Let $C_0: y^{2}=x^4-7$ be a genus $1$ curve.
Let another affine curve be
$C_1: v^{2}=u^{4}(1/u^4-7)=1-7u^4$.
The glueing maps between the two charts are given by
$(x,y)\mapsto (1/x,y/x^{2})$
and
$(v,v)\mapsto (1/u,v/u^{2})$
Gluing these two curves gives gives projective variety $C$.
I have question about this curve.
$1.$ Why is $C$ a curve in $\Bbb{P}^3$, not in $\Bbb{P}^2$?
$2$. Why is $C$ smooth ? (This is a main question)
I'm wondering what the equation of $C$ is, but these question need no equation of $C$, Regarding $1$. Regarding $1$, $C_0$ and $C_1$ are affine charts of $C$, and $C_0$ and $C_1$ is singular in its projective closure, so $C$ seems to be singular also, I'm wondering where singular point was removed.
Best Answer
As you start from a polynomial of degree 4, it seems hard to me to describe the elliptic curve $C$ as a cubic in $\mathbb P^2$. I think it is simpler to give it as complete intersection of 2 quadrics in $\mathbb P^3$ as follows. First consider the closure of $C_0$ (or $C_1$, it is the same) in the wheighted projective space $W:=\mathbb P (1,1,2)$: $y^2=x^4-7w^4$ ($y$ is the variable of degree 2). Now, you can embedd $W$ in $\mathbb P^3$ (with coordinates $(a,b,c,d)$) via $(x:w:y)\mapsto (x^2:xw:w^2:y)$, whence the equations describing your curve are $$ac=b^2\quad d^2=a^2-7c^2$$ It should be now easy to see that $C$ is smooth.
This curve $C$ is the normalization of the projective closure of $C_0$, and the singular point was removed and replace by $C_1$.