## Section17.1Polynomial Rings

Throughout this chapter we shall assume that $$R$$ is a commutative ring with identity. Any expression of the form

\begin{equation*} f(x) = \sum^{n}_{i=0} a_i x^i = a_0 + a_1 x +a_2 x^2 + \cdots + a_n x^n\text{,} \end{equation*}

where $$a_i \in R$$ and $$a_n \neq 0\text{,}$$ is called a polynomial over $$R$$ with indeterminate $$x\text{.}$$ The elements $$a_0, a_1, \ldots, a_n$$ are called the coefficients of $$f\text{.}$$ The coefficient $$a_n$$ is called the leading coefficient. A polynomial is called monic if the leading coefficient is 1. If $$n$$ is the largest nonnegative number for which $$a_n \neq 0\text{,}$$ we say that the degree of $$f$$ is $$n$$ and write $$\deg f(x) = n\text{.}$$ If no such $$n$$ exists—that is, if $$f=0$$ is the zero polynomial—then the degree of $$f$$ is defined to be $$-\infty\text{.}$$ We will denote the set of all polynomials with coefficients in a ring $$R$$ by $$R[x]\text{.}$$ Two polynomials are equal exactly when their corresponding coefficients are equal; that is, if we let

\begin{align*} p(x) & = a_0 + a_1 x + \cdots + a_n x^n\\ q(x) & = b_0 + b_1 x + \cdots + b_m x^m\text{,} \end{align*}

then $$p(x) = q(x)$$ if and only if $$a_i = b_i$$ for all $$i \geq 0\text{.}$$

To show that the set of all polynomials forms a ring, we must first define addition and multiplication. We define the sum of two polynomials as follows. Let

\begin{align*} p(x) & = a_0 + a_1 x + \cdots + a_n x^n\\ q(x) & = b_0 + b_1 x + \cdots + b_m x^m\text{.} \end{align*}

Then the sum of $$p(x)$$ and $$q(x)$$ is

\begin{equation*} p(x) + q(x) = c_0 + c_1 x + \cdots + c_k x^k\text{,} \end{equation*}

where $$c_i = a_i + b_i$$ for each $$i\text{.}$$ We define the product of $$p(x)$$ and $$q(x)$$ to be

\begin{equation*} p(x) q(x) = c_0 + c_1 x + \cdots + c_{m + n} x^{m + n}\text{,} \end{equation*}

where

\begin{equation*} c_i = \sum_{k = 0}^i a_k b_{i - k} = a_0 b_i + a_1 b_{i -1} + \cdots + a_{i -1} b _1 + a_i b_0 \end{equation*}

for each $$i\text{.}$$ Notice that in each case some of the coefficients may be zero.

###### Example17.1.

Suppose that

\begin{equation*} p(x) = 3 + 0 x + 0 x^2 + 2 x^3 + 0 x^4 \end{equation*}

and

\begin{equation*} q(x) = 2 + 0 x - x^2 + 0 x^3 + 4 x^4 \end{equation*}

are polynomials in $${\mathbb Z}[x]\text{.}$$ If the coefficient of some term in a polynomial is zero, then we usually just omit that term. In this case we would write $$p(x) = 3 + 2 x^3$$ and $$q(x) = 2 - x^2 + 4 x^4\text{.}$$ The sum of these two polynomials is

\begin{equation*} p(x) + q(x)= 5 - x^2 + 2 x^3 + 4 x^4\text{.} \end{equation*}

The product,

\begin{equation*} p(x) q(x) = (3 + 2 x^3)( 2 - x^2 + 4 x^4 ) = 6 - 3x^2 + 4 x^3 + 12 x^4 - 2 x^5 + 8 x^7\text{,} \end{equation*}

can be calculated either by determining the $$c_i$$s in the definition or by simply multiplying polynomials in the same way as we have always done.

###### Example17.2.

Let

\begin{equation*} p(x) = 3 + 3 x^3 \qquad \text{and} \qquad q(x) = 4 + 4 x^2 + 4 x^4 \end{equation*}

be polynomials in $${\mathbb Z}_{12}[x]\text{.}$$ The sum of $$p(x)$$ and $$q(x)$$ is $$7 + 4 x^2 + 3 x^3 + 4 x^4\text{.}$$ The product of the two polynomials is the zero polynomial. This example tells us that we can not expect $$R[x]$$ to be an integral domain if $$R$$ is not an integral domain.

Our first task is to show that $$R[x]$$ is an abelian group under polynomial addition. The zero polynomial, $$f(x) = 0\text{,}$$ is the additive identity. Given a polynomial $$p(x) = \sum_{i = 0}^{n} a_i x^i\text{,}$$ the inverse of $$p(x)$$ is easily verified to be $$-p(x) = \sum_{i = 0}^{n} (-a_i) x^i = -\sum_{i = 0}^{n} a_i x^i\text{.}$$ Commutativity and associativity follow immediately from the definition of polynomial addition and from the fact that addition in $$R$$ is both commutative and associative.

