I was reading Beauville's Complex algebraic surfaces, at page 5 there is an example in which curves in $\mathbb{P}^1 \times \mathbb{P}^1$ are classified by the bidegree up to linear equivalence.

I'm happy with that, and it makes sense that the defining polynomial determines the linear equivalence class of the curve. However, the bidegree is also the degree of the morphism defined by the projections to the lines $\lbrace 0 \rbrace \times \mathbb{P}^1$ and $\mathbb{P}^1 \times \lbrace 0 \rbrace$ . Thence, I was wondering if the bidegree of the curve also gives some contraints on the other possible morphisms from the curve to the two lines above.

I heard somewhere that the degree of the curve is somehow related to its gonality, for instance, is there such a relation?

I'm sorry if my question can sound confused, I just started facing algebraic geometry topics and still don't know precisely how to work with these concepts practically.

  • $\begingroup$ I apologize for what might be an extremely novice question, but what does bidegree mean (besides the degree of the two projection morphisms)? Thanks! $\endgroup$ Mar 30, 2014 at 16:20
  • 1
    $\begingroup$ Analogous to degree (counting intersections with a generic hyperplane), you count intersections with the generic lines $\mathbb P^1\times \text{pt}$ and $\text{pt}\times\mathbb P^1$. These are, as you surmised, @Alex, the degrees of the corresponding projections. $\endgroup$ Mar 30, 2014 at 16:52


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