Let $A$ be a commutative ring with unity and $(f,g)\in A[X]^2$ is unimodular with $f$ unitary then the resultant of $(f,g)$ is not in the maximal ideal $\mathfrak{m}$ of $A$, i.e., $R(f,g)\notin \mathfrak{m}$.
I have some idea how to proceed if someone can help me to go further in that direction it would be amazing or if you have better way then please let me know. This is how am I thinking:
We know that $R(f,g)=ff_1+gg_1$, with $\deg f_1 <m=\deg g$ and $\deg g_1 <n=\deg f.$ Since $R(f,g)\in A$, we only need to show that $\overline{R(f,g)}\neq 0 $ in $\frac{A}{\mathfrak{m}}$. We also know that resultant of two polynomials is $0$ iff they have common factor, provided $(a_n,b_m)\neq 0$, where $a_n,b_m$ are leading coefficients of polynomials. So if we can show that $(\bar{f},\bar{g})$ unimodular element with $\bar{f}$ unitary does not have any common factor then we are done.
But how to show this?
Any help is appreciated.
I was almost there but not able to see at that moment.
$(\bar{f},\bar{g})$ is unimodular and assume the resultant is $0$. Since $f$ is unitary therefore leading coefficient of $\bar{f}$ is non-zero and so by using the stated theorem in the question $\bar{f},\bar{g}$ will have common factor but unimodularity of $(\bar{f},\bar{g})$ implies that $1 = 0$.
Hence, $R(f,g)\notin \mathfrak{m}$.