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(i) With justification, give an example of a ring R and a prime ideal A of R such that A is not a maximal ideal in R.

(ii) With justification, give an example of a ring R and a maximal ideal B of R such that B is not a prime ideal in R.

I know the definitions of Prime and Maximal ideals

Maximal ideal: Let R be a ring. A two-sided ideal I of R is called maximal if $I\neq R$ and no proper ideal of R properly contains I.

Prime ideal: Let R be a commutative ring. An ideal I of R is called prime if $I\neq R$ and whenever $ab \in I$ for elements $a$ and $b$ of R, either $a \in I$ or $b \in I$.

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    $\begingroup$ In a commutative ring with identity, every maximal ideal is prime. For an example where such a phenomena does not hold, consider $4 \mathbb Z$ as an ideal of $2 \mathbb Z$. Why is it maximal? Why is it not prime? Also, the zero ideal is prime in an integral domain, but not always maximal, right? $\endgroup$ – астон вілла олоф мэллбэрг May 7 '18 at 5:02
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For a commutative ring with unity $R$ and an ideal $I$, we have the following:

  1. $I$ is prime if and only if $R/I$ is an integral domain.
  2. $I$ is maximal if and only if $R/I$ is a field.

This shows that every maximal ideal is a prime ideal since all fields are integral domains (every unit is not a zero-divisor).

The isomorphism $R[x]/(x) \cong R$ also provides a way to construct a prime ideal which is not maximal. For instance, pick any polynomial ring over a domain which is not a field and look at the ideal $(x)$. In particular, $\mathbb{Z}[x]$ for instance, $(x)$ is prime but not maximal since $\mathbb{Z} \cong \mathbb{Z}[x]/(x)$ is an integral domain, but not a field.

If the ring does not have an identity, it is actually possible. For instance $2 \mathbb{Z}$, the even integers. The ideal $(4)$ is maximal but not prime since $2\cdot 2 \in (4)$ but $2\notin (4)$.

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