My question regards a perplexity I have on how to apply the genus-degree formula for irreducible, projective, complex plane curves. Consider first the affine complex plane curve given by the equation $$C: \ (x-2)(x-1)(x+1)(x+2) -y^2 = x^4-5x^2+4 -y^2=0.$$ The Jacobian is given by $$(x(2x+\sqrt{10})(2x-\sqrt{10}), -2y)$$, so it never vanishes on $$C$$. Let us now look at the compactification $$\hat{C}$$, which has the same points as $$C$$ plus a point at infinity with coordinates $$[x:y:z]=[0:1:0]$$. This point lies in the affine chart with $$y=1$$, and the affine equation for $$\hat{C}$$ in terms of $$x$$ and $$z$$ in that chart is $$x^4-5z^2x^2+4z^4-z^2 = 0.$$ The differential $$(4x^3-10xz^2, -10x^2z+16z^3-2z)$$ vanishes at $$(x,z)=(0,0)$$, hence $$\hat{C}$$ has one singularity: the point at infinity. If we pretend for a moment that it doesn't (i.e., that it is smooth), one can do the "usual construction" to see that it is topologically a torus: one can draw two cuts along $$[-2,-1]$$ and $$[1,2]$$ on two copies of the Riemann sphere and glue them together with the right orientation. So if $$\hat{C}$$ were regular, it would be an elliptic curve, and in particular have genus $$1$$. However, the genus-degree formula for projective plane curves predicts genus $$3$$, since the equation of $$\hat{C}$$ has degree $$4$$. But the Wikipedia article on the genus-degree formula also mentions that the formula actually gives the arithmetic genus and that for every ordinary singularity of multiplicity $$r$$, the geometric genus is smaller than the arithmetic genus by $$\frac{1}{2}r(r-1)$$. Now, I am not really sure about how to measure the multiplicity of a singularity, but in this case it seems that for any value of $$r \ge 0$$, we never have that $$3-\frac{1}{2}r(r-1)=1$$. So the geometric intuition and the formula seem to disagree. The only other thing that comes to my mind is that I have not checked yet that $$\hat{C}$$ is irreducible, but this can be checked on $$C$$ using Eisentein's criterion applied to the polynomial ring $$(\mathbb{C}[x])[y]$$ using the prime ideal $$\mathfrak{p}=(x+1)$$.

Reassuming, my question is: what is the genus of $$\hat{C}$$? If it is $$1$$, why is the genus-degree formula wrong? If it is $$3$$, why is the geometric intuition wrong? After all, also the article of Wikipedia on elliptic curves seems to confirm that $$\hat{C}$$ should have genus $$1$$.

• the genus is certainly $1$, but the formula in wikipedia seems not fully explained - see mathoverflow.net/questions/122725/… Nov 16, 2019 at 22:24
• Just in case the nice answer below didn't completely clear things up: the arithmetic genus is 3, which only depends on the degree of the polynomial cutting out your curve; this is what the genus-degree formula calculates. The geometric genus is 1, which takes in to account the geometric information about the singularity at $[0:1:0]$. In the case that the curve is smooth, the numbers must agree, but if the curve is singular, they may differ like you've found here. Nov 17, 2019 at 1:20

I think the reason the formula does not apply is that the singularity is not an "ordinary singularity of multiplicity r" (a.k.a. $$r$$ distinct lines crossing at a point).

From your second chart, we have a cusp at the origin, and if we blow it up (basically substitute $$xz$$ for $$z$$ and factor out an $$x^2$$ -- this is equivalent to enlarging the coordinate ring by adjoining $$z/x$$, which is integral over it), we're left with $$4 x^2 z^4-5 x^2 z^2+x^2-z^2=0$$ And the lowest degree part is $$x^2 - z^2 = 0 = (x-z)(x+z)$$, so locally the singularity is now ordinary of multiplicity 2. If this were the entire singularity, we would just be subtracting $$1$$, but since we also needed the function $$z/x$$ to resolve the cusp, we also subtract one more.

More details to justify the last part. For any smooth projective curve $$C$$, let $$f: \tilde{C} \to C$$ be the normalization. There is a short exact sequence

$$0 \to \mathcal{O}_C \to f_*\mathcal{O}_{\tilde{C}} \to F \to 0$$,

where $$F$$ is supported only along the singularities of $$C$$, and is a finite-length $$\mathcal{O}_C$$-module. Let its length be $$\ell$$. The cohomology gives

$$0 \to H^0(\mathcal{F}) \to H^1(\mathcal{O}_C) \to H^1(f_*\mathcal{O}_{\tilde{C}}) \to 0$$,

so in particular $$g_a(C) + \ell = g(\tilde{C})$$, where $$g_a(C)$$ means the arithmetic genus and $$g(\tilde{C})$$ is the geometric genus. So $$\ell$$ is basically "how many extra regular functions" the normalization has.

The claim is that $$\ell = 2$$. Resolving the cusp introduced $$z/x$$ to the coordinate ring. I think $$z/x$$ satisfies a quadratic polynomial (this is true at least locally, you can check that $$(1-4z^2) (z/x)^2 + (5z^2 - x^2) = 0,$$ and since the leading coefficient doesn't vanish at the origin, this is as good as a monic polynomial.)

So we've only introduced one more function in this step. Then, the second step introduces the $$\tfrac{1}{2}r(r-1) = 1$$ additional function to separate the two lines.