Show that no finite field is algebraically closed

Question

otoplilp1 · Accepted Answer

Step 1: Given to Show no finite field is algebraically closed.
Statement: Algebraically closed fields must be infinite.
Step 2: Proof:
Suppose a field has n elements which are finite, then its multiplicative group has

n - 1

elements, that satisfy the equation

x^{(n - 1)} = 1

Therefore, All the fields elements including zero satisfy the equation

x^{n} = x

So, the equation,

x^{n} = x + 1

Has no roots.
Thus, the field is not algebraically closed.
Hence proved.

Navreaiw · Accepted Answer

No.  Suppose a finite field has elements a1,a2,…,an. Then the polynomial P(x)=(x−a1)(x−a2)…(x−an)+1 has no roots, because P(x) is always 1 and not 0.

user_27qwe · Accepted Answer

Nope. An algebraically closed field must, for example, contain all roots of unity of all orders: the roots of the polynomial $X^{N} - 1$ . For any N not divisible by the characteristic, this polynomial must have N distinct roots.
Therefore, algebraically closed fields must be infinite. You can start with a finite field such as $F_{p}$ , the field of residues modulo a prime p, and then you can take an algebraic closure of it $F_{p}$ . You’ll get a wonderful algebraically closed field of characteristic p, but it’s infinite.

Show that no finite field is algebraically closed

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