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.
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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 $19 \ \text {(iv)}$