Discriminant of elliptic curves as 1-dimensional abelian variety

Question

We define an abelian variety over a field k as a connected and complete algebraic group over k. And we define an elliptic curve 1-dimensional(as schemes) abelian variety. This should equals a normal elliptic curve, so this should have the discriminant. However the definition of an elliptic curve as an abelian variety don't have Weierstrass forms, hence the discriminant can't be defined as usual. Can we define the discriminant?

Answer 1

I am expanding a little bit on my comment.

Let $E$ be an elliptic curve over $k$, i.e., a $1$-dimensional Abelian variety over $k$. Let us suppose for simplicity that $k$ is algebraically closed. Denote the identity of the group law of $E$ by $P_0$.

Being an Abelian variety, the canonical divisor $K_E$ is trivial. Moreover, the genus of $E$ is $1$. Hence, if $D_n := n[P_0]$ for $n \in \mathbb Z$, Riemann-Roch (cf. Hartshorne, Chapter IV, Theorem 1.3) for the divisor $D_n$ reads as

$$l(D_n) - l(-D_n) = n.$$

Here, $l(D) = \dim L(D)$, where $L(D) = \{ f \in k(E)^* \ | \ \mathrm{div}(f) \geq -D\}$. Since the degree of a principal divisor is $0$, we obtain that $l(D) = 0$ if $\deg(D) < 0$. Thus, the formula obtained by Riemann-Roch yields

$$l(D_n) = n, \ \ \ \ \text{ if } n \geq 0.$$

Having obtained this, we may choose elements

$$y \in L(D_3) \setminus L(D_2), \ \ \ \ x \in L(D_2) \setminus L(D_1).$$

The $7$ elements

$$1, \ x, \ x^2, \ x^3, \ y, \ xy, \ y^2$$

are all contained in $L(D_6)$, which is a $6$-dimensional space. Hence, they are linearly dependent, so that there are $\lambda_0, ..., \lambda_6 \in k$ such that

$$\lambda_0 \cdot 1 + \lambda_1 x + \lambda_2 x^2 + \lambda_3 x^3 + \lambda_4 y + \lambda_5 xy + \lambda_6 y^2 = 0.$$

Note that $x^3$ and $y^2$ have a pole at $P_0$, which has order $6$, and all of the other five remaining elements have poles of order $< 6$ in $P_0$. Hence, in order for the above linear combination to equal $0$, the poles of $x^3$ and $y^2$ have to cancel out. Thus $\lambda_3 \neq 0 \neq \lambda_6$.

Without loss of generality, $\lambda_6 = 1$ and so one gets an equation of the form

$$y^2 = p(x) + yq(x),$$

where $p(x)$ is a cubic polynomial in $x$ and $q(x)$ is a linear polynomial in $x$. By elementary changes of coordinates, this equation can be simplified to an equation in Weierstraß form.