Question

If R is a ring such that$\mathit{R}\left[\mathit{x}\right]$ is a UFD, prove that R is a UFD.

Short answer

It is proved that R is a UFD.

Step-by-step solution

Use the fact and Example 1

The degree of products of polynomials is the sum of the degrees of the factors.

So, any factorization into irreducible of an element$r\in R$in $R\left[x\right]$has factorscontained in R.

If role="math" localid="1653723630510" $r=\prod _{i=1}^m{p}^i\left(x\right)$inrole="math" localid="1653723636497" $R\left[x\right]$then$p^i\left(x\right)=p_0^i\in R$.

Example 1

Let R be any integral domain and role="math" localid="1653723583454" $p\in R$. Prove that p is irreducible in R if and only if the constant polynomial p is irreducible inrole="math" localid="1653723614449" $R\left[x\right]$.

Given that R is a ring such that role="math" localid="1653723544237" $R\left[x\right]$ is a UFD.

Prove R is UFD

By Example 1, we can say that $p_0^i$ are also irreducible in R.

Therefore, R inherits factorizations into irreducible from $R\left[x\right]$.

On the other hand, any factorization into irreducible in R is also in $R\left[x\right]$.

Therefore, factorizations in $R$are unique since they are unique in $R\left[x\right]$.

Therefore, R is a UFD.