# plane hyperelliptic curves

I'm aware that h/w problems are frowned upon (understandably) here. However - this really is just inspired by some h/w related confusion, so hopefully that's ok. Anyway, can one have a smooth projective plane curve be hyperelliptic (i.e admitting a double cover of the projective line and of genus greater than 1)? It is easy enough to get affine curves that double cover the affine line but upon taking the closure in the projective plane I always encounter singularities. It is my understanding that there is a way to resolve such singularities but I imagine that this needn't result in something isomorphic to a plane curve (merely something birationally equivalent to one.) Thanks in advance for any help given.

## 1 Answer

I don't know how to answer this question at homework level. If you have a plane curve of degree $d$, it has lots of maps to $P^1$ of degree $d-1$ by projecting from points. If the curve is also hyperelliptic, it has a map of degree two to $P^1$. For at least one of the maps of degree $d-1$, the conditions of the Castelnuovo genus bound (you'll have to look that up) is satisfied and we get that the genus satisfies $g \le d-2$. Now, if your plane curve is smooth (which I had not previously assumed) then $g = (d-1)(d-2)/2$, which combined with the previous bound gives, not surprisingly, $d \le 3$.

• I beg to differ. This is a trivial homework problem (which the student should do herself). Hint: What is the definition of a hyperelliptic curve in terms of $|K_C|$? What is $K_C$ of a plane curve?
– VA.
Commented Dec 19, 2009 at 4:55
• @ VA - this is not in fact a h/w problem. This arose due to a h/w problem that seemed to suggest that there may have been such curves for d >4 (of course it did not in fact suggest this, I was being very foolish.) Perhaps the problem should just be removed if it really is of no interest. (Also, Emile is a guy's name!) Sorry for the waste of time.
– user2697
Commented Dec 19, 2009 at 6:11