In Example 3.3.1 Chapter II Hartshorne, he considers a morphism of schemes $f : X \to Y$, where $X = \operatorname{Spec}k[x,y,t]/(ty – x^2)$ and $Y =\operatorname{Spec}k[t]$. $k$ is algebraically closed. And he says $f$ is a surjective morphism.
1) How can I prove that $f$ is surjective?
(definition was clarified here Definition of a surjective morphism of schemes)
Then he takes a point $0 \not = a \in k$ and says that $X_a$, the fibred product of $f$ over the point $(t- a)$ is the plane curve $ay = x^2 $ in $\mathbb{A}_k^2$.
2) What does he mean by the scheme $X_a$ is the plane curve $ay = x^2 $ in $\mathbb{A}_k^2$?
Any explanation would be appreciated. Thank you.
Best Answer
1) One way to show this is to show that the fiber over every point is nonempty.
2) Well, he means what he says! In this case, where everything is affine, the fiber product is given by Spec of the tensor product on rings. That is, $X\otimes_Y \{a\}=\operatorname{Spec} (k[x,y,t]/(ty-x^2)\otimes_{k[t]} k[t]/(t-a))$, and as $M\otimes_R R/I\cong M/IM$ for any $R$-module $M$ and any ideal $I\subset R$, we have that this ring is isomorphic to $k[x,y,t]/(ty-x^2,t-a)\cong k[x,y]/(ay-x^2)$. So our fiber product is the plane curve $ay-x^2$.