Fermat's Little Theorem and polynomials

Question

We know that in $F_p[y]$, $y^p-y=y(y-1)(y-2)\cdots (y-(p-1))$. Let $g(y)\in F_p[y]$. Why is it valid to set $y=g(y)$ in the above equation to obtain $g(y)^p-g(y)=g(y)(g(y)-1)\cdots (g(y)-(p-1))$. This is done in Theorem 1 of Chapter 22 of A Concrete Introduction to Higher algebra by Lindsay Childs.

Answer 1

For any field $F$, and element $a\in A$ of an (associative) $F$-algebra $A$, the substitution $X:=a$ defines a ring morphism $F[X]\to A: P\mapsto P[a]$. This works in particular when $A=F[X]$, which is what happens in the question (with $F=\Bbb F_p$, and $X=y$). The morphism property implies that the substitution $y:=g(y)$ can be performed separately in the factors of the product $y^p-y=\prod_{r\in\Bbb F_p}(y-r)$, giving as result the identity $g(y)^p-g(y)=\prod_{r\in\Bbb F_p}(g(y)-r)$.

Answer 2

The identity $$ z^p-z=z(z-1)(z-2)\cdots (z-p+1) $$ holds for all elements $z$ of any commutative ring $R$ of characteristic $p$. This follows from the corresponding identity in the polynomial ring by the universal property of univariate polynomial rings.

In this example the selections $R=F_p[y]$, $z=g(y)$ were made.