Question

LetR be any integral domain and $p\in R$. Prove that p is irreducible in R if and only if the constant polynomial p is irreducible in $R\left[x\right]$. [Hint: Corollary 4.5 may be helpful.]

Short answer

It is proved that $p$is irreducible in R if and only if the constant polynomial p is irreducible in$R\left[x\right]$.

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 .

Step 2: p is irreducible in Rx

Assume that p is irreducible in R .

Then, there exists a polynomial $a\left(x\right)=\sum _{i=0}^m{a}_i{x}_i$and $b\left(x\right)=\sum _{j=0}^n{b}_j{x}_j$in $R\left[x\right]$such that $p=a\left(x\right)b\left(x\right)$.

Since R is an integral domain and ${\left(m+n\right)}^{th}$coefficient of $a\left(x\right)b\left(x\right)$is $a_m{b}_n\ne 0,m=n=0$implies $a\left(x\right)=a_0$and $b\left(x\right)=b_0$.

Thus, $p=a_0{b}_0$and since p is irreducible, either $a\left(x\right)$and $b\left(x\right)$must be a unit.

Therefore, the constant polynomial p is irreducible in$R\left[x\right]$ .

Step 3: p is irreducible in R

Assume that the constant polynomial p is irreducible in $R\left[x\right]$and $a,b\in R$ implies $p=ab$.

Here, p is a constant polynomial which implies a and b are constant polynomials where either a or b must be a unit in R .

Thus, p is irreducible in R .

Therefore, p is irreducible in R if and only if the constant polynomial p is irreducible in $R\left[x\right]$.