Calculating the genus of Quartic model $y^2=x^4+bx^3+cx^2+dx+e$ of elliptic curves. This link says that $y^2=x^4+bx^3+cx^2+dx+e$ is an elliptic curve. 


*

*How do we compute its genus (which should be 1)? 

*Under what conditions is $y^2=p(x)$ an elliptic curve where $p(x)$ is polynomial in $x$ of degeree $\ge 5$?

 A: This is called a hyperelliptic curve.
Basically your questions are already answered in the above wiki link, so please read that first.
I just want to add several important details as complements:


*

*One usually should assume that the polynomial $f(x) = x^4+bx^3+cx^2+dx+e$ doesn't have multiple roots, i.e. that $f$ is coprime to its derivative $f'$. Any multiple root of $f$ will lead to a singularity of the curve, hence will decrease the genus (in this case, necessarily to $0$).

*Even when we assume this, the plane curve $y^2 = f(x)$ does not represent an elliptic curve, even if we look at its completion in $\Bbb P^2$ (that is, the projective curve defined by $Y^2Z^2 = X^4 + bX^3Z + cX^2Z^2 + dXZ^3 + eZ^4$). This is because the point at infinity $(X, Y, Z) = (0, 1, 0)$ becomes a singularity. The correct statement is that the unique complete smooth curve which is birational to the curve $y^2 = f(x)$ has genus $1$. This curve can be obtained by blowing up (possibly multiple times) the previous projective curve at the point at infinity. The point at infinity eventually separates as two "points at infinity".

*Finally, that complete smooth curve is still one step from being an elliptic curve, since one should fix a rational point on the curve (the "point at infinity" for the Weierstrass model, or the neutral element for the group law) to get an elliptic curve. This step however is easy: a common choice is any one of the two "points at infinity" obtained via blow up.

