Finite field is algebraically closed under constraint?

[gepolv]

A field K is called algebraically closed field if any no-zero polynomial has at least one root in K.

Given finite field F_q, q=p^m, p is a prime and m is non-negative integer. A famous property of finite field is any element in F_q satisfies: x^q=x.

Then I have such an assumption:
F_q[x_1,x_2,...x_n] has at least one root with the constraint "degrees less than q".

I have two questions:
1: Is this assumption true?
2: If it is true, can I say "F_q" is algebraically closed under condition "degrees less than q"?

Thanks a lot.
Gepo

 

[VKint]

Unfortunately, no. For example, the polynomial x^2 + 1 has no root in \mathbb{F}_p whenever p is a prime congruent to 3 (mod 4). In fact, the question of which polynomials have roots in \mathbb{F}_p--let alone \mathbb{F}_{p^k}--is quite difficult in general. A typical approach to understanding degree-n equations in \mathbb{F}_p is to look for so-called reciprocity laws that generalize the law of quadratic reciprocity, which allows for a complete understanding (in principle) of quadratic equations mod p. Laws of cubic, quartic, and quintic reciprocity are known, as well as some higher-degree versions as well. As far as I know, the most general reciprocity law currently known is the law of Artin reciprocity, which is a major theorem in class field theory.

Interestingly enough, there is a version of quadratic reciprocity for polynomial rings over finite fields (i.e., a law relating the "polynomial Legendre symbol" (f/g) to the symbol (g/f) for coprime polynomials f and g).