Let $g(x) \in \mathbb{F}[x]$ be a polynomial of degree $\geq 1$. The residual class of $a(x) \in \mathbb{F}[x]$ modulo $g(x)$ is the set.

$ \overline{a(x)} = \{b(x) | b(x) \equiv a(x) \mod g(x) \} $

The set of all possible residual classes is denoted by $\frac{\mathbb{F}[x]}{g(x)}$.

$\frac{\mathbb{F}[x]}{g(x)} = \{\overline{a(x)} | a(x) \in \mathbb{F}[x] \}$

Exercise: Let $g(x) = x^3 + x + 1 \in \mathbb{Z}_3[x]$. Find $\frac{\mathbb{Z}_3[x]}{g(x)}$.

My attempt...

Let $a(x) = x^2$. Because $x^2 = 0(x^3+x+1) + x^2$ and $x^3 + x^2 + x + 1 = 1(x^3 + x + 1) + x^2$ and $2x^3+x^2+2x+2 = 2(x^3+x+1) + x^2$, $x^2$ and $x^3 + x^2 + x + 2$ and $2x^3+x^2+2x+2$ are members of $\overline{x^2}$ and $\overline{x^2}$ is a member of $\frac{\mathbb{Z}_3[x]}{g(x)}$.

How do I find all members of $\overline{x^2}$ (if I hadn't already)? How do I find all members of $\frac{\mathbb{Z}_3[x]}{x^3 + x + 1}$? What is a general constructive way to answer these type of questions?


1 Answer 1


One way to look at $\dfrac{\mathbb{Z}_3[x]}{g(x)}$ is as a ring $\mathbb{Z}_3[u]$ such that $g(u)=0$.

This implies that $\mathbb{Z}_3[u]$ is formed by polynomial expressions in $u$ of degree at most $2$, an expression of the fact that every polynomial $f(x) \in \mathbb{Z}_3[x]$ leaves a remainder of degree at most $2$ when divided by $g(x)$.

As a consequence, $\mathbb{Z}_3[u]$ is a vector space over $\mathbb{Z}_3$ of dimension $3$. This takes care of the additive structure of $\mathbb{Z}_3[u]$.

The multiplicative structure of $\mathbb{Z}_3[u]$ follows from $g(u)=0$. In particular, since $g(x)=(x+2) (x^2+x+2)$, the ring $\mathbb{Z}_3[u]$ has zero divisors. The structure of the ring $\mathbb{Z}_3[u]$ can be deduced from the Chinese Remainder theorem from this factorization.

This argument can be generalized for any base field instead of $\mathbb{Z}_3$.


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