The Ring of Polynomials with Ring Coefficients

# The Ring of Polynomials with Ring Coefficients

Recall from the Rings page that if $+$ and $*$ are binary operations on the set $R$, then $R$ is called a ring under $+$ and $*$ denoted $(R, +, *)$ when the following are satisfied:

• 1. For all $a, b \in R$ we have that $(a + b \in R)$ (Closure under $+$).
• 2. For all $a, b, c \in R$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $R$ under $+$).
• 3. There exists an $0 \in R$ such that for all $a \in R$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $R$ under $+$).
• 4. For all $a \in R$ there exists a $-a \in R$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $R$ under $+$).
• 5. For all $a, b \in R$ we have that $a + b = b + a$ (Commutativity of elements in $R$ under $+$).
• 6. For all $a, b \in R$ we have that $a * b = b * a$ (Closure under $*$).
• 7. For all $a, b, c \in R$, $a * (b * c) = (a * b) * c$ (Associativity of elements in $R$ under $*$).
• 8. There exists a $1 \in R$ such that for all $a \in R$ we have that $a * 1 = a$ and $1 * a = a$ (The existence of an identity element $1$ of $R$ under $*$).
• 9. For all $a, b, c \in R$ we have that $a * (b + c) = (a * b) + (b * c)$ and $(a + b) * c = (a * c) + (b * c)$ (Distributivity of $*$ over $+$).

On The Ring of Polynomials with Real Coefficients page we noted that the set of polynomials with real coefficients $\mathbb{R}[x]$ under the operations of $+$ and $*$ which denote standard polynomial addition and polynomial multiplication form a ring, $(\mathbb{R}[x], +, *)$.

Now consider any ring $(R, +_1, *_1)$ and let $R[x]$ denote the set of all polynomials whose coefficients are from $R$, that is, for $a_0, a_1, ..., a_n \in R$:

(1)
\begin{align} \quad (a_0 + a_1x + a_2x^2 + ... + a_nx^n) \in R[x] \end{align}

The ring $R[x]$ may be very abstract or concrete depending on the ring $R$. For example, if $R$ is the set of $n \times n$ matrices with real coefficients then it's somewhat odd to think of a polynomial with matrix coefficients. Similarly, if $R$ is simply the set of real numbers then $R[x]$ reduces to $\mathbb{R}[x]$ which is a set we've already looked at and are familiar with.

If $(R, +_1, *_1)$ is a ring under the operations of $+_1$ and $*_1$, then showing that the set $R[x]$ under the operations of polynomial addition $+$ and polynomial multiplication $*$ form a ring is intuitively simple. Let $p, q, r \in R[x]$ where we denote the coefficients of $p$ as $a_i$s, the coefficients of $q$ by $b_i$s, and the coefficients of $r$ by $c_i$s.

Recall that if $p$ and $q$ are polynomials then the polynomial $p + q$ is the polynomial obtained by taking terms with the same degree and summing their coefficients. The coefficients of the sum $p + q$ will all be elements in $R$ and so $R[x]$ is closed under $+$.

Furthermore, the sum of corresponding coefficients in $p + q$ is not affected by their order since $+$ is associative from $R$ being a ring. That is, if $(a_i +_1 (b_i +_1 c_i))x^i$ is an arbitrary term in the sum of polynomials $p + (q + r)$ then since $R$ is a ring and $a_i, b_i, c_i \in R$ we have that:

(2)
\begin{align} \quad a_i +_1 (b_i +_1 c_i) = (a_i +_1 b_i) +_1 c_i \end{align}

Therefore $p + (q + r) = (p + q) + r$ and so $+$ is associative.

Now let $0_R \in R$ be the identity of $R$ under the operation $+$. Then we can define an identity of $R[x]$ under $+$ as the "zero" polynomial given by:

(3)
\begin{align} \quad z(x) = 0_R \end{align}

For all polynomials $p \in R[x]$ we have that the coefficients of $p + z$ will be the same as the coefficients of $p$ since any coefficient of $p + z$ is of the form $a_i +_1 0_R = a_i$, and any coefficient of $z + p$ is of the form $0_R + a_i = a_i$. Therefore $z(x)$ is the identity element for $+$.

For each polynomial $p \in R[x]$ we can define the inverse $-p \in R[x]$ under $+$ to be the polynomial whose coefficients are the corresponding inverses under $+$ of the coefficients of $p$. So any coefficient of $p + (-p)$ is of the form $a_i + (-a_i) = 0_R$ and so $p + (-p) = z$. Note that the polynomial $- p$ does indeed exist due to the inverses of elements in $R$ under $+_1$.

Like mentioned before, the sum of corresponding coefficients in $p + q$ is not affected by order. Hence, for any term $(a_i +_1 b_i) x^i$ of $p + q$ we have that since $(R, +_1, *_1)$ is a ring that:

(4)
\begin{align} \quad a_i +_1 b_i = b_i +_1 a_i \end{align}

Therefore $p + q = q + p$ so $+$ is commutative.

The reader should verify that the ring axioms for $*$ and the distributivity axiom for $*$ over $+$ also holds using the same reasoning as above.