Question

If a,b are nonzero elements of an integral domain and a is a nonunit, prove that

(ab) φ (b)1

.

Short answer

It is proved that$\left(ab\right)⊈\left(b\right)$

.

Step-by-step solution

As given in the question

Consider R as an integral domain, and two non-zero elements of R are a and b, such that a is not a unit in R.

Proving that (ab)⊈  ( b ) 

Let c be an arbitrary element of (ab) and r is an element of R such that, c=abr .

Simplify c=abr as:

c = abr

=b (ar)

=b r₁ (r₁ = ar ∈ R)

This implies .c ∈ b .

Therefore, ab ⊂ b .

Let’s assume . (ab) = b

According to this assumption, every element which is present in b must be an element in ab.

So, b ∈ b implies b ∈ ab .

And b ∈ (ab) implies b=abd for some d ∈ R .

Therefore, abd - b=0r , where 0_r is a non-zero element in R.

$\Rightarrow b(ad-1_R)=1_R,1_R$is an identity element in R.

$\Rightarrow (ad-1_R)=0_R,$$R$is a n integral domain and $b\ne 0_R$.

$\Rightarrow ad=1_R$

From the above result, it is clear that a is a unit in R, this contradicts our assumption.

Therefore, $\left(ab\right)\ne b$

Since ab ⊂ b and , $\left(ab\right)\ne b$ . $\left(ab\right)⊈b$

Hence, it is proved that $\left(ab\right)⊈b$ .