y(Y)$ if both $X$ and $Y$ are affine varieties, but this lecture notes does not tell if this isomorphism is also true in the general case. I also read this MO post but it does not satisfy me as well.

Any helps are really appreciated. Happy New Year 2024 to everyone!

Answer 1

Your solution to (i) is correct, but your solution to (ii) is incorrect.

You have morphisms $Y\xrightarrow \varphi X\times Y\xrightarrow{\pi_Y} Y$ which composes to the identity, so you have morphisms $$\mathcal O_Y(Y)\xrightarrow{\pi_Y^*}\mathcal O_{X\times Y}(X\times Y)\xrightarrow{\varphi^*}\mathcal O_Y(Y)$$ which composes to the identity. However, this does not necessarily mean $\pi_Y^*$ or $\varphi^*$ is an isomorphism!

Example For any field $k$ , $$k[s]\to k[s,t]\to k[t]$$ composes to the identity, where the first map is the natural inclusion and the second map sends $s\mapsto0$. But $k[s]$ and $k[s,t]$ are not isomorphic, as seen by for example comparing their Krull dimensions. (In fact, this is what you obtain when $X=Y=\mathbb A^1$ and $P_0=0$.)

To check that $\varphi^*$ is an isomorphism, you also need that the other composition $\pi_Y^*\circ \varphi^*$ is also the identity!

So, here is actually how to prove the statement:

Recall that $\mathcal O_X(X)$ is the ring of regular functions $X\to k$. Thus, $\mathcal O_X(X)\simeq k$ exactly means the only regular functions $X\to k$ are constants. We hope to check $\pi_Y^*\colon\mathcal O_Y(Y)\to\mathcal O_{X\times Y}(X\times Y)$ is an isomorphism. Since the injectivity of $\pi_Y^*$ is obvious, we must prove $\pi_Y^*$ is surjective, which is equivalent to: $$\begin{align*}&\text{any regular function $f\colon X\times Y\to k$ factors through $\pi_Y$,} \\&\text{i.e., $f(x_1,y)=f(x_2,y)$ for any $x_1,x_2\in X$ and $y\in Y$.}\end{align*}$$ But this follows from the observation that $f(\bullet,y)\colon X\to k$ is a regular function, hence constant by assumption.