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.


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 $ay-x^2$.

  • $\begingroup$ Is $M\otimes_R R/I\cong M/IM$ isomorphism as $R$-modules? How does it imply isomorphism as rings in the step that follows? $\endgroup$ – Johnny T. Oct 20 '19 at 21:15
  • $\begingroup$ It's straightforwards (basic first year graduate abstract algebra) to check that it's also an isomorphism of rings if both sides are rings. $\endgroup$ – KReiser Oct 20 '19 at 21:19
  • 1
    $\begingroup$ Since I see you've posted a new question related to this content, I should mention that the map which exhibits the isomorphism $M\otimes_R R/I\cong M/IM$ is also a ring isomorphism - it's just defined by sending $m\otimes (r+I)\to rm+IM$ and extending linearly. You can check this yourself, or refer to this answer if you get stuck. $\endgroup$ – KReiser Oct 21 '19 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.