Classification of connected unipotent two-dimensional algebraic groups over an algebraically closed field

Statement

In characteristic zero

Suppose $K$ is an algebraically closed field of characteristic zero and $G$ is a connected unipotent two-dimensional algebraic group over $K$. Then, $G$ must be an abelian algebraic group and in fact $G$ is a direct product of two copies of the additive group of $K$. Another way of putting this is that $G$ is isomorphic to the additive group of a two-dimensional vector space over $K$.

Note that in characteristic zero, any unipotent algebraic group is connected, so the adjective "connected" can be dropped from the above statement.

In prime characteristic

Suppose $K$ is an algebraically closed field of characteristic equal to a prime number $p$ and $G$ is a connected unipotent two-dimensional algebraic group over $K$. Then, $G$ must be an abelian algebraic group and there are two possibilities for it:

1. The direct product of two copies of the additive group of $K$
2. The additive group of the truncated ring of Witt vectors of length two over $K$

Related classifications

• Classification of groups of prime-square order: This is the finite group analogue for the prime characteristic version. The proof in both cases is fairly similar.
• Classification of connected one-dimensional algebraic groups over an algebraically closed field
• Classification of connected unipotent abelian algebraic groups over an algebraically closed field

Related cohomology computations

• Algebraic second cohomology group for trivial group action of additive group of a field on additive group of a field

Facts used

1. Classification of connected unipotent abelian algebraic groups over an algebraically closed field

Proof

The proof has two steps. The first step is showing that any connected unipotent two-dimensional algebraic group must be abelian. The second step is to invoke the classification in the abelian case from Fact (1).