An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.

An ideal $I$ of $R$ is maximal if and only if $R/I$ is a field.

Example 1: $\Z$ and $(0)$

The first example is the ring of integers $R=\Z$ and the zero ideal $I=(0)$.
Note that the quotient ring is $\Z/(0)\cong \Z$ and it is integral domain but not a field.
Thus the ideal $(0)$ is a prime ideal by Fact 1 but not a maximal ideal by Fact 2.

Remark

Note that $(0)$ is the only prime ideal of $\Z$ that is not a maximal ideal.
Nonzero ideals of $\Z$ are $(p)$ for some prime number $p$.

Example 2: $\Z[x]$ and $(x)$

The second example is the ring of polynomials $R=\Z[x]$ over $\Z$ and the principal ideal $I=(x)$ generated by $x\in \Z[x]$.
The quotient ring is $\Z[x]/(x)\cong \Z$, which is an integral domain but not a field.
Thus the ideal $(x)$ is prime but not maximal by Fact 1, 2.

Example 3: $\Q[x,y]$ and $(x)$

The third example is the ring of polynomials in two variables $R=\Q[x, y]$ over $\Q$ and the principal ideal $I=(x)$ generated by $x$.
The quotient ring $\Q[x,y]/(x)$ is isomorphic to $\Q[y]$.
(The proof of this isomorphism is given in the post Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$.)

Note that $\Q[y]$ is an integral domain but it is not a field since, for instance, the element $y\in \Q[y]$ is not a unit.
Hence Fact 1, 2 implies that the ideal $(x)$ is prime but not maximal in the ring $\Q[x, y]$.

Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$
Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\]
be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]
Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?
Solution.
We […]

Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal […]

Every Maximal Ideal of a Commutative Ring is a Prime Ideal
Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.
We give two proofs.
Proof 1.
The first proof uses the following facts.
Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral […]

A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]

Is the Quotient Ring of an Integral Domain still an Integral Domain?
Let $R$ be an integral domain and let $I$ be an ideal of $R$.
Is the quotient ring $R/I$ an integral domain?
Definition (Integral Domain).
Let $R$ be a commutative ring.
An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]

In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal
Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
Definition
A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element […]

If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian?
Let $R$ be a commutative ring with $1$.
Suppose that the localization $R_{\mathfrak{p}}$ is a Noetherian ring for every prime ideal $\mathfrak{p}$ of $R$.
Is it true that $A$ is also a Noetherian ring?
Proof.
The answer is no. We give a counterexample.
Let […]

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