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,\end{equation*} where $a_i \in R$ and $a_n \neq 0$, is called a polynomial over $R$ with indeterminate $x$. The elements $a_0, a_1, \ldots, a_n$ are called the coefficients of $f$. 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 $a_n \neq 0$, we say that the degree of $f$ is $n$ and write $\deg f(x) = n$. If no such $n$ exists—that is, if $f=0$ is the zero polynomial—then the degree of $f$ is defined to be $-\infty$. We will denote the set of all polynomials with coefficients in a ring $R$ by $R[x]$. 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, \end{align*} then $p(x) = q(x)$ if and only if $a_i = b_i$ for all $i \geq 0$.

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. \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,\end{equation*} where $c_i = a_i + b_i$ for each $i$. 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},\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 $i$. 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 ${\mathbb Z}[x]$. 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 $q(x) = 2 - x^2 + 4 x^4$. The sum of these two polynomials is \begin{equation*}p(x) + q(x)= 5 - x^2 + 2 x^3 + 4 x^4.\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,\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 ${\mathbb Z}_{12}[x]$. The sum of $p(x)$ and $q(x)$ is $7 + 4 x^2 + 3 x^3 + 4 x^4$. 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.

##### Proof

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

##### Proof

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