Question

Give an example of polynomials $f\left(x\right),g\left(x\right)\in R\left[x\right]$such that $f\left(x\right)$and $g\left(x\right)$ are associates in $F\left[x\right]$but not in $R\left[x\right]$. Does this contradict Corollary l0.36?

Short answer

The polynomials x - 2 and 3x - 6 are associates in $\mathrm{\mathbb{Q}}\left[x\right]$but not in $\mathrm{\mathbb{Z}}\left[x\right]$ and it does not contradict Corollary 10.36.

Step-by-step solution

Corollary 4.5 and 10.36

Corollary 4.5

Let R be an integral domain and $f\left(x\right)\in R\left[x\right]$ . Then, f (x) 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 .

Corollary 10.36

Let R be a unique factorization domain and F its field of quotients. Let $f\left(x\right),g\left(x\right)$be primitive polynomials in $R\left[x\right]$. If $f\left(x\right)$and $g\left(x\right)$are associates in $F\left[x\right]$, then they are associates in $R\left[x\right]$.

Example of polynomials

Consider two polynomials $x-2$and $3x-6$in $\mathrm{\mathbb{Z}}\left[x\right]$.

Since one polynomial can be written as the product of other polynomial as $x-2=3\left(x-2\right)$implies they are associates in $\mathrm{\mathbb{Q}}\left[x\right]$but not in $\mathrm{\mathbb{Z}}\left[x\right]$.

Therefore, the polynomials $x-2$and $3x-6$are associates in $\mathrm{\mathbb{Q}}\left[x\right]$but not in $\mathrm{\mathbb{Z}}\left[x\right]$.

Contradiction of Corollary 10.36

The polynomial $3x-6$is divisible by 3, which implies it is not primitive.

Therefore, this example does not contradict Corollary 10.36.