Patricia Crane

2021-12-26

Show that no finite field is algebraically closed

otoplilp1

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
${x}^{\left(n-1\right)}=1$
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

No.
Suppose a finite field has elements ${a}_{1},{a}_{2},\dots ,{a}_{n}$. Then the polynomial $P\left(x\right)=\left(x-{a}_{1}\right)\left(x-{a}_{2}\right)\dots \left(x-{a}_{n}\right)+1$ has no roots, because P(x) is always 1 and not 0.

user_27qwe

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.

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