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Abstract Algebra Theory and Applications

Thomas W. Judson

Section 17.1 Polynomial 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 \(i \geq 0\text{.}\)
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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.
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Example 17.1.

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.
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Example 17.2.

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.
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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.
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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.
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Proof.

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.
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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 \(R[x_1, x_2, \ldots, x_n]\text{.}\)
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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*}
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The map \(\phi_{\alpha} : R[x] \rightarrow R\) is called the evaluation homomorphism at \(\alpha\text{.}\)
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