TITLE: Problem in Rick Miranda: finding genus of a Projective curve
QUESTION [29 upvotes]: I asked the following question in stack exchange (https://math.stackexchange.com/questions/21164/problem-in-rick-miranda-finding-genus-of-a-projective-curve) a few days ago, but didn't get any solution.  Somebody please help me with it.
I have just started learning Riemann Surfaces and I am using the book by Rick Miranda: Algebraic curves and Riemannn Surfaces. #F in section 1.3 asks to determine the genus of the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2 + x_1^2 +x_2^2 +x_3^2 = 0$.  #G also has a similar question in which he asks to determine the genus of the twisted cubic.  Please explain how to approach this type of question.

REPLY [5 votes]: This was meant to be a comment on the ending remark of Sándor Kovács' answer, but it got too long to fit:
In a student seminar today, some people had the old edition of Miranda, and some had the new edition, so we had both the original problem and your degenerate version.  (The old edition has the degenerate version).
The way we ended up seeing that solution set $X$ of $x_0x_1 = x_2x_3$ is $\mathbb{P}^1\times\mathbb{P}^1$ was to observe that we could rewrite it as $det\begin{pmatrix} x_0 & x_2 \\\\ x_3 & x_1 \end{pmatrix} = 0$ or  $det\begin{pmatrix} x_0 & x_3 \\\\ x_2 & x_1 \end{pmatrix} = 0$.  So both matrices must be rank 1.  So we have two maps from $X$ to $\mathbb{P}^1$, namely the maps which send a point of $X$ to the corresponding element of the nullspace of one or the other  matrix.  This gives a map $X \to \mathbb{P}^1\times\mathbb{P}^1$, which is not too hard to compute explicitly (in fact $(\langle a,b\rangle, \langle c,d\rangle) \mapsto \langle ac,ad, bc, bd \rangle$ just as you say).  So we are really looking at the zero set of  $a^2c^2+a^2d^2+b^2c^2+b^2d^2=0$ in $\mathbb{P}^1\times\mathbb{P}^1$.  At this point we basically followed the rest of your post.  I just thought someone might like the observation about determinants!