Are elliptic curve irreducible?

Question

Suppose we have a typical elliptic curve (smooth, projective curve with a distinguished point $\mathcal{O} = [0,1,0]$) over some algebraically closed field $\overline{K}$ $$E: F(X,Y,Z) = Y^2Z + a_1XYZ + a_3YZ^2 - X^3 - a_2X^2Z - a_4XZ^2 - a_6Z^3 =0$$ is $F$ irreducible in $\overline{K}[X,Y,Z]$? I cant find anything definitive online.

Answer 1

Just to get this off the unanswered list.

If $F$ is reducible (e.g. $F(x,y,z)=(x-y)(z-w)(y-z)$) then it cannot be smooth. The reason is the following nice observation:

Observation 1: Let $X$ be a Noetherian regular connected scheme. Then, $X$ is integral.

Proof: We need to show that $X$ is irreducible and reduced. Since $\mathcal{O}_{X,x}$ is regular for all $x$, and regular rings are integral (e.g see [1, Tag00NP]) we see that $X$ is clearly reduced. To see that it’s irreducible note that we can decompose $X$ into finitely many irreducible components $C_i$. Note then that if $X$ is not irreducible we must have that $C_i\cap C_j\ne \varnothing$ for some $i\ne j$ else then $\displaystyle X=\bigsqcup_i C_i$ which contradicts that $X$ is connected. But, let $x$ be a closed point of $C_i\cap C_j$ (for the $i,j$ where they intersect). Note then that $\mathcal{O}_{X,x}$ cannot be integral since $\mathrm{Spec}(\mathcal{O}_{X,x})$ contains $\eta_{C_i}$ which are distinct minimal primes. Thus, we arrive at a contradiction. Thus, $X$ is irreducible as desired. $\blacksquare$

This seems eminently useful if you know, a priori, that your hypersurface $E:=V(F)\subseteq \mathbb{P}^2_k$ is connected. But, again, this is actually automatic!

Observation 2: Let $F\in k[x,y,z]$ be a non-zero homogenous polynomial. Then, $C:=V(F)\subseteq \mathbb{P}^2_k$ is connected.

Proof: There are several proofs of this fact. But, one is to note that if $n=\deg(F)$ then we have a short exact sequence

$$0\to \mathcal{O}(-n)\to \mathcal{O}_{\mathbb{P}^2_k}\to i_\ast \mathcal{O}_C\to 0$$

where $i:C\hookrightarrow \mathbb{P}^2_k$ is the inclusion. From the LES in cohomology we see that we have an exact sequence

$$0\to \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\to (i_\ast \mathcal{O}_{\mathbb{P}^2_k})(\mathbb{P}^2_k)\to H^1(\mathbb{P}^2_k,\mathcal{O}(-n))$$

But, this last group is zero (e.g. see [1, Tag01XT]) and so we see that

$$k\cong \mathcal{O}_{\mathbb{P}^2_k}(\mathbb{P}^2_k)\cong (i_\ast\mathcal{O}_C)(\mathbb{P}^2_k)=\mathcal{O}_C(C)$$

and so $\mathcal{O}_C(C)$ has no non-trivial idempotents, and so is connected. $\blacksquare$

EDIT: This is somewhat tangential, but alternatively to Observation 1 if you want to assume that your $E$ is a group, then it’s connected if and only if it’s irreducible (e.g. see Lemma 8 of this blog post of mine).

References:

[1] Stacks project authors. The Stacks Project. https://stacks.math.columbia.edu