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

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) uses::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).