Polynomial Multiplication over $\mathbb Z_4$

Detailed Solution

We work in the polynomial ring $$ R=\mathbb Z_4[x], $$ where every coefficient is calculated modulo $4$. Given $$ f(x)=2x+1,\qquad g(x)=2x-1, $$ their product is $$ \begin{aligned} f(x)g(x) &=(2x+1)(2x-1)\\ &=4x^2-2x+2x-1\\ &=4x^2-1. \end{aligned} $$ Since $4\equiv0\pmod4$, we obtain $$ 4x^2-1\equiv-1\equiv3\pmod4. $$ Therefore, $$ \boxed{f(x)g(x)=3}. $$ This proves that option B is correct. We examine the remaining options. Option A is false. Both $f$ and $g$ have degree $1$, but their product is the nonzero constant polynomial $3$, which has degree $0$. Thus $$ \deg(fg)=0\neq2=\deg f+\deg g. $$ The usual degree formula can fail when the coefficient ring has zero divisors. Option C is false. In fact, both $f$ and $g$ are units. We have $$ \begin{aligned} f(x)^2 &=(1+2x)^2\\ &=1+4x+4x^2\\ &=1 \end{aligned} $$ in $\mathbb Z_4[x]$. Hence $f^{-1}=f$. Similarly, $$ \begin{aligned} g(x)^2 &=(2x-1)^2\\ &=4x^2-4x+1\\ &=1. \end{aligned} $$ Hence $g^{-1}=g$. Therefore, both polynomials are units, not nonunits. Option D is false. The coefficient ring $\mathbb Z_4$ has a nonzero zero divisor because $$ 2\cdot2=4\equiv0\pmod4. $$ The same relation holds in $\mathbb Z_4[x]$, where the constant polynomial $2$ is nonzero but satisfies $2\cdot2=0$. Thus $\mathbb Z_4[x]$ has zero divisors. Consequently, the correct option is $$ \boxed{\mathrm{B}}. $$