Degree of the Product of Polynomials

Detailed Solution

Let $$ f(x)=a_mx^m+\cdots+a_1x+a_0 $$ and $$ g(x)=b_nx^n+\cdots+b_1x+b_0, $$ where $a_m\neq0$ and $b_n\neq0$ are the leading coefficients of $f$ and $g$. When we multiply $f(x)$ and $g(x)$, the coefficient of the highest possible power $x^{m+n}$ is $$ a_mb_n. $$ Since $R$ is an integral domain, it has no zero divisors. Therefore, the product of two nonzero elements is nonzero: $$ a_m\neq0,\quad b_n\neq0 \quad\Longrightarrow\quad a_mb_n\neq0. $$ Consequently, the term $a_mb_nx^{m+n}$ cannot vanish. No term of degree greater than $m+n$ can occur in the product. Hence $$ \boxed{\deg(fg)=m+n}. $$ Thus option A is correct. Why the other options are incorrect: Option B: The degree formula does not hold in every polynomial ring. For example, consider $R=\mathbb Z_4$, which has a nonzero zero divisor because $2\cdot2=0$. In $\mathbb Z_4[x]$, let $$ f(x)=2x+1,\qquad g(x)=2x+1. $$ Then $$ \begin{aligned} f(x)g(x) &=4x^2+4x+1\\ &=1 \end{aligned} $$ in $\mathbb Z_4[x]$. Therefore, $$ \deg(fg)=0\neq2=\deg f+\deg g. $$ Option C: False. If $R$ is an integral domain, then $R[x]$ is also an integral domain. Indeed, the product of any two nonzero polynomials has a nonzero leading coefficient. Option D: False. Zero divisors in the coefficient ring can cause the leading coefficient of a product to vanish, reducing its degree. The example over $\mathbb Z_4[x]$ above demonstrates this. Therefore, the correct option is $$ \boxed{\mathrm{A}}. $$ Key takeaway: For nonzero polynomials $f,g\in R[x]$, $$ \deg(fg)=\deg f+\deg g $$ holds whenever $R$ is an integral domain. The absence of zero divisors guarantees that the product of the leading coefficients is nonzero.