Question

If n is a composite integer, prove that (n) is not a prime ideal in$\mathrm{\mathbb{Z}}$.

Short answer

It is proved as, n is not prime ideal in$\mathrm{\mathbb{Z}}$.

Step-by-step solution

Determine n is not integral domain

Consider that n is a composite integer, that means n=pq for $p,q\in \mathrm{\mathbb{Z}}$so that

$1<\left|p\right|,\left|q\right|<\mathbf{\left(}n\mathbf{\right)}$.

Here, ${\mathbf{\left[}\mathbf{p}\mathbf{\right]}}_n$and ${\mathbf{\left[}\mathbf{q}\mathbf{\right]}}_n$are the non-zero element in ${\mathrm{\mathbb{Z}}}_{\left|n\right|}$, so that,

${\mathbf{\left[}\mathbf{p}\mathbf{\right]}}_n{\mathbf{\left[}\mathbf{q}\mathbf{\right]}}_n={\mathbf{\left[}\mathbf{p}\mathbf{q}\mathbf{\right]}}_n={\mathbf{\left[}\mathbf{n}\mathbf{\right]}}_n={\mathbf{\left[}\mathbf{0}\mathbf{\right]}}_n$

Therefore,${\mathrm{\mathbb{Z}}}_{\left|n\right|}$is not an integral domain.

Determine n is not a prime ideal

By theorem 6.14, that is, Consider P be an ideal in a commutative ring R with the identity. This implies that P is a prime ideal if and only if the quotient ring R /P is an integral domain.

As, $\mathrm{\mathbb{Z}}/\mathbf{\left(}\mathbf{n}\mathbf{\right)}\cong {\mathrm{\mathbb{Z}}}_{\left|n\right|}$

Therefore, n is not prime ideal in $\mathrm{\mathbb{Z}}$.