Field with 4 Elements has only Order 2 Elements/Proof 2

Proof
Let $\struct {F, +, \times}$ be a field of order $4$ whose zero is $0_F$.

By definition, $F$ is a Galois field.

The additive group $\struct {F, +}$ of $F$ can be one of two:


 * $(1): \quad$ the cyclic group of order $4$, generated by the identity of $\struct {F, +}$ which is $0_F$

or:
 * $(2): \quad$ the Klein $4$-group, whose elements are all of the form $a + a = 0_F$.

$(1)$ is the additive group of a field.

Then the characteristic of $F$ is $4$.

But from Characteristic of Galois Field is Prime it follows that $F$ is not a field.

From that contradiction it follows that the additive group $\struct {F, +}$ of $F$ cannot be the cyclic group of order $4$.

Hence for $F$ to be a field at all, it is necessary for $\struct {F, +}$ to be the Klein $4$-group.