1

Given $n∈N$ and $p$ a prime, why is $Z_p[x]/(x-n)≅Z_p$?

I think that the reason is that $x∈(x-n)$.

If this is the case, given $f=a_0+a_1x+...+a_nx^n∈Z_p[x]$

$a_ix^i∈(x-n)$ $∀i∈\{1,...,n\}$ and so $Z_p[x]/(x-n)=\{a+(x-n)|a∈Z_p\}$.

But I can't understand how can $x∈(x-n)$.

Dolly Parton
  • 33
  • 1
    Generally, $x\notin(x-n)$. Let $y=x-n$. Then, $\Bbb Z_p[x]=\Bbb Z_p[y]$ (even if $p$ is not prime). – Anne Bauval Jul 27 '24 at 18:14
  • $x$ won't be an element of the ideal generated by $x-n$. Rather, $x$ is congruent to $n$ modulo that ideal. In other words, you can think of $x$ as the constant $n$. BTW $n$ has several distinct meanings in your post. It also appears as the degree of $f(x)$. But the first occurence as $(x-n)$ already nailed down the meaning of $n$, so you need another variable for $\deg f$. – Jyrki Lahtonen Jul 27 '24 at 18:24
  • 2
    Same as in linked dupe, except use eval map $,x\mapsto n,,$ & Factor Theorem $,f(n) = 0 \iff x-n\mid f\ \ $ – Bill Dubuque Jul 27 '24 at 20:19

2 Answers2

2

Define $\phi: \mathbb{Z}_p[x] \to \mathbb{Z}_p$ by $$\phi(f(x))= f(n) \pmod p.$$ This is an onto homomorphism and the kernel is $\langle x-n \rangle$. Hence the isomorphism.

$\textbf{Note.}$ $x$ need not being to $\langle x-n\rangle$.

Yathi
  • 2,859
0

Apply the following theorem for $f(x)=x-m$, and $n=deg(f)=1$:

Theorem: If $f\in \Bbb F_p[x]$ is a monic irreducible polynomial of degree $n$, then the quotient ring $\Bbb F_p[x]/(f(x))$ is isomorphic to the finite field $\Bbb F_{p^n}$.

This result part of (almost) every course in abstract algebra, and the proof is using the first isomorphism theorem. See also the following post, for example:

Proving $\mathbb{F}_p[x]/\langle f(x)\rangle$ with $f(x)$ irreducible of degree $n$ is a field with $p^n$ elements

Dietrich Burde
  • 140,055