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in what follows all the rings are commutative, nontrivial, with unit. Recall the following definitions:

1) $\pi\in A$ is prime if $(\pi)$ is a nonzero prime ideal

2) $\pi\in A$ is irreducible if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $a$ or $b$ is a unit.

3) $\pi\in A$ is indivisible if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $\pi\mid a$ or $\pi\mid b$.

The following facts are known:

  • if $A$ is a domain, prime$\Rightarrow$ irreducible.

This is not true anymore if $A$ has zero divisor (e.g. $A=\mathbb{Z}/6\mathbb{Z}$: $A$ has prime elements, but no irreducible elements)

  • irreducible $\Rightarrow$ indivisible

  • there exist indivisible elements which are not irreducible: $A=\mathbb{Z}/6\mathbb{Z}$, $\pi=3$ . However, this one is prime.

  • if $A$ is a domain, irreducible $\iff$ indivisible

  • if $A$ is noetherian, $A$ has indivisible elements

  • if $A$ is a noetherian domain which is not a field, $A$ has irreducible elements.

After this lengthy introduction, let me ask the following questions:

  • Q1: can we find an example of an indivisible element which is neither prime or irreducible ? If possible, I would like $A$ to be noetherian or, even better, finite.

  • Q2: can we find an example of a noetherian ring which is not a field, which has no prime elements AND no irreducible elements ? (so necessarily, $A$ has zero divisors)

  • Q3: can we find an example of a noetherian domain which is not a field which has no prime elements ?

  • Q4: if the answer to Q3 is NO, can we find an example of a domain which has irreducible elements, but which has no prime elements ?

Thanks for your time.

Greg

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  • $\begingroup$ Q2 and Q3: Fields are noetherian domains and have no primes and no irreducible elements. However, they have no zero divisors, so your parenthetical comment is false. In particular, it is false that Noetherian domains must have indivisble elements. $\endgroup$ – Arturo Magidin Jan 20 '15 at 20:50
  • $\begingroup$ In a commutative ring, "indivisible" and "prime" are the same thing for nonzero elements. For in a commutative ring, prime and completely prime are equivalent for ideals. If $p$ is prime, and $p|ab$, then $ab\in (p)$, hence $a\in (p)$ or $b\in (p)$, hence $p|a$ or $p|b$. Conversely, if $p$ is indivisible and $ab\in (p)$, then $p|ab$, hence $p|a$ or $p|b$, hence $a\in (p)$ or $b\in (p)$, thus $(p)$ is completely prime, and hence prime; therefore, $p$ is prime. $\endgroup$ – Arturo Magidin Jan 20 '15 at 20:52
  • $\begingroup$ Hi,sorry , I meant "indivisible" in Q1, and I forgot to get rid of the trivial case of a field. I've modified the text now. However, I don't understand how you can say that $p\mid ab\Rightarrow p\mid a$ or $p\mid b$ when $p$ is indivisible. $\endgroup$ – GreginGre Jan 20 '15 at 21:02
  • $\begingroup$ I misread the condition for "indivisible". $\endgroup$ – Arturo Magidin Jan 20 '15 at 21:02
  • $\begingroup$ On question 3, take the ring $A=K[[t^2,t^3]]$. The only non-zero prime ideal in this ring is the maximal ideal $(t^2,t^3)$, which is not principal. Thus, there are no prime elements. It is noetherian, according to this wiki: en.wikipedia.org/wiki/Cohen%E2%80%93Macaulay_ring $\endgroup$ – Pace Nielsen Jan 21 '15 at 1:41
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To answer Q2: Every noetherian (commutative) ring that is not a finite product of fields admits irreducible elements. Since any finite product of $\ge 2$ fields admits prime elements, this shows that the answer to Q2 is no.

In an arbitrary commutative ring $A$, define $x\in A$ to be quasi-irreducible if $Ax$ is maximal among principal ideals $\neq A$. Thus if $A$ is noetherian and not a field, then it admits nonzero quasi-irreducible elements. (Note that 0 is quasi-irreducible iff $A$ is a field.) Moreover, irreducible implies quasi-irreducible.

The converse to holds for nonzero elements when $A$ is a domain, but also more generally when $A$ is connected, in the sense that it admits no nontrivial idempotent: (*) if $A$ is connected and $x\in A$ with $x\neq 0$, $x$ quasi-irreducible, then $x$ is irreducible.

Indeed, suppose that $x\neq 0$ is quasi-irreducible. Suppose by contradiction that $x=yz$ with $y,z$ non-invertible. Since $x$ is quasi-irreducible, this implies that $y,z\in Ax$. Write $y=ax$ and $z=bx$. Hence $x=abx^2$. That is, $x(1-abx)=0$. Then $(abx)^2=ab(abx^2)=abx$. Thus $abx$ is idempotent. Since $A$ is connected, either this implies $abx=1$, so $x$ is invertible (contradiction), or $abx=0$, whence $abx^2=x=0$, another contradiction.

Note that the converse to (*) does not hold in the non-connected case: for instance in a product of 2 fields, $(1,0)$ is quasi-irreducible but not irreducible.

Let us now prove the main claim. Assume that $A\neq 0$ is an arbitrary noetherian ring. Then it can be written as a product $A=A_1\times \dots \times A_k$ with each $A_i$ connected.

Suppose that one of the $A_i$ is not a field. After reindexing, we can suppose $i=1$. Let $x_1$ be a quasi-irreducible element in $A_1$; by (*) it is irreducible in $A_1$. Then $(x_1,1,\dots,1)$ is an irreducible element in $A$.

