5th Cyclotomic Ring is not a Unique Factorization Domain

From ProofWiki
Jump to navigation Jump to search


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:

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


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

For all $x \in D$ such that $x$ is non-zero and not a unit of $D$:
$(1): \quad x$ 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.

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.