# Algebraic Closure

1. Feb 15, 2015

### PsychonautQQ

I'm confused on why exactly the following two statements are equivalent for a finite field K:
-If K has no proper finite extensions, then K is algebraically closed.
-If every irreducible polynomial p with coefficients in K is linear then K is closed.

Can somebody help shed some light on this?

Last edited: Feb 15, 2015
2. Feb 15, 2015

### Fredrik

Staff Emeritus
It's likely that more people would be able to answer if you include some definitions and tell us what sort of object K is.

3. Feb 15, 2015

### PsychonautQQ

Haha sorry, I'm obviously very knew to this stuff. F is a finite field.

4. Feb 16, 2015

### Erland

Let K be a field.
Form the polynomial ring K[x]. Let p(x) be an irreducible polynomial over K with degree n>0. Then, the qoutient ring L=K[x]/[p(x)] is a field, which is a finite extension of K with degree n (basis over K: {1, x, x2, ... xn-1}).
Thus: if there exists an irreducible polynomial over K with degree n>1, then K has a finite proper extension.
On the other hand, if K has a finite proper extension L of degree n>1, and if c ∈ L - K, then the set {1, c, c2, ... cn} is linearly dependent over K (since it contains n+1 elements, and the vector space L over K has dimension n).This means that there are a0, a1, ... an ∈ K such that if we put q(x) = a0 + a1x + .... + anx^n, then q(c)=0. The polynomial q(x) has an irreducibel factor p(x) over K such that p(c)=0. deg p(x) > 1, since otherwise, c ∈ K, which contradicts our assumption,
Thus: if K has a finite proper extension, then there exists an irreducible polynomial over K with degree > 1.