what is genus of complete intersection for: $F_1 = x_0 x_3 – x_1 x_2 , F_2 = x_0^2 + x_1^2 + x_2^2 + x_3^2$

In the below problem I want to answer second part:

Show that the curve in $\mathbb{P}^3$ defined by the two equations $x_0 x_3 = x_1 x_2$ and
$x_0^2 + x_1^2 + x_2^2 + x_3^2 = 0$ is a smooth complete intersection curve. What is its
topological genus?

but I don’t know how can I calculate it.

Solutions Collecting From Web of "what is genus of complete intersection for: $F_1 = x_0 x_3 – x_1 x_2 , F_2 = x_0^2 + x_1^2 + x_2^2 + x_3^2$"

The jacobian $2\times 4$ matrix $Jac_P(F_1,F_2)=(\frac {\partial F_i}{\partial x_j}(P))$ has rank $2$ at every point $P\in C$ of the intersection curve $C=V(F_1,F_2)$.
This follows from an easy calculation and has as a consequence that $C$ is smooth.
The topological genus of $C$ is then equal to its arithmetic genus.
The arithmetic genus of the complete intersection of two surfaces of degrees $a,b$ in $\mathbb P^3_\mathbb C$ is given by the formula $$p_a(C)=\frac {1}{2}ab(a+b-4)+1$$ [cf. Hartshorne, Chapter I, exercise 7.2 (d), page 54]
In your case $a=b=2$ and the formula shows that the required genus of your curve is $1$.

The above reasoning shows that more generally the smooth complete intersection of two quadrics in projective $3$-space always has genus one i.e. is an elliptic curve : this a well-known and fundamental result in the theory of space curves.