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Removed a redundancy.
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Oscar Lanzi
Oscar Lanzi
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You may want to read this answer, in which a domain with class number 2 is modified by overlaying the lattice of its integers with a second lattice, whose elements are sums involving two square-root terms (like $(1/2)(\sqrt{2}+\sqrt{-10})$ in the augmentation of $\mathbb Z[\sqrt{-5}]$). When multiplied pairwise, these give products that lie in the original domain. The unique factorization with this augmentation is then converted to nonunique factorizations in the original domain by pairing up the augmented-domain factors in various ways. For instance,

$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

yields $2×3$ if the above factors are paired in one way but $(1+\sqrt{-5})(1-\sqrt{-5})$ with a different pairing.

This works with class number 2 where the class group must have a single 2-cycle, and it may be generalized for higher class numbers if the class group consists only of multiple 2-cycles two-cycles (see here for an example involving class number 4). With any class group involving larger cycles, such as all class-3 domains where the class group has to have a 3-cycle or class-4 domains where the class group is $\mathbb Z/4\mathbb Z$ instead of $\mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z$, this square root extraction does not work and if unique factorization is to be repaired at all, it requires more complex embedding.

You may want to read this answer, in which a domain with class number 2 is modified by overlaying the lattice of its integers with a second lattice, whose elements are sums involving two square-root terms (like $(1/2)(\sqrt{2}+\sqrt{-10})$ in the augmentation of $\mathbb Z[\sqrt{-5}]$). When multiplied pairwise, these give products that lie in the original domain. The unique factorization with this augmentation is then converted to nonunique factorizations in the original domain by pairing up the augmented-domain factors in various ways. For instance,

$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

yields $2×3$ if the above factors are paired in one way but $(1+\sqrt{-5})(1-\sqrt{-5})$ with a different pairing.

This works with class number 2 where the class group must have a single 2-cycle, and it may be generalized for higher class numbers if the class group consists only of multiple 2-cycles two-cycles (see here for an example involving class number 4). With any class group involving larger cycles, such as all class-3 domains where the class group has to have a 3-cycle or class-4 domains where the class group is $\mathbb Z/4\mathbb Z$ instead of $\mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z$, this square root extraction does not work and if unique factorization is to be repaired at all, it requires more complex embedding.

You may want to read this answer, in which a domain with class number 2 is modified by overlaying the lattice of its integers with a second lattice, whose elements are sums involving two square-root terms (like $(1/2)(\sqrt{2}+\sqrt{-10})$ in the augmentation of $\mathbb Z[\sqrt{-5}]$). When multiplied pairwise, these give products that lie in the original domain. The unique factorization with this augmentation is then converted to nonunique factorizations in the original domain by pairing up the augmented-domain factors in various ways. For instance,

$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

yields $2×3$ if the above factors are paired in one way but $(1+\sqrt{-5})(1-\sqrt{-5})$ with a different pairing.

This works with class number 2 where the class group must have a single 2-cycle, and it may be generalized for higher class numbers if the class group consists only of multiple 2-cycles (see here for an example involving class number 4). With any class group involving larger cycles, such as all class-3 domains where the class group has to have a 3-cycle or class-4 domains where the class group is $\mathbb Z/4\mathbb Z$ instead of $\mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z$, this square root extraction does not work and if unique factorization is to be repaired at all, it requires more complex embedding.

Source Link
Oscar Lanzi
Oscar Lanzi
  • 41.4k
  • 2
  • 45
  • 107

You may want to read this answer, in which a domain with class number 2 is modified by overlaying the lattice of its integers with a second lattice, whose elements are sums involving two square-root terms (like $(1/2)(\sqrt{2}+\sqrt{-10})$ in the augmentation of $\mathbb Z[\sqrt{-5}]$). When multiplied pairwise, these give products that lie in the original domain. The unique factorization with this augmentation is then converted to nonunique factorizations in the original domain by pairing up the augmented-domain factors in various ways. For instance,

$6=(\sqrt{2})^2\left(\dfrac{\sqrt2+\sqrt{-10}}2\right)\left(\dfrac{\sqrt2-\sqrt{-10}}2\right)$

yields $2×3$ if the above factors are paired in one way but $(1+\sqrt{-5})(1-\sqrt{-5})$ with a different pairing.

This works with class number 2 where the class group must have a single 2-cycle, and it may be generalized for higher class numbers if the class group consists only of multiple 2-cycles two-cycles (see here for an example involving class number 4). With any class group involving larger cycles, such as all class-3 domains where the class group has to have a 3-cycle or class-4 domains where the class group is $\mathbb Z/4\mathbb Z$ instead of $\mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z$, this square root extraction does not work and if unique factorization is to be repaired at all, it requires more complex embedding.