Basic properties of quotients of a polynomial ring

Let $R$ be a commutative ring with $1 \neq 0$. Let $f(x) \in R[x]$ be a monic polynomial of degree $n \geq 1$ (that is, the leading coefficient of $f(x)$ is 1), and let bars denote the natural projection $R[x] \rightarrow R[x]/(f(x))$.

1. Show that every element of $R[x]/(f(x))$ is of the form $\overline{p(x)}$ for some polynomial $p(x) \in R[x]$ of degree less than $n$.
2. Prove that if $p(x),q(x) \in R[x]$ are distinct and have degree less than $n$, then $\overline{p(x)} \neq \overline{q(x)}$.
3. Suppose $f(x) = a(x)b(x)$ for some $a(x),b(x) \in R[x]$ of degree less than $n$. Prove that $\overline{p(x)}$ is a zero divisor in $R[x]/(f(x))$.
4. If $f(x) = x^n-a$ for some nilpotent element $a \in R$, prove that $\overline{x}$ is nilpotent in $R[x]/(f(x))$.
5. Let $p$ be a prime, let $R = \mathbb{Z}/(p)$, and let $f(x) = x^p - a$ for some $a \in \mathbb{Z}/(p)$. Prove that $\overline{x-a}$ is nilpotent in $R[x]/(f(x))$.

We begin with a lemma.

Lemma: Let $R$ be a commutative ring with $1 \neq 0$. If $f(x) \in R[x]$ is monic, then for all $g(x) \in R[x]$, $\mathsf{deg}(fg) = \mathsf{deg}(f) + \mathsf{deg}(g)$. Proof: The inequality $\mathsf{deg}(fg) \leq \mathsf{deg}(f) + \mathsf{deg}(g)$ always holds. Write $f(x) = x^n + f^\prime(x)$ and $g(x) = b_mx^m + g^\prime(x)$, where $f^\prime$ and $g^\prime$ have degree less than $f$ and $g$, respectively. Then $(fg)(x) = b_mx^{n+m} + x^ng^\prime(x) + x^mf^\prime(x) + (g^\prime f^\prime)(x)$; since $b_m \neq 0$, the degree of $fg$ is the sum of the degrees of $f$ and $g$. $\square$

1. Let $\overline{g(x)} \in R[x]/(f(x))$, and write $f(x) = x^n - f^\prime(x)$, where the degree of $f^\prime$ is less than $n$. We proceed by induction on the degree of $g$.

For the base case, if the degree of $g$ is less than $n$, then we may use $g(x)$ itself to represent $\overline{g(x)}$. For the inductive step, suppose that for some $m \geq n-1$, if $h(x)$ is a polynomial of degree at most $m$ then there exists $p(x)$ of degree at most $n-1$ such that $\overline{h(x)} = \overline{p(x)}$. Let $g(x)$ have degree $m+1$. We may write $g(x) = b_{m+1}x^{m+1} + g^\prime(x)$, where the degree of $g^\prime$ is at most $m$. Note that since $\overline{f(x)} = 0$, $\overline{x^n} = \overline{f^\prime}$. Now $\overline{g(x)} = \overline{b_{m+1} x^{m+1} + g^\prime(x)}$ $= \overline{b_{m+1}x^{m-n+1}}\overline{x^n} + \overline{g^\prime(x)}$ $= \overline{b_{m+1}x^{m-n+1}}\overline{f^\prime(x)} + \overline{g^\prime(x)}$ $= \overline{b_{m+1}x^{m-n+1}f^\prime(x) + g^\prime(x)}$. Note that the degree of $f^\prime$ is at most $n$, so that the degree of $x^{m-n+1}f^\prime(x)$ is at most $m$. Thus by the induction hypothesis, we have $\overline{g(x)} = \overline{p(x)}$ for some polynomial $p(x)$ of degree less than $n$.

2. Suppose $p(x),q(x) \in R[x]$ have degree less than $n$ and that $p(x) \neq q(x)$. Suppose $\overline{p(x)} = \overline{q(x)}$. Then $p(x) - q(x) = f(x)g(x)$ for some $g(x) \in R[x]$. Note, however, that $p(x)-q(x)$ has degree less than $n$, while $f(x)g(x)$ has degree at least $n$. Thus we have a contradiction.
3. Suppose $f(x) = a(x)b(x)$, where both $a$ and $b$ have degree less than $n$. Then $\overline{a(x)}\overline{b(x)} = \overline{f(x)} = 0$ in $R[x]/(f(x))$, but by the previous part, neither $\overline{a(x)}$ nor $\overline{b(x)}$ is zero. Thus $a(x)$ is a zero divisor in $R[x]/(f(x))$..
4. Suppose $a^m = 0$ in $R$. Now $\overline{x}^nm = \overline{(x^n)^m}$ $= \overline{a^m} = 0$, so that $\overline{x}$ is nilpotent in $R[x]/(f(x))$.
5. Using Fermat’s Little Theorem, we have $\overline{x-a}^p = \overline{(x-a)^p}$ $= \overline{x^p-a^p}$ $= \overline{a-a} = 0$. Thus $\overline{x-a}$ is nilpotent in $R[x]/(f(x))$.