Is $X^2 + Y^2 = U^2, X^2 - Y^2 = V^2$ in $\mathbb{P}^3_k$ an elliptic curve? If so, please show the explicit transformation to the Weierstrass equation.
Now genus-degree formula fails, so I can't calculate the genus.
 A: The quadric $X^2+Y^2=U^2$ is parametrised by
$$(X,Y,Z)=(T^2-W^2,2TW,T^2+W^2)$$
and so your curve is birational to that with equation
$$V^2=(T^2-W^2)^2-(2TW)^2=T^4-6T^2W^2+W^4.$$
If you can find a rational point on this, you can send it to infinity and get a Weiestrass equation.
A: We will assume that the field $k$ is algebraically closed and call our quadrics $$Q_1=V(X^2+Y^2-U^2), \: Q_2=V(X^2-Y^2-V^2)\subset \mathbb P_{X:Y:U:V}^3(k)$$ If $char.k=2$ the $Q_i$'s are double planes whose intersection is a non-reduced line and thus certainly not an elliptic curve. So let us assume from now on that $char.k \neq 2$.    
Our quadrics are both singular: they are  quadric cones with only singularity at their vertex $$p_1=[0:0:0:1],\quad p_2=[0:0:1:0]$$ This looks like a bad omen, but the good news is that since $p_1\notin Q_2$ and $p_2\notin Q_1$, the intersection $E=Q_1\cap Q_2$ is included in the smooth locus of both quadrics: $E\subset Q_1^{\operatorname {smooth}}\cap Q_2 ^{\operatorname {smooth}}$.
Fine,  but is $E$ itself smooth?
YES, E is smooth!
Indeed, at a point $e=[x:y:u:v]\in E$ the gradients of the polynomials  $X^2+Y^2-U^2, \:X^2-Y^2-V^2$ defining our quadrics  are $2(x,y,-u,0)$ and $2(x,-y,0,-v)$.
Since they are  linearly independent, the good old criterion of advanced calculus says that our quadrics intersect transversally at $e$ and thus we see that $E$ is smooth at any of its points $e$.
From now on it is plain sailing to the elliptic coast :  the intersection of two surfaces of degree $2$ in $\mathbb P^3$ has arithmetic genus $p_a(E)=1$ (Hartshorne: Chapter I, Exercise 7.2 (d), page 54).
Since $E$ is smooth its geometric genus $p_g(E)=p_a(E)$ equals $1$ too and $E$ is an elliptic curve!
