17.1: Polynomial Rings
Throughout this chapter we shall assume that \(R\) is a commutative ring with identity. Any expression of the form
\[ f(x) = \sum^{n}_{i=0} a_i x^i = a_0 + a_1 x +a_2 x^2 + \cdots + a_n x^n\text{,} \nonumber \]
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\\[4pt] 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
\[ p(x) + q(x) = c_0 + c_1 x + \cdots + c_k x^k\text{,} \nonumber \]
where \(c_i = a_i + b_i\) for each \(i\text{.}\) We define the product of \(p(x)\) and \(q(x)\) to be
\[ p(x) q(x) = c_0 + c_1 x + \cdots + c_{m + n} x^{m + n}\text{,} \nonumber \]
where
\[ c_i = \sum_{k = 0}^i a_k b_{i - k} = a_0 b_i + a_1 b_{i -1} + \cdots + a_{i -1} b _1 + a_i b_0 \nonumber \]
for each \(i\text{.}\) Notice that in each case some of the coefficients may be zero.
Example \(17.1\)
Suppose that
\[ p(x) = 3 + 0 x + 0 x^2 + 2 x^3 + 0 x^4 \nonumber \]
and
\[ q(x) = 2 + 0 x - x^2 + 0 x^3 + 4 x^4 \nonumber \]
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.
Solution
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
\[ p(x) + q(x)= 5 - x^2 + 2 x^3 + 4 x^4\text{.} \nonumber \]
The product,
\[ p(x) q(x) = (3 + 2 x^3)( 2 - x^2 + 4 x^4 ) = 6 - 3x^2 + 4 x^3 + 12 x^4 - 2 x^5 + 8 x^7\text{,} \nonumber \]
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.
Example \(17.2\)
Let
\[ p(x) = 3 + 3 x^3 \qquad \text{and} \qquad q(x) = 4 + 4 x^2 + 4 x^4 \nonumber \]
be polynomials in \({\mathbb Z}_{12}[x]\text{.}\)
Solution
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.
Theorem \(17.3\)
Let \(R\) be a commutative ring with identity. Then \(R[x]\) is a commutative ring with identity.
- Proof
-
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.
Proposition \(17.4\)
Let \(p(x)\) and \(q(x)\) be polynomials in \(R[x]\text{,}\) where \(R\) is an integral domain. Then \(\deg p(x) + \deg q(x) = \deg( p(x) q(x) )\text{.}\) Furthermore, \(R[x]\) is an integral domain.
- Proof
-
Suppose that we have two nonzero polynomials
\[ p(x) = a_m x^m + \cdots + a_1 x + a_0 \nonumber \]
and
\[ q(x) = b_n x^n + \cdots + b_1 x + b_0 \nonumber \]
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 \(R[x_1, x_2, \ldots, x_n]\text{.}\)
Theorem \(17.5\)
Let \(R\) be a commutative ring with identity and \(\alpha \in R\text{.}\) Then we have a ring homomorphism \(\phi_{\alpha} : R[x] \rightarrow R\) defined by
\[ \phi_{\alpha} (p(x) ) = p( \alpha ) = a_n \alpha^n + \cdots + a_1 \alpha + a_0\text{,} \nonumber \]
where \(p( x ) = a_n x^n + \cdots + a_1 x + a_0\text{.}\)
- Proof
-
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 \(\alpha\text{.}\)