genus of $y^2＝1＋x^4$ I heard $E:y^2＝1＋x^4$ is Elliptic curve because it is birational to $y^2＝x^3-4x$.
But I have a question.
The projective closure of $E$ is $Y^2Z^2＝Z^4＋X^4$, and this is smooth curve（if it is not smooth, then the curve is not elliptic curve）, so its genus is
$（4-1）（4-2）/2＝3$.
But elliptic curve has genus 1.
Where am I missing?
 A: The confusion comes from the fact that you've heard the definition of an elliptic curve require $E$ to be smooth. In fact, it's often convenient to allow $E$ to be singular, but asserting that the base point $O$ is nonsingular.
From the genus degree formula you are getting the wrong answer because $C$ is singular - and the formula only holds for smooth plane curves. You assert your curve is smooth but this is not the case - on the affine patch $y = 1$ we have $f(x,z) = z^2 - x^4 - z^4$ which has vanishing first partial deriviatives at $(0,0)$, hence $C$ is not smooth at the point at infinity.
To put $E$ in Weierstrass form it is suggested in the comments that you can blowup the singular point (at infinity). However in the spirit of Silverman Chapter III, let's just think about what $x$ and $y$ "are" in a Weierstrass equation, and follow the proof that one must exist.
You have your special point $O$, and $\{1,x'\}$ is a basis for $L(2O)$ and $\{1,x',y'\}$ is a basis for $L(3O)$ - then $x'$ and $y'$ will satisfy some Weierstrass equation. So one should just go out and find $x'$ and $y'$ in the wild.
In our case, if we take $O = (0,1)$ we can just take $x' = \frac{y + 1}{x^2}$ and $y' = \frac{y + 1}{x^3}$. Now $L(6O)$ is overloaded with stuff - namely $1, x', y', (x')^2, (y')^2, x'y',(x')^3$, so we just do linear algebra looking for a relation amongst these, one sees that
$$2(y')^2 + x'^3 - x' = 0$$
modulo the equation for your curve. Replacing $x'' = -2x'$ and $y'' = 2y'$ we see that
$$(y'')^2 = (x'')^3 - 4x''$$
and we have our Weierstrass equation.
