Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In one of the proof in the book "Abstract Algebra'' by Dummit and Foote (Theorem 41, pg. 554) we have a monic polynomial $g(x)\in\mathbb{Z}[x]$, and the book claims that $g(x^{p})=(g(x))^{p}\mod p$

Can someone please explain why this is true ? I know that $\forall a\in\mathbb{F}_{p}:a=a^{p}$, but I don't see how this imply the equality as polynomials

share|cite|improve this question
up vote 5 down vote accepted

First, note that the claim is clearly true for the polynomial $x\in\mathbb{Z}[x]$, as well as for any constant polynomial $a\in\mathbb{Z}$ (the latter is just Fermat's Little Theorem).

Now note that any element of $\mathbb{Z}[x]$ can be obtained as a combination of sums and products of the polynomial $x$ and constant polynomials.

For any $f,g\in\mathbb{Z}[x]$, we have that $$(f+g)^p=f^p+\binom{p}{1}f^{p-1}g+\cdots+\binom{p}{p-1}fg^{p-1}+g^p\equiv f^p+g^p\bmod p$$ and $(fg)^p=f^pg^p$, so we certainly have that $(fg)^p\equiv f^pg^p\bmod p$.

Thus, the claim is true for all polynomials in $\mathbb{Z}[x]$ (the condition that $g$ be monic is unnecessary).

share|cite|improve this answer
So infact I have: $g(x^p)=(g(x))^p=g(x)$ in the reduction mod $p$, right ? – Belgi Jul 19 '12 at 4:48
No, the indeterminate $x$ is still an indeterminate; we didn't have $x^p = x$ in $\mathbb{Z}[x]$, so we won't have $\bar{x}^p=\bar{x}$ in $\mathbb{F}_p[x]$. This is a common sticking point. To give an explicit example, if $g(x)=2x+1\in\mathbb{Z}[x]$, then when considered in $\mathbb{F}_3[x]$, $$g(x)^3=(2x+1)^3=2x^3+1=g(x^3)$$ but $2x^3+1\neq 2x+1$ in $\mathbb{F}_3[x]$. – Zev Chonoles Jul 19 '12 at 4:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.