Question

3. If $c_1\dots c_{m}f\left(x\right)=g\left(x\right)$with $c_1\in R$and $g\left(x\right)$primitive in role="math" localid="1653720732267" $R\left[x\right]$, prove that each $c_i$is a unit.

Short answer

It is proved that each $c_i$ is a unit.

Step-by-step solution

Corollary 4.5

Let R be an integral domain and $f\left(x\right)\in R\left[x\right]$. Then, $f\left(x\right)$is a unit in $R\left[x\right]$if and only if $f\left(x\right)$ is a constant polynomial that is a unit in R .

ci is a unit

Consider two polynomial $f\left(x\right)=\sum _{i=0}^n{f}_i{x}^i$and $g\left(x\right)=\sum _{i=0}^n{g}_i{x}^i$where, $c_1,\dots ,c_m\in R$such that role="math" localid="1653721154629" $\left(\prod _{i=1}^n{c}_i\right)f\left(x\right)=g\left(x\right)$implies that $\left(\prod _{i=1}^n{c}_i\right)f_j=g_j$ for each $0\le j\le n$.

Since $g\left(x\right)$is primitive, $gcd\left(g_i\right)=1$.

Thus, the equation $\left(\prod _{i=1}^n{c}_i\right)f_j=g_j$can be written as $\prod _{i=1}^n{c}_i|1$.

Here, each $c_i$divides the unit implies $c_i$is an unit.

Therefore, if $c_1\dots c_{m}f\left(x\right)=g\left(x\right)$with $c_1\in R$and $g\left(x\right)$primitive in $R\left[x\right]$then each $c_i$is a unit.