Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to know some examples with the following properies.

Let $R$ be a domain such that every non unit element $x$ is a product of finite irreducible elements,but $R$ is not a UFD, and there is some element $y\in R$ such that $y$ has two distinct factorizations with different lengths.

Textbooks tell me, $\mathbb{Z}[\sqrt{-5}]$ is a non-UFD since $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$. Since $\mathbb{Z}[\sqrt{-5}]$ is Noetherian, then it is easy to show that every non unit element is a product of finite irreducible elements.

But I donot know if every two factorizations of any given element of $\mathbb{Z}[\sqrt{-5}]$ have the same lengths ? That is to ask if this is an example? More, what about general algebraic integer domains ?

What is the famous (easy understood) example that a atomic domain is not a HFD(any two factorizations of any given x have the same length)?

Thanks.

share|improve this question

3 Answers 3

up vote 9 down vote accepted

Take $R$ to be the ring of polynomials in which the $x$ term does not appear, i.e. polynomials of the form $$a_0+a_2x^2+...$$ Then by induction on the degree we have that $R$ is an atomic domain, since every element is prime or product of two polynomials with less degree. And if we take the polynomial $$x^6$$ we have that $$x^2\cdot x^2 \cdot x^2$$ and $$x^3\cdot x^3$$ have different length as desired. (One can easily prove that $x^2$ and $x^3$ are prime in $R$.)

share|improve this answer
    
Nice answer! Thank you. –  wxu Sep 12 '11 at 1:48
1  
Let $A=k+x^2k[x]$, then $A$ is Noetherian and not an HFD. Let $B=A[y_1,y_2,\ldots]$, then $B$ is atomic and $B$ is not an HFD but non-Noetherian. So we get the Noetherian and Non-Noetherian cases. Nice! –  wxu Sep 12 '11 at 8:54
    
That's new for me. Nice, thanks. –  Josué Tonelli-Cueto Sep 12 '11 at 18:39

By a classic result of Carlitz, a number ring is a half-factorial domain iff it has class number $1$ or $2$. In particular this is true for $\:\mathbb Z[\sqrt{-5}]\:.\:$ For a proof see Coykendall, Half-factorial domains, a survey.

For a very simple example of a non-HFD consider $\rm\: x\: =\: 2^n\: (x/2^n) \in \mathbb Z + x\ \mathbb Q[x]\:.\:$ More generally it is easy to show that if $\rm\:R\:$ is a subring of a field $\rm\:K\:$ then $\rm\:R + x\ K[x]\:$ is an HFD $\rm\iff\:R\:$ is a field.

Note also that if a domain is not a UFD then it has an equal-length non-unique factorization , see Coykendall and Smith, Unique factorization domains.

share|improve this answer
    
Thanks, thank you for the paper link. –  wxu Sep 12 '11 at 1:52
    
And the paper "Unique factorization domains" gives us some surprising and interesting result. It is funning. –  wxu Sep 12 '11 at 1:59
    
But $A=\mathbb{Z}+x\mathbb{Q}[x]$ is not an atomic domain as you pointed out, so of course is not an HFD. And the fact that $A$ is not atomic implies $A$ is not Noetherian. –  wxu Sep 12 '11 at 8:48
    
@wxu Yes, that was my point, to mention a simple non-HFD of a different type than the number rings in my first paragraph. –  Bill Dubuque Sep 12 '11 at 14:34

The ring of integers of any number field will have the property that elements factor as a finite product of irreducibles. To see this, let $R$ be such a ring of integers and $x$ a nonzero, noninvertible element of $R.$ Then, as $R$ is noetherian, we may choose an ideal maximal among the principal ideals containing $x.$ Any generator of this ideal is irreducible. Let $u_0$ be such a generator. It follows that the ideal $(x)$ factors $(u_0)I$ for some nonzero ideal $I.$ Considering the image of these ideals in the class group of $R,$ we obtain $I$ is principal generated by some element $x_1.$ But the prime factorization of the ideal $(x_1)$ is finite and shorter than the prime factorization of the ideal $(x).$ Thus by an inductive argument, we obtain that every element of $R$ factors as the product of irreducibles.

Two factorizations of any element will have the same length if and only $R$ has class number 1 or 2. This answers your question about $\mathbb{Z}[\sqrt{-5}].$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.