# 5th Cyclotomic Ring is not a Unique Factorization Domain

## Theorem

Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.

Then $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is not a unique factorization domain.

The following elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are irreducible:

- $2$
- $3$
- $1 + i \sqrt 5$
- $1 - i \sqrt 5$

## Proof

By definition, a unique factorization domain $D$ is an integral domain with the properties that:

- $(1): \quad x$ possesses a complete factorization in $D$

- $(2): \quad$ Any two complete factorizations of $x$ are equivalent.

A complete factorization is a tidy factorization

- $x = u \circ y_1 \circ y_2 \circ \cdots \circ y_n$

such that:

- $u$ is a unit of $D$
- all of $y_1, y_2, \ldots, y_n$ are irreducible.

A particular theorem is missing.It needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an integral domain.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding the theorem.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{TheoremWanted}}` from the code. |

From Units of 5th Cyclotomic Ring, the only units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.

From Irreducible Elements of 5th Cyclotomic Ring, all of the elements of the set $S$, where:

- $S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$

are irreducible in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$

Then we have:

\(\ds 6\) | \(=\) | \(\ds 1 \times 2 \times 3\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 1 \times \paren {1 + i \sqrt 5} \paren {1 - i \sqrt 5}\) |

So there are two tidy factorizations of $6$ in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ which are not equivalent.

Hence the result.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $9$: Rings: Exercise $19 \ \text {(iv)}$