# Smooth affine curve with no immersion in projective plane

(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!

• 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. – Ben C Mar 26 '20 at 22:17
• 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 $\mathbb{P}^2$. – Ben C Mar 26 '20 at 22:22
• No worry. I appreciate your response. – Ben C Mar 26 '20 at 22:40
• This may be relevant: mathoverflow.net/questions/9751/… – KhashF Mar 26 '20 at 23:57