Rick Miranda exercise complete intersection curve. Prove it and find genus. The book by Rick Miranda asks to prove that the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2+x_1^2+x_2^2+x_3^2=0$ is a smooth complete intersection curve. Also asks to determine its genus. 
For the first part, how can I prove that the Jacobian has maximal rank at every point at the common zero locus $X$?
For the second part, do you have any ideas to determine the genus?
 A: Why don't you just write down the Jacobian? It's
$$J=\begin{pmatrix}
x_3 & -2x_2 & -2x_1 & x_0 \\
2x_0 & 2x_1 & 2x_2 & 2x_3
\end{pmatrix}$$
Now, we can look at the points where $J$ does not have maximal rank. It's easy to see that it's the points where the first row is the same as the second row, up to sign (I am assuming we are not in characteristic two).
So, if $x_3=\pm 2x_0$ and $x_1=\pm x_2$, if this point was to be on the curve, you'd have $2x_0^2 = 2x_1^2 = 2x_2^2$ and therefore, $3x_0^2 + x_3^2 = 0$, implying $4x_0^2 = x_3^2 = -3x_0^2$. This means $x_0=0$, and this will imply that all coordinates are zero. But we're in projective space, so there's no such point. 
Let $S=\Bbbk[x_0,\ldots,x_3]$. Let $f$ be homogeneous of degree $e$ and let $g$ be homogeneous of degree $d$ in $S$. You can compute the Hilbert function polynomial of $S/(f,g)$ with the free resolution
$$ 0 \to  S(-e)\oplus S(-d) \xrightarrow{\quad(f,g)\quad} S \twoheadrightarrow S/(f,g)$$
You'll get 
$$ P(x) = \binom{3+x}3 - \binom{3+x-d}3 - \binom{3+x-e}3$$
Because I am no good at calculating, I put your case ($e=d=2$, this means) into a CAS and the constant term is $1$. So, it's of genus $1$.
Also, I feel kinda stupid writing this all down now, because this question is a duplicate.
