Question

If R is a ring such that R[x] is a principle ideal domain, prove that R is a field.

Short answer

It is proved that R is a field.

Step-by-step solution

Unique Factorization Domain

An integral domain R with unit element is a unique factorization domain if any non-zero element in R is either a unit or can be written as a product of finite number of irreducible elements of R.

Therefore, we have R as an Integral domain with unity.

We also have $R\left[x\right]$as a principal ideal domain.

To prove: R is a field

We will prove is a field

Now, we know that $R/I$ is a field if and only if I is the maximal ideal.

Therefore, we will prove is maximal ideal of$R\left[x\right]$ .

Prove that x is a maximal ideal and R is a filed

Let I be an ideal of $R\left[x\right]$such that .

Now, to prove: either or $I=R\left[x\right]$.

Since $R\left[x\right]$is principle ideal domain, for some $f\left(x\right)\in R\left[x\right]$.

Now, and it is simplified as:

Also, $x=f\left(x\right)g\left(x\right)$for some $f\left(x\right)g\left(x\right)\in R\left[x\right]$.

Now, we have 3 possibilities:

If $f\left(x\right)=x\Rightarrow g\left(x\right)=1$.

If $f\left(x\right)=ax\Rightarrow g\left(x\right)={\alpha }^{-1}$.

If $f\left(x\right)=1\Rightarrow g\left(x\right)=x$.

Now, if .

If .

If $f\left(x\right)$= 1 then .

Simplify as:

Hence, either or $I=R$.

Hence, is a maximal ideal of $R\left[x\right]$.

Therefore, is a field.

Also,role="math" localid="1653725386603" .

Therefore, R is a field.