How many elements does $\mathbb F = \mathbb Z_{7}[x]/I$ contain?

Question

Let $p(x) \in \mathbb Z_{7}[x]$, given by $p(x) = x^{2}+3x+1$ and let $I = <p(x)>$ be the ideal in $\mathbb Z_{7}[x]$ constructed by $p(x)$.

How many elements does $\mathbb F = \mathbb Z_{7}[x]/I$ contain?

Is this $7^{2}$?

Since $f(x)+<x^{2}+3x+1> = f_0+f_1 x+<x^{2}+3x+1>$, with $f_0,f_1 \in \mathbb Z_7$, so there can only be 7 options for $f_0$ and 7 for $f_1$.

Or is $f(x)+<x^{2}+3x+1> = f_0+f_1 x+f_2x^{2}+<x^{2}+3x+1>$ and is the number of elements $7^3$?

Answer 1

From the context I'm assuming by $\mathbb{Z}_7$ you mean $\mathbb{Z}/7\mathbb{Z}$ rather than $7$-adic integers.

$x^2$ is effectively same as $-3x-1$ in your new ring F so the first argument is valid. But you had to check p(x) is irreducible as well for this argument to work which is not very difficult.

Answer 2

What we do is, we consider what happens to $\mathbb{Z}_7[x]$, if we let $x^2+3x+1$ be the identity element, so if $x^2+3x+1=0\Rightarrow x^2=-3x-1=4x+6\,(mod\,7)$ so now:

$\mathbb{Z}_7[x]/I=\{f\in\mathbb{Z}_7[x]\,|x^2=4x+6\}=\{a+bx\,|a,b\in\mathbb{Z}_7\}$ which contains $7\times 7=7^2$ elements.