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In the integral domain every prime is irreducible. But the converse is not true, for example, $1+\sqrt{-3}$ is an irreducible but not a prime in ${\Bbb Z}[\sqrt{-3}]$. In a UFD, "prime" and "irreducible" are equivalent.

Here is my question:

Is there a non-UFD integral domain such that prime is equivalent to irreducible?

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    $\begingroup$ If you restrict to the Noetherian case, a domain is a UFD if and only if every irreducible element is prime. $\endgroup$ – Brandon Carter Feb 27 '13 at 19:42
  • $\begingroup$ Related (but different): math.stackexchange.com/questions/1716294 $\endgroup$ – Watson Feb 8 '17 at 23:34
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If $R$ is a GCD domain, then prime is equivalent to irreducible. Since any Bezout domain is a GCD domain, it is enough to find a Bezout domain which is not an UFD. Such examples can be found here.

A classical example is $R=\mathbb Z+X\mathbb Q[X]$. $R$ is not an UFD since we have the following (strictly) ascending chain of principal ideals: $(X)\subset (\frac{1}{2}X)\subset\cdots\subset(\frac{1}{2^n}X)\subset\cdots$. Note that $R$ is not a "trivial" example, that is, $R$ has irreducible elements, since any prime number is irreducible in $R$. I leave you as an exercise to prove that $R$ is a GCD domain.

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Yes, e.g. its vacuously true in the ring of all algebraic integers, which has no irreducibles (so no primes), since $\rm\: a = \sqrt{a} \sqrt{a}.\:$ Such domains are known as antimatter domains, since they have no atoms (irreducibles).

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  • $\begingroup$ I didn't notice this "vacuously true" example. Thank you! $\endgroup$ – Jack Feb 28 '13 at 13:40
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Set $\mathbb{Z}_{(2)} = \{\frac{x}{y} \mid x,y\in\mathbb{Z}, 2\nmid y\}$, $R_n = \mathbb{Z}_{(2)}[\sqrt[2^n]{2}]$ and $R = \bigcup_{n\in\mathbb{N}} R_n$. Then $R$ contains no prime elements and no irreducible elements.

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  • $\begingroup$ Thanks. Is there a name for ${\Bbb Z}_{(2)}$. I haven't seen this before. $\endgroup$ – Jack Feb 28 '13 at 13:41
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    $\begingroup$ @Jack It is $\mathbb Z_{2\mathbb Z}$, the localization of $\mathbb Z$ at the prime ideal $2\mathbb Z$. $\endgroup$ – user18119 Feb 28 '13 at 17:43
  • $\begingroup$ @Jack Exactly. The $(2)$ in the index stands for the ideal $2\mathbb{Z}$. $\endgroup$ – azimut Mar 1 '13 at 14:40
  • $\begingroup$ @azimut: So it is different from ${\Bbb Z}/2{\Bbb Z}$, isn't it? $\endgroup$ – Jack Mar 1 '13 at 14:45
  • $\begingroup$ Right. $\mathbb{Z}_{(2)}$ has infinitely many elements, while $\mathbb{Z}/2\mathbb{Z}$ has only two. $\endgroup$ – azimut Mar 1 '13 at 14:46

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