I think I learned about this from a comment on MathOverflow.

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.

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!

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

April Fool !

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!

Hm something funny is going on, because another lemma says every algebraically closed field must be infinite. For example, see this proof https://proofwiki.org/wiki/Algebraically_Closed_Field_is_Infinite.

This method is known as proof by variation of definition: prove that the polynomial f is constant (meaning constant-valued) and claim that it is constant (meaning degree zero).

The proof above is wrong because it views a polynomial over a finite field as a function over that field, and this was the problem. The maps that assigns to a polinomial a function is not injective for finite fields!