F[x]/p(x): A ring

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Let $F$ be a field. Let $p(x) \in F[x]-\{0\}$. Let $F[x]/p(x) = \{a(x)\%p(x): a(x) \in F[x]\}$.

Then $(F[x]/p(x), +, \circ)$ is a commutative ring, where $a(x) \circ b(x) = (a(x)b(x))\% p(x)$.

Proof

\begin{align} & p \in F - \{0\} \\ &\Rightarrow \forall a(x) \in F[x], a(x) \% p(x) = 0 \\ &\Rightarrow F[x]/p(x) = \{0\} \\ &\Rightarrow (F[x]/p(x), +, \circ) \textrm{ is a commutative ring} \end{align}

So let's assume $p(x) \in F[x]-F \Rightarrow \deg(p) \ge 1$.

Lemma 1: $\deg(a) < \deg(p) \iff a(x) \in F[x]/p(x)$

\[ a(x) \in F[x]/p(x) \Rightarrow \exists f(x) \in F[x], a(x) = f(x) \% p(x) \Rightarrow \deg(a) < \deg(p) \]

\begin{align} & \deg(a) < \deg(p) \\ &\Rightarrow a(x) = 0p(x) + a(x) \\ &\Rightarrow a(x) = a(x) \% p(x) \\ &\Rightarrow a(x) \in F[x]/p(x) \end{align}

$(F[x]/p(x), +)$ is an abelian group

$F[x]/p(x)$ is a subset of $F[x]$. $0 \in F[x]/p(x) \Rightarrow F[x]/p(x) \neq \{\}$.

\begin{align} & a(x), b(x) \in F[x]/p(x) \\ &\Rightarrow (\deg(a) < \deg(p) \wedge \deg(b) < \deg(p)) \tag{by lemma 1} \\ &\Rightarrow (\deg(a) < \deg(p) \wedge \deg(-b) < \deg(p)) \\ &\Rightarrow \deg(a-b) \le \max(\deg(a), \deg(-b)) < \deg(p) \\ &\Rightarrow a(x) - b(x) \in F[x]/p(x) \tag{by lemma 1} \end{align}

Therefore, $(F[x]/p(x), +)$ is a subgroup of $(F[x], +)$.

$F[x]/p(x)$ inherits commutativity from $F[x]$, so $(F[x]/p(x), +)$ is an abelian group.

Lemma 2: $a\%p = b\%p \iff p \mid (a - b)$ in $F[x]$

Let $a = k_ap + a\%p$ and $b = k_bp + b \% p$.

\begin{align} & a\%p = b\%p \\ &\Rightarrow a-k_ap = b - k_bp \\ &\Rightarrow a - b = k_ap - k_bp = (k_a - k_b)p \\ &\Rightarrow p \mid (a - b) \end{align}

\begin{align} & p \mid (a - b) \\ &\Rightarrow a - b = kp \\ &\Rightarrow (a\%p + k_ap) - (b\%p + k_bp) = kp \\ &\Rightarrow (a\%p - b\%p) = (k_b - k_a + k)p \end{align}

\begin{align} & \deg(a\%p) < \deg(p) \wedge \deg(b\%p) < \deg(p) \\ &\Rightarrow \deg(a\%p - b\%p) < \deg(p) \\ &\Rightarrow \deg(p(k_b - k_a + k)) < \deg(p) \\ &\Rightarrow \deg(p) + \deg(k_b - k_a + k) < \deg(p) \tag{$F$ has no zero-divisors} \\ &\Rightarrow \deg(k_b - k_a + k) < 0 \\ &\Rightarrow k_b - k_a + k = 0 \\ &\Rightarrow p(k_b - k_a + k) = 0 \\ &\Rightarrow a\%p - b\%p = 0 \\ &\Rightarrow a\%p = b\%p \end{align}

Multiplicative properties

\[ \deg(a \circ b) = \deg((ab)\%p) < \deg(p) \Rightarrow a \circ b \in F[x]/p(x) \] Therefore, $F[x]/p(x)$ is closed under $\circ$.

Since $F$ is commutative, $F[x]$ is commutative. \[ a \circ b = (ab)\% p = (ba) \% p = b \circ a \] Therefore, $\circ$ is commutative.

\begin{align} & a \circ b = (ab) \% p \\ &\Rightarrow ab = pq + (a \circ b) \\ &\Rightarrow p \mid (ab - a \circ b) \\ &\Rightarrow p \mid (ab - a \circ b)c \\ &\Rightarrow p \mid (abc - (a \circ b)c) \\ &\Rightarrow (abc)\%p = ((a \circ b)c)\%p = (a \circ b) \circ c \tag{by lemma 2} \end{align}

\begin{align} & b \circ c = (bc) \% p \\ &\Rightarrow bc = pq + (b \circ c) \\ &\Rightarrow p \mid (bc - b \circ c) \\ &\Rightarrow p \mid a(bc - b \circ c) \\ &\Rightarrow p \mid (abc - a(b \circ c)) \\ &\Rightarrow (abc)\%p = (a(b \circ c))\%p = a \circ (b \circ c) \tag{by lemma 2} \end{align}

Therefore, $a \circ (b \circ c) = (a \circ b) \circ c$. Therefore, $\circ$ is associative.

Lemma 3: $(a + b) \% p = a\%p + b\%p$

Let $a(x), b(x) \in F[x]$. Let $a = k_ap + a\%p$ and $b = k_bp + b\%p$. Then $a + b = (k_a + k_b)p + (a\%p + b\%p)$. $\deg(a\%p + b\%p) \le \max(\deg(a\%p), \deg(b\%p)) < \deg(p)$. Therefore, by polynomial division theorem, $(a+b)\%p = a\%p + b\%p$.

Distributive properties

\begin{align} & a \circ (b + c) \\ &= (a(b+c)) \% p \\ &= (ab+ac) \% p \\ &= (ab)\%p + (ac)\%p \\ &= (a \circ b) + (a \circ c) \end{align}

Similarly, $(a + b) \circ c = (a \circ c) + (a \circ b)$.

Therefore, $(F[x]/p(x), +, \circ)$ is a commutative ring.

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Depth: 6
Number of transitive dependencies: 18