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 $1$.