To show that polynomial multiplication is associative, let

\begin{align*} p(x) & = \sum_{i = 0}^{m} a_i x^i,\\ q(x) & = \sum_{i = 0}^{n} b_i x^i,\\ r(x) & = \sum_{i = 0}^{p} c_i x^i\text{.} \end{align*}

Then

\begin{align*} [p(x) q(x)] r(x) & = \left[ \left( \sum_{i=0}^{m} a_i x^i \right) \left( \sum_{i=0}^{n} b_i x^i \right) \right] \left( \sum_{i = 0}^{p} c_i x^i \right)\\ & = \left[ \sum_{i = 0}^{m+n} \left( \sum_{j = 0}^{i} a_j b_{i - j} \right) x^i \right] \left( \sum_{i = 0}^{p} c_i x^i \right)\\ & = \sum_{i = 0}^{m + n + p} \left[ \sum_{j = 0}^{i} \left( \sum_{k=0}^j a_k b_{j-k} \right) c_{i-j} \right] x^i\\ & = \sum_{i = 0}^{m + n + p} \left(\sum_{j + k + l = i} a_j b_k c_l \right) x^i\\ & = \sum_{i = 0}^{m+n+p} \left[ \sum_{j = 0}^{i} a_j \left( \sum_{k = 0}^{i - j} b_k c_{i - j - k} \right) \right] x^i\\ & = \left( \sum_{i = 0}^{m} a_i x^i \right) \left[ \sum_{i = 0}^{n + p} \left( \sum_{j = 0}^{i} b_j c_{i - j} \right) x^i \right]\\ & = \left( \sum_{i = 0}^{m} a_i x^i \right) \left[ \left( \sum_{i = 0}^{n} b_i x^i \right) \left( \sum_{i = 0}^{p} c_i x^i \right) \right]\\ & = p(x) [ q(x) r(x) ] \end{align*}

The commutativity and distribution properties of polynomial multiplication are proved in a similar manner. We shall leave the proofs of these properties as an exercise.

Suppose that we have two nonzero polynomials

\begin{equation*} p(x) = a_m x^m + \cdots + a_1 x + a_0 \end{equation*}

and

\begin{equation*} q(x) = b_n x^n + \cdots + b_1 x + b_0 \end{equation*}

with $$a_m \neq 0$$ and $$b_n \neq 0\text{.}$$ The degrees of $$p(x)$$ and $$q(x)$$ are $$m$$ and $$n\text{,}$$ respectively. The leading term of $$p(x) q(x)$$ is $$a_m b_n x^{m + n}\text{,}$$ which cannot be zero since $$R$$ is an integral domain; hence, the degree of $$p(x) q(x)$$ is $$m + n\text{,}$$ and $$p(x)q(x) \neq 0\text{.}$$ Since $$p(x) \neq 0$$ and $$q(x) \neq 0$$ imply that $$p(x)q(x) \neq 0\text{,}$$ we know that $$R[x]$$ must also be an integral domain.

We also want to consider polynomials in two or more variables, such as $$x^2 - 3 x y + 2 y^3\text{.}$$ Let $$R$$ be a ring and suppose that we are given two indeterminates $$x$$ and $$y\text{.}$$ Certainly we can form the ring $$(R[x])[y]\text{.}$$ It is straightforward but perhaps tedious to show that $$(R[x])[y] \cong R([y])[x]\text{.}$$ We shall identify these two rings by this isomorphism and simply write $$R[x,y]\text{.}$$ The ring $$R[x, y]$$ is called the ring of polynomials in two indeterminates $$x$$ and $$y$$ with coefficients in $$R\text{.}$$ We can define the ring of polynomials in $$n$$ indeterminates with coefficients in $$R$$ similarly. We shall denote this ring by $$R[x_1, x_2, \ldots, x_n]\text{.}$$

Let $$p(x) = \sum_{i = 0}^n a_i x^i$$ and $$q(x) = \sum_{i = 0}^m b_i x^i\text{.}$$ It is easy to show that $$\phi_{\alpha}(p(x) + q(x)) = \phi_{\alpha}(p(x)) + \phi_{\alpha}(q(x))\text{.}$$ To show that multiplication is preserved under the map $$\phi_{\alpha}\text{,}$$ observe that

\begin{align*} \phi_{\alpha} (p(x) ) \phi_{\alpha} (q(x)) & = p( \alpha ) q(\alpha)\\ & = \left( \sum_{i = 0}^n a_i \alpha^i \right) \left( \sum_{i = 0}^m b_i \alpha^i \right)\\ & = \sum_{i = 0}^{m + n} \left( \sum_{k = 0}^i a_k b_{i - k} \right) \alpha^i\\ & = \phi_{\alpha} (p(x) q(x))\text{.} \end{align*}

The map $$\phi_{\alpha} : R[x] \rightarrow R$$ is called the evaluation homomorphism at $$\alpha\text{.}$$