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Algebra-II: Companion notes

Abhay Soman

Section 4.3 Polynomial rings

Let \(R\) be a ring. A polynomial ring in one variable over \(R\) is the following:
\begin{equation*} R[X]=\{a_0+a_1X+\cdots+a_nX^n:a_i\in R\text{ and }n\in\mathbb{N}\} \end{equation*}
with \(a_0+a_1X+\cdots+a_nX^n=0\) if and only if \(a_i=0\) for \(0\leq i\leq n\text{.}\) If \(f=a_0+a_1X+\cdots+a_nX^n\) and \(g=b_0+b_1X+\cdots+b_mX^m\) with \(m\leq n\) then the addition is defined to be commutative and is given as follows.
\begin{equation*} f+g=\sum_{i=0}^{m}(a_i+b_i)X^i+\sum_{j=m+1}^{n}b_jX^j=g+f \end{equation*}
and the multiplication is defined by
\begin{equation*} fg=\sum_i\left(\sum_{j=0}^{i}a_{j}b_{i-j}\right)X^i. \end{equation*}
Under the binary operations defined above \(R[X]\) is a ring with unity. The additive identity is \(0=0_R+0_RX+0_RX^2+\cdots\) and the multiplicative identity is \(1=1_R+0_RX+0_RX^2+\cdots\text{.}\) The ring \(R[X]\) is called a polynomial ring over \(R\text{.}\) It is a commutative ring if and only if \(R\) is a commutative ring. There is a ring monomorphism
\begin{equation*} i\colon R\to R[X]\quad\text{given by}\quad r\mapsto r+0_RX+0_RX^2+\cdots \end{equation*}
Suppose \(f=a_0+a_1X+\cdots+a_nX^n\in R[X]\text{.}\) We call \(a_0\) the constant term of \(f\) and \(a_n\) the leading coefficient of \(f\text{.}\) We denote the leading coefficient of \(f\) by \(LC(f)\text{.}\) If \({\rm LC}(f)=a_n\neq 0\) then the we define the degree of \(f\), \(\deg(f)=n\text{.}\) If \(LC(f)=1\) then \(f\) is called a monic polynomial. If \(f=0\) or \(\deg(f)=0\) then we call \(f\) a constant polynomial. We let \(LC(0)=0_R\text{.}\) We define the degree of the zero polynomial \(\deg(0)=-\infty\text{.}\) We follow the convention that for any \(n\in\Z\) we have \(-\infty\lt n\) and \(-\infty+n=-\infty\text{.}\)
A polynomial ring in \(n\) variables, \(R[X_1,X_2,\ldots,X_n]\) is defined as the polynomial ring in \(X_n\) over \(R[X_1,\ldots,X_{n-1}]\text{,}\) i.e.,
\begin{equation*} R[X_1,\ldots,X_n]=R[X_1,\ldots,X_{n-1}][X_n]. \end{equation*}
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