The curve $y^2 = f(x)$ where $f$ has degree $d$ and no repeated roots has genus $[(d-1)/2]$?

Let $f$ be a polynomial in $x$ of degree $d$ (over $\mathbb{C}$, say) without repeated roots. I've heard that the curve $y^2 = f(x)$ has genus $[(d-1)/2]$, but I can't find a proof.

To be more precise, I'm asking about the genus of the projective curve with function field $\mathbb{C}(x,y)$ where $y^2 = f(x)$.

For $d > 4$ this is a hyperelliptic (for which I am also unable to locate a proof of the genus), but I'm most interested in the case $d = 4$.

Edit: For $d \geq 4$, the point at infinity of the projective plane curve given by the projective closure of $y^2 = f(x)$ is singular by the Jacobian criterion, and this is the only singular point of the curve. Now I think it might be a good idea to use one of those genus-degree formulas for singular projective plane curves, $g = (d-1)(d-2)/2 - C$ for some correction term $C$ involving properties of the singular points of the curve. In the case $d = 4$ we should have $g = 1$, so we must find a formula that assigns $C = 2$ to the singular point at infinity. Now I'm asking for a nice definition of the genus for which such a formula is available. I would prefer not to use any of the theory of Riemann surfaces.

• The degree-genus formula applies to smooth projective curves. When is the projective closure of the curve $y^2=f(x)$ smooth at the point at infinity? – user64687 Jul 30 '13 at 17:32
• A great reference is Miranda's Algebraic Curves and Riemann Surfaces (Chapter III, Lemma 1.7) – edo arad Jul 30 '13 at 17:40
• Thanks. Indeed I missed that the projective closure is singular at the point at infinity for $d \geq 4$ (by the Jacobian criterion). I've updated the question. – Ricardo Buring Jul 31 '13 at 18:30
• Let me repeat that it is not a good idea to embed your curve in the projective plane and resolve the singularities. Consider rather your (projective) smooth curve as a double cover of $\mathbb P^1$ and use Riemann-Hurwitz formula to compute the genus. – user18119 Jul 31 '13 at 21:33

Disclaimer: This is not an answer, rather a long comment. Essentially, I am only backing up QiL'8's comment above.

The upshot:

1. I know of no extension (does someone know of one?) of the degree-genus formula for projective plane curves, which would encompass the special case you are asking about.

2. In principle you could resolve the singularities of the projective closure of $y^2-f(x)$ and then appy a variant of the degree-genus formula, but this is unwieldy in practice, and relies in your case on the exact form of $f(x)$ (so does probably not work in a reasonable way in the generality you desire).

3. On the other hand, the curve $y^2-f(x)$ has the very special property that its smooth, projective model $C$ can be realized as a double cover of $\Bbb{P}^1$, and then you can use the Riemann-Hurwitz formula to compute the genus of $C$ (and hence of $y^2-f(x)$). Admittedly, this is Riemann surface theory in disguise, but my impression is that it would be unwise to completely ignore this very special aspect of the geometry of $C$, and instead try to apply a totally general mechanism, which would in principle work for every other projective plane curve just as "well".

1.
If I understand you correctly, you would like to have a generalization of the degree-genus formula for projective plane curves, admitting more or less arbitrary singularities.
Now there are extensions of the degree-genus formula for curves admitting at most ordinary multiple points (cf. Fulton, Proposition 8.3.5), and with the Plücker formulas you can even compute the genus of a curve having in addition some ordinary cusps as singularities (cf. Griffiths-Harris, p.277f, in particular p.280). But beyond that, I don't know of any general formula for the genus of an arbitrary projective plane curve.

The problem is now that the projective closure $C$ of $y^2-f(x)$ (with $\deg(f) \geq 4$) has a non-ordinary cusp at infinity, so that neither of the above formulas apply.

2.
So if you really want to use some variant of the degree-genus formula, then you first have to find a projective plane curve $C'$, which is birational to $C$ and has "better" singularities as $C$.
Ideally of course, you'd like $C'$ to have no singularities at all, but this does not work in general. There are projective plane curves, whose non-singular model is not a plane curve (e.g. the projective closure of $y^2-f(x)$ with $\deg(f)=9$).

The best you can get in general is that $C'$ has only ordinary multiple points as singularities (cf. Fulton, Theorem 7.4.2). This of course sounds great, because you can then apply the above-mentioned extension of the degree-genus formula to $C'$ to compute its genus, and hence a fortiori to compute the genus of $C$ as well.
But the problem is that computing $C'$ explicitly, while possible in principle, can be extremely arduous, and then you still don't know anything about the multiple points of $C'$, which you have to find on top of that (of course together with its respective multiplicities!). So this procedure simply seems to be impractical for your purposes.

3.
So you should better find another way, which really uses the very particular geometry of $y^2-f(x)$. Indeed, the smooth, projective model $C$ for $y^2-f(x)$ can be realized as a double cover \begin{split} \pi: C \rightarrow \Bbb{P}^1 \end{split} of the projective line, and from then on you can compute the genus of $C$ from the Riemann-Hurwitz formula. For this you only need to know that $\Bbb{P}^1$ has genus zero, and you have to know the branch points of $\pi$, but these are simply the zeros of $f$ and additionally $\infty \in \Bbb{P}^1$, if $\deg(f)$ is odd. Hence you get, since every ramification point has necessarily multiplicity $2$ \begin{split} 2g_C-2=(2g_{\Bbb{P}^1}-2) \cdot \deg(\pi)+\sum_{P \in C}e_P-1=-4+d^* \end{split} where $d^*=\deg(f)$ if $\deg(f)$ is even, and $d^*=\deg(f)+1$ if $\deg(f)$ is odd, from which the result follows.
I recommend you have a look at Exercise A.4.2. of Hindry-Silverman, Diophantine Geometry, which asks you to derive exactly the result you are asking for, giving very generous hints along the way.

• Thank you very much for your comment, Nils. Indeed, I was hoping to find a more elementary proof at least in the case $n = 4$ (I don't understand Riemann-Hurwitz yet), but as per your comment this might not be possible. Since you do answer the question with your third point, I'll accept your comment as an answer. – Ricardo Buring Aug 4 '13 at 10:44
• You're welcome, Ricardo. The Riemann-Hurwitz theorem is an extremely convenient tool for genus computations, and the best thing is that you don't need to fully understand its proof in order to use it profitably. So not understanding it completely shouldn't prevent you from applying it, in my opinion. – Nils Matthes Aug 4 '13 at 14:19