Polynomials over a finite field

Question

Let $\mathbb{F}_p$ be a finite field where $p$ is a prime. Consider the following set of polynomials over $\mathbb{F}_p$: $$G_n(p)=\{{x+a_2x^2+\cdots+a_nx^n\mid a_i\in \mathbb{F}_p}\}.$$ Is $G_n(p)$ a group under the operation of composition of functions modulo $x^{n+1}$?

Example: the product of the elements $P=x+x^2$ and $G=x+x^3$ in $G_4(p)$ is $G\circ P=(x+x^3)+(x+x^3)^2=x+x^2+x^3+2x^4$.

Any ideas? I would appreciate! :)

Answer 1

Yes. One may note that since associativity, closure, and identity are obvious, it would suffice to prove that, for every element $G$, all we need is that there is some $k$ such that $G^k$ is the identity. (In fact this is equivalent to it being a group, as the set is finite)

This proves to be reasonably easy, however. Firstly, consider any polynomial of the form: $$G=x+\alpha x^n.$$ It may be proved easily that $$G^k=x+k \alpha x^n$$ and that therefore $G^p=x$ in a field of characteristic $p$. However, we can use similar reasoning for $$G=x+\alpha x^{n-1}+\beta x^n$$ and see that $$G^k=x+k\alpha x^{n-1}+\gamma x^n$$ where $\gamma$ is some element we don't care to discuss. Thus, $G^p$ is of the form $x+\gamma x^n$, meaning $G^{p^2}$ is the identity. We may generalize this line of reasoning to show that if $c_G\geq 2$ is the lowest number such that the coefficient of $x^c$ in $G$ is non-zero then $c_G$ is less than $c_{G^p}$ - implying that $G^{p^{n-1}}$ is always the identity for any polynomial - so $G^{p^{n-1}-1}$ is the inverse of $G$.

(More generally, we find that this result holds in any commutative ring of finite characteristic)

Answer 2

Closure$$f(x)=\sum_{j=1}^na_jx^j,\qquad a_1=1,a_r=0\text{ if }r\notin\{1,2,..,n\}\\ \text{coeff of } x^r\text{ in }f(x)^j={1\over r!}{d^r\over dx^r}{f(x)^j}\bigm{|}_{x=0}=\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{f^{(n_k)}(x)\over n_k!}\bigm{|}_{x=0}=\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{a_{n_k}}\\ \text{coeff of } x\text{ in }f(x)^j=\delta_{1j}a_1=\delta_{1j}\\ \therefore\text{coeff of } x\text{ in }(g\circ f)(x)=1\quad\forall f,g\in G_n(p)$$

Identity

$$e(x)=x$$

Inverse

$$g(x)=\sum_{j=1}^nb_jx^j,\qquad b_1=1,b_r=0\text{ if }r\notin\{1,2,..,n\}\\ g(f(x))=x\Rightarrow \text{ coeff of }x^r\text{ in }g(f(x))=0\quad\forall r>1\\ \sum_{j=1}^nb_j\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{a_{n_k}}=\sum_{j=1}^rb_j\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{a_{n_k}}=0\quad\forall r>1\\ r=2\Rightarrow b_1a_2+b_2a_1^2=0\Rightarrow b_2=-a_2\\ r=3\Rightarrow b_1a_3+2b_2a_1a_2+b_3a_1^3=0\Rightarrow b_3=-a_3+2a_2^2\\ \cdots$$

$a_1\neq0$ guarantees that we can figure out $b_j$ for each $j$ uniquely, so $f^{-1}$ exists. In particular $(x+3x^2+2x^3)^{-1}=x-3x^2+x^3$ in $G_3(5)$.