Question

Question 11: Show that the principal ideal(x - 1)in $\mathbf{\mathbb{Z}}\left[x\right]$ is prime but not maximal.

Short answer

Answer:

It is proved that the principal ideal (x - 1) in $\mathrm{\mathbb{Z}}\left[x\right]$ is prime but not maximal.

Step-by-step solution

Statement of theorem 6.14 and theorem 6.15 

Theorem 6.14states that if P is an ideal in a commutative ringR with identity. Then, P would be a prime ideal such that in the quotient ring, R/P would be an integral domain.

Theorem 6.15 states that if M is an ideal in acommutative ring R with identity. Then, M would be a maximal idealif and only if the quotient ring R/M would be a field.

Show that the principal ideal (x - 1)  in ℤx is prime but not maximal

First isomorphism theoremstates that if$f:R\to S$is a surjective homomorphismof rings with kernel K . Then, the quotient ring R/K would be isomorphic to S.

Considerthe homomorphism $\phi :\mathrm{\mathbb{Z}}\left[x\right]\to \mathrm{\mathbb{Z}};\phi \left(f\left(x\right)\right):=f\left(1\right)$. This is a surjective homomorphism. There is $\phi \left(\left(x-1\right)p\left(x\right)\right)=0$, therefore $\left(x-1\right)\subseteq \ker \phi$.

Also, $f\left(x\right)\in \ker \phi$ implies that 1 would be the root of f(x) and then (x-1) divides f(x) . It implies that $f\left(x\right)\in \left(x-1\right)$.

As a result, $\ker \phi \subseteq \left(x-1\right)$ and$\ker \phi =\left(x-1\right)$ .

It concludes that $\raisebox{1ex}{$\mathrm{\mathbb{Z}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$(x-1)$}\right.\cong \mathrm{\mathbb{Z}}$ would be an integral domain but not a field according to the first isomorphism theorem.

As a result, (x - 1) would be a prime but not maximal according to theorem 6.14 and theorem 6.15.

Hence, it is proved that the principal ideal (x-1) in $\mathrm{\mathbb{Z}}\left[x\right]$is prime but not maximal.