# Algebraic closure of the field of two elements

Recall that the field of two elements is the ring of integers modulo . In other words, it consists of the elements and with addition and the obvious multiplication. Clearly every nonzero element is invertible, so is a field.

Lemma. The field is algebraically closed.

Proof. We need to show that every non-constant polynomial has a root. Suppose does not have a root, so that and . Then , so is the constant polynomial . This contradicts the assumption that is non-constant. ## 5 thoughts on “Algebraic closure of the field of two elements”

1. Owen Biesel on said:

Amazing! I’ve had trouble picturing the algebraic closure of finite fields in the past, but it’s great to have this simple example of a finite field that’s already algebraically closed!

Can you do F1 in a future post? I’ve wondered about that one too!

2. Gianluca on said:

Hi, the polynomial seems to me a counterexample. Am I missing something?

• A. B. on said:

April Fool !

3. Sebastian Walawili on said:

F_2 is not algebraically closed because x² + x + 1 has no roots over it. This is actually how you construct GF(2ⁿ) for any n (quotient F_2[x] by the ideal generated by an irreducible polynomial of degree n over F_2). The algebraic closure of F_2 is the union of all GF(2ⁿ) for all n, which is infinite as a set.
Cheers!