(1) I am trying to find an example of a smooth affine curve $C$ over $k$ with no immersion $C \to \mathbb{P}^2_k$ (for me a curve is an integral separated dimension one scheme of finite type over $k$).

Here I will discuss my attempt so far.

Since the immersion factors, $C \to U \to \mathbb{P}^2_k$ as a closed immersion followed by an open immersion I first tried considering special cases of the open $U$. If we restrict to principal affine opens $U = D(q) \subset \mathbb{A}^2_k \subset \mathbb{P}^2_k$ for some $q \in k[x,y]$ then I can find an example. What I mean is I can exhibit a curve that has no closed immersion into any $D(q) \subset \mathbb{A}^2_k$. The idea is to show that a smooth curve embedded in $D(q)$ must have a trivial canonical bundle and then produce an example of an affine curve with a nontrivial canonical bundle which I have done. This is an easy computation using that $C \cong \mathrm{Spec}{(k[x,y,q^{-1}]/(f))}$ and $(f, \partial_x f, \partial_y f)$ is the unit ideal since $C$ is smooth. Then one can explicitly compute that $$ \Omega_{C/k} = R \mathrm{d}{x} \oplus R \mathrm{d}{y} / (\partial_x f \mathrm{d}{x} + \partial_y f \mathrm{d}{y}) \cong R $$ where $R = k[x,y,q^{-1}]/(f)$. This leads me to my second question:

(2) Is the fact that $\Omega_{C/k} = \mathcal{O}_{C}$ for affine plane curves $C \subset D(q)$ a consequence of a more general fact or is this computation the ``correct'' way of seeing why this is true?

One possible example is to take a degree 4 curve in $\mathbb{P}^2_k$ such that the tangent line at a point $P$ intersects the curve in at least one other point. Then the canonical bundle is the pullback of $\mathcal{O}_{\mathbb{P}^2_k}(1)$ but we know that $\{ P \}$ is not a hyperplane section by construction so $K_C \not\sim \ell \cdot [P]$ and thus removing the point $P$ gives an affine curve with a nontrivial canonical bundle. Notice that this example is immersed in $\mathbb{P}^2_k$ since it is an open subset of a complete smooth curve in $\mathbb{P}^2_k$, therefore, the vanishing of the canonical bundle cannot be an obstruction to such an immersion.

To go beyond this, I tried to find an example $C$ with no immersion $C \to \mathbb{A}^2_k$ which may be easier. We can take the closure $C \to \overline{C} \to \mathbb{A}^2_k$ and since $C$ is smooth by assumption $C \subset \overline{C}_{\text{smooth}}$. Furthermore, $\overline{C}_{\text{smooth}} = \overline{C} \cap (D(\partial_x f) \cup D(\partial_y f))$ where $\overline{C} = V(f)$ for some $f \in k[x,y]$. Therefore, I have affine opens $C \cap D(\partial_x f)$ and $C \cap D(\partial_y f)$ on which the canonical bundle must vanish (these are affine since $\overline{C} \to \mathbb{A}^2_k$ is affine (closed immersion) and $C \to \overline{C}$ is affine since $C$ is affine and $\overline{C}$ is separated). But I am not sure how to proceed from here since it seems hard / impossible to come up with a curve which has no size 2 affine open cover trivializing its canonical bundle.

I would be most interested in an example with $k = \mathbb{C}$ but in the unlikely case that a nice example requires some not algebraically closed field that would be very interesting.

Finally, I was wondering about the converse of the partial result I had found:

(3) Is $\Omega_{C/k} = \mathcal{O}_C$ the only obstruction to having a closed immersion $C \to D(q) \subset \mathbb{A}^2_k$ for some $q \in k[x,y]$? That is, for such a curve does there always exist such a $q \in k[x,y]$ and a closed immersion $C \to D(q) \subset \mathbb{A}^2_k$.

Many thanks!

  • 2
    $\begingroup$ Can you say why $X \to \mathbb{P}^2$ has to be a closed immersion (in particular why an immersion since closed is obvious)? E.g. if you took a curve $C$ in $\mathbb{P}^2$ with one cusp $P$ and removed the cusp point then $C \setminus P$ should admit a smooth projective model $X$ (the normalization of $C$) but the image of $X$ must contain $C$ since it is closed but that is not smooth. What has gone wrong with my reasoning? Thanks for the answer. $\endgroup$ – Ben C Mar 26 '20 at 22:17
  • 2
    $\begingroup$ For example, I thought that genus 2 curves all have a hyperelliptic affine model $y^2 = f(x)$ giving an immersion of this affine part in $\mathbb{A}^2$ but we know that genus 2 complete curves are not plane curves so their smooth projective model can't embed in $\mathbb{P}^2$. $\endgroup$ – Ben C Mar 26 '20 at 22:22
  • 2
    $\begingroup$ No worry. I appreciate your response. $\endgroup$ – Ben C Mar 26 '20 at 22:40
  • 1
    $\begingroup$ This may be relevant: mathoverflow.net/questions/9751/… $\endgroup$ – KhashF Mar 26 '20 at 23:57

Your Answer

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

Browse other questions tagged or ask your own question.