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 ![f \in \mathbf F_2[x]](https://lovelylittlelemmas.rjprojects.net/wp-content/ql-cache/quicklatex.com-4a513b79a98813c2aa8183054921960a_l3.svg "Rendered by QuickLaTeX.com")
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!