# Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals

## Problem 520

Give an example of a commutative ring $R$ and a prime ideal $I$ of $R$ that is not a maximal ideal of $R$.

## Solution.

We give several examples. The key facts are:

1. An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
2. 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 $\Q[x, y]$.

### 1 Response

1. 07/26/2017

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##### The Quadratic Integer Ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD)

Prove that the quadratic integer ring $\Z[\sqrt{5}]$ is not a Unique Factorization Domain (UFD).

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