Why $\Bbb Z_{p}[x]$ is finite field?

Question

I'm learning cryptography and today I learned Galoi's Field, but I cannot understand why $\Bbb Z_p[x]$ is finite field.

I know $\Bbb Z_p$ is finite filed because $p$ is prime number and by EEA, but $Z_p[x]$ has too many polynomials which degree is $x^0$ ~ $x^{\infty}$.

To understand Galoi's field, I should understand why that is finite field. How can I prove it?

$\mathbb{Z}_p[x]$ is neither finite nor a field. But maybe there's some misunderstanding here. Can you give a reference to the source of this claim? — Alex Kruckman, Sep 20 '18 at 13:32
Actually if $f(x)$ is an irreducible polynomial of degree $n$ over $\Bbb{Z}_p$, then $\Bbb{Z}_p[x] / \langle f(x) \rangle$ is a finite field of order $p^n$ — Chinnapparaj R, Sep 20 '18 at 13:38
Originally, what I wanted to ask is why $\Bbb Z_2[x] / (x^2+x+1)$ is {$0,1,x,x+1$} — baeharam, Sep 20 '18 at 13:40
The $p$-adic integers $\mathbb{Z}_p=\varprojlim\mathbb{Z}/(p^n)$ is not a field. — user10354138, Sep 20 '18 at 13:45

score 3 · Answer 1 · answered Sep 20 '18 at 13:34

3

$\Bbb Z_p[x]$ is not a field because $x$ is not invertible.

$\Bbb Z_p[x]$ is not finite because $x^n$ for $n\in\Bbb N$ are infinitely many different elements.

answered Sep 20 '18 at 13:34

lhf

1 Answers1