Algebraic closure of the field of two elements

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

Recall that the field of two elements \mathbf F_2 is the ring \mathbf Z/2\mathbf Z of integers modulo 2. In other words, it consists of the elements 0 and 1 with addition 1+1 = 0 and the obvious multiplication. Clearly every nonzero element is invertible, so \mathbf F_2 is a field.

Lemma. The field \mathbf F_2 is algebraically closed.

Proof. We need to show that every non-constant polynomial f \in \mathbf F_2[x] has a root. Suppose f does not have a root, so that f(0) \neq 0 and f(1) \neq 0. Then f(0) = f(1) = 1, so f is the constant polynomial 1. This contradicts the assumption that f is non-constant. \qedsymbol

7 thoughts on “Algebraic closure of the field of two elements

  1. 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. 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.

  3. 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).

  4. 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!

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