Question

Let p be a fixed prime and let Jbe the set of polynomials in $\mathrm{\mathbb{Z}}\left[\mathit{x}\right]$ whose constant terms are divisible by p. Prove that J is a maximal ideal in $\mathrm{\mathbb{Z}}\left[\mathit{x}\right]$.

Short answer

It is proved that J is a maximal ideal in $\mathrm{\mathbb{Z}}\left[x\right]$.

Step-by-step solution

Statement of theorem 6.15

Theorem 6.15states that consider that M as an ideal in a commutative ring with identity. Then M would be amaximal idealif and only if the quotient ring R/M would be a field.

Show that J is a maximal ideal in ℤx

A subring I of a ring R would be an idealsuch that when $r\in R$and $a\in I$, then $ra\in$and $ar\in I$.

It follows that J would be an ideal from the definition of ideal and the fact that the constant term of the sum and the product of two polynomial would be the sum and product of the constant term of every polynomial respectively.

Introduce the homomorphism $\phi :\mathrm{\mathbb{Z}}\left[x\right]/J\to {\mathrm{\mathbb{Z}}}_{p,}\phi \left(\sum _{i=0}^m{a}_i{x}^i+J\right):=\left[a_0\right]$

This would be surjective because $\phi \left(a+J\right)=\left[a\right]$for every $\left[a\right]\in {\mathrm{\mathbb{Z}}}_p$. Moreover, this would be injective because when $\phi \left(\sum _{i=0}^m{a}_i{x}^i+J\right)=\left[0\right]$ then $p|a_0$and therefore, $\sum _{i=0}^m{a}_i{x}^i\in J.J$is maximal because ${\mathrm{\mathbb{Z}}}_p$would be a field according to theorem 6.15.

Hence, it is proved that J is a maximal ideal in $\mathrm{\mathbb{Z}}\left[x\right]$.