I am reading through Contemporary Abstract Algebra by Gallian.
In an example they state that multiplicative group $$E = \mathbb{Z}_3[x]/\langle x^2+1\rangle$$ contains 9 elements.
It is not immediately clear to me where they get this number from. My best intuition is that you have 3 possible coefficients in $\mathbb{Z}_3$ for an $x^2$ term and $1$ term, respectively, so $3\times 3 = 9$. Would someone be able to clarify?
(Naturally I am assuming that $\mathbb{Z}_3$ means the integers modulo $3$, and not the $3$-adics...)
The ring $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ consists of equivalence classes modulo $x^2+1$. The class of a polynomial $p(x)\in\mathbb{Z}_3[x]$ is represented by the remainder when you divide $p(x)$ by $x^2+1$; since the remainder is either $0$ or of degree at most $1$, every class is represented by a polynomial of the form $ax+b$ with $a,b\in\mathbb{Z}_3$, and any two such classes are equal if and only if the corresponding polynomials are equal. That means that as a field $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ has $9$ elements, since you have $3$ choices for $a$ and $3$ choices for $b$.
Because $x^2+1$ is irreducible over $\mathbb{Z}_3$, the quotient is in fact a field (because the ideal $\langle x^2+1\rangle$ is maximal). The multiplicative group of this field consists of the nonzero elements. So the multiplicative group has eight elements, not nine. The multiplicative monoid (if you include $0$) has nine elements.