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I overcame a theorem which says:
Let R be an Integral Domain. If every Irreducible element is Prime then it satisfies UFD 2. Converse (i.e, UFD 2 implies every Irreducible element is Prime) is true if it is UFD 1.

where,
UFD 1 is basically the Existence of a Factorization of every element.
UFD 2 is the Uniqueness of the Factorization for every factorization of every element.
We know that being UFD satisfies both UFD 1 and UFD 2.

Now I was having trouble finding examples for the situation when UFD 2 satisfies but NOT UFD 1.

Can anyone help me with this? Thanks in advance. ( Please write in a bit details if possible)

My basic idea was to figure out examples for the case that UFD 2 does not imply that Every Irreducible element is Prime by avoiding UFD 1 ( if factorization exists then it has to be unique but it is not mandatory that every element will have factorization).

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  • $\begingroup$ UFD 2 is, in fact, equivalent to every irreducible element being prime. $\endgroup$ – Geoffrey Trang Aug 22 '20 at 16:32
  • $\begingroup$ But that is for UFD.....UFD2 does not guarantee the existence. $\endgroup$ – tom_choudhurry Aug 22 '20 at 16:36
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Note that a ring has to be non-Noetherian to fail UFD 2, since if $x$ fails to have a factorization into irreducibles, then it must be that $x$ does factor somehow as $x=ab$ where $a$ and $b$ are not units and either $a$ or $b$ lacks a factorization into irreducibles - and we can then look at a factorization of that an so on to yield an infinite chain $x_1,x_2,x_3,\ldots$ where each term strictly divides the last - and then $(x_1)\subsetneq (x_2) \subsetneq (x_3) \subsetneq\ldots$ would be an infinite ascending chain of ideals.

So, we can just start by looking at our favorite non-Noetherian ring and see what happens - the example that comes to my mind would be letting $R$ be the set of polynomial expressions with rational coefficients in terms of the form $x^{n/2^k}$ for $n,k\in\mathbb N$ - or, equivalently, the direct limit of the rings $$\mathbb Q[x]\rightarrow\mathbb Q[x^{1/2}] \rightarrow \mathbb Q[x^{1/4}]\rightarrow\ldots .$$ The element $x$ lacks a factorization into irreducibles in this ring. This follows from noting that the factors of $x^{\alpha}$ are just the terms of the form $x^{\beta}$ where $\beta \leq \alpha$ - but none of these are irreducible.

However, it is true that every irreducible element $p$ is prime. In particular, suppose that we had some $a,b\in R$ such that $p|ab$. Then, it must be that all of $p$ and $a$ and $b$ are in some ring $\mathbb Q[x^{1/2^k}]$ and that $p$ is irreducible in this ring - but this ring is isomorphic to $\mathbb Q[x]$ which is a PID meaning $p$ is prime in it, thus $p$ must divide either $a$ or $b$ in this ring and therefore also in $R$. Thus, this ring has uniqueness of prime factorizations, but not existence.

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One idea would be to find a ring where no element has a factorisation. An example is the ring of algebraic numbers $\overline{\mathbb Z}$, which has no irreducible elements at all: if $a$ is algebraic non-unit, then so is $\sqrt a$; since $a = \sqrt a\cdot\sqrt a$, $a$ is not irreducible.

For a less vacuous example, consider the ring $\overline{\mathbb Z}[X]$ of polynomials with algebraic coefficients. In this ring, elements of the form $aX + b$ are irreducible, where $a, b\in \overline{\mathbb Z}[X]$ are coprime. If an element $f\in \overline{\mathbb Z}[X]$ has a factorisation, then it must be primitive: its coefficients must have no common factor. Such an element can be factorised uniquely as a product of primitive linear factors.

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  • $\begingroup$ If I have understood correctly ...your first example is basically UFD 1 but not UFD 2 $\endgroup$ – tom_choudhurry Aug 22 '20 at 18:31
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    $\begingroup$ The opposite. There are no irreducibles, so no element can be factorised as a product of irreducibles. UFD2 is vacuously true. $\endgroup$ – Mathmo123 Aug 22 '20 at 18:44
  • $\begingroup$ ohh okkk ...I see your point ...Thanks...The examples were nice $\endgroup$ – tom_choudhurry Aug 22 '20 at 19:52

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