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## Basic properties of quotients of a polynomial ring

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.14

Solution:

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 \begin{align*}\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)}.\end{align*} 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 $R[x]/(f(x))$.

#### Linearity

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