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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.

Pele
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  • Let $g(y)=z$, then $z^p-z=... (\bmod p)$, then let $z=g(y)$ – Empy2 Oct 23 '13 at 09:32
  • Yes, that is what they have done, but my question is why is this ok or valid? For example if you take a concrete $g(y)$ and try to prove it, it is not quite obvious to get what we need. – Pele Oct 23 '13 at 09:35
  • $z^p-z=z(z-1)...(z-p+1)+pQ(z)$ where Q(z) is some polynomial of $z$. Let $z=g(y)$, and $g(y)^p-g(y)=g(y)(g(y)-1)...(g(y)-p+1)+pQ(g(y))$. Now $Q(g(y))$ is a polynomial of $y$, so the difference of $pQ(g(y))$ is zero in $F_p[y]$ – Empy2 Oct 23 '13 at 09:38
  • So in terms of a morphism from $F_p[x]\to F_p[x]$, can this substitution be interpreted as something? – Pele Oct 23 '13 at 09:44
  • Sorry, I'm not familiar with category theory. – Empy2 Oct 23 '13 at 09:47

2 Answers2

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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 $g(y)^p-g(y)=\prod_{r\in\Bbb F_p}(g(y)-r)$.

Marc van Leeuwen
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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.

Jyrki Lahtonen
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    I have let my answer convince myself that "commutative" in the above is unnecessary: any (unitary) ring of characteristic $p$, commutative or not, is an $\Bbb F_p$-algebra. – Marc van Leeuwen Oct 23 '13 at 10:26
  • @Marc, You're right. Commutativity is needed only for the Freshman's dream: $(x+y)^p=x^p+y^p$ to hold. Here commutativity of the subring of $R$ generated by $z$ is actually automatic, because $z$ commutes with itself and the elements of the base field. – Jyrki Lahtonen Oct 23 '13 at 10:33