Finally, if all $A_i$ are fields and $k\ge 2$, then $(0,1,\dots,1)$ is a prime element.

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  • $\begingroup$ The product of <= 1 field does not admit prime elements. $\endgroup$ – ACL May 11 '15 at 7:23
  • $\begingroup$ Do you have a reference for the definition of quasi-irreducible? (I've tried the usual tactic without much success.) Is quasi-irreducible a standard term? $\endgroup$ – Jan Grabowski Nov 17 '15 at 10:51
  • $\begingroup$ @JanGrabowski No, I coined the word to fit this answer. $\endgroup$ – YCor Nov 17 '15 at 14:15
  • $\begingroup$ Oh :-( More work for me then... It's a good idea, though, IMHO - assuming I've understood correctly, it gives a way to say that $x\in k[x,x^{-1}]$ can't be factorised, except as $x^{-1}x^{2}$ or some consequence of this. It's obviously related to irreducibility in the overring $k[x]$ but I wanted an internal notion (actually in a noncommutative setting...). But thanks for the swift reply and (non-)pointer, saves me spending too long looking elsewhere. $\endgroup$ – Jan Grabowski Nov 17 '15 at 15:07
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On question 1: Let $A=\mathbb{F}_2[a,b,d\ :\ a^2=b^2,ad=a,bd=b]$. The element $a$ is not prime since $A/(a)$ has a nonzero nilpotent element $\overline{b}$. The element $a$ also is not irreducible since $a=ad$ is a product of non-units. All that remains is to show that when $a=xy$ with $x,y\in A$, then either $a|x$ or $a|y$.

Working modulo the ideal generated by $b$, we find that (after switching $x$ and $y$ if necessary) we have $x=a+x'$, and $y=f(d)+y'$ for some $x',y'\in (b)$ and some polynomial $f$ with an odd number of terms. Thinking of the relations on the elements of $A$ as a reduction system, we can treat $x'$ and $y'$ as a sum of monomials involving no $d$ and no $a^2$. Subtracting $a$ from both sides of $a=xy$ we have $$(\ast)\qquad 0=ay'+x'+x'y'.$$ When we talk about the reduced form of $(\ast)$, we mean that on the RHS we replace all instances of $a^2$ by $b^2$ repeatedly. (Note, there are no $d$'s left over.)

If every monomial in $x'$ is divisible by $a$ (or $b^2=a^2$) then we are done since then $a|x$. So assume otherwise. Thus $b$ is in the support of $x'$. But this contradicts $(\ast)$ since there is no other monomial on the RHS of $(\ast)$ that can reduce to $b$, because the only relation $a^2=b^2$ respects degrees.

This gives us the necessary contradiction, and shows that $a$ is indivisible.

Note that $A$ is Noetherian (being a quotient of a Noetherian ring).

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  • 1
    $\begingroup$ @GreginGre, in a finite ring, indivisible elements are either prime or irreducible. Sketch: Work by contradiction, with $a$ indivisible but not prime or irreducible. Get $a=ad$ for some nonzero, non-unit $d$. Then $d$ is a zero-divisor (since $A$ is finite) so fix $d'\neq 0$ with $dd'=0$. Consider $a=(a+d')\cdot d$. Case 1: $a|d'$, then $d'=ax$ and $0=dd'=dax=ax=d'$, a contradiction. Case 2: $a|d$, then $d=ax$ and $a=ad=a^2x$. So $e:=ax=e^2$. But $R/(a)=R/(e)\cong (1-e)R$ is not a domain. Fixing $st=0$, $s,t\in (1-e)R\setminus\{0\}$, we have $(a+s)(e+t)=a$. But $a$ divides neither factor. $\endgroup$ – Pace Nielsen Jan 24 '15 at 17:31
  • $\begingroup$ Thanks! I wouldn't mind having a detailed proof for the existence of $d$. Clearly, this is not true for any finite ring and any element: for example, if $A=\mathbb{Z}/4\mathbb{Z}$ and $a=\bar{2}$, then all the elements $d$ satisfying $ad=a$ are units. So, it probably uses the fact that $a$ is not prime or irreducible, but I can't see the argument. $\endgroup$ – GreginGre Feb 26 '15 at 18:33
  • $\begingroup$ Ok, the way i formulated my question was confusing. I was talking about your sketch of proof of the fact that in a finite ring, any indivisible element is prime or irreducible. If $a$ is indivisible but not prime nor irreducible, your first step is to pick a nonzero and nonunit element $d$ such that $ad=a$. I don't understand how you find such a $d$, because for an arbitrary finite ring and an arbitrary $a$, this is not true (take $A=\mathbb{Z}/4\mathbb{Z}, a=\bar{2}$. Then, if $a=ad$, we necessarily have $d=\bar{1}$ or $\bar{3}$, which are units). $\endgroup$ – GreginGre Feb 26 '15 at 21:11
  • $\begingroup$ @GreginGre: Okay, I understand now. In the sketch $a$ is assumed to be indivisible but not irreducible. Since it is not irreducible, write $a=xy$ for some $x,y$ non-units. By indivisibility, we may assume $a|x$, so write $x=a\cdot r$. Setting $d=ry$, which is not a unit since $y$ is not a unit, we have $a=ad$. $\endgroup$ – Pace Nielsen Feb 26 '15 at 21:33
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It looks like these questions have been taken care of, but I just thought I would mention a few references and talk about other similar concepts in between the definitions you provide here! I find this kind of thing very interesting, so I am always excited to see others interested in it as well.

If you are interested in even more of these kinds of irreducible properties, I would suggest taking a look at the paper by D.D. Anderson and S. Valdez-Leon http://projecteuclid.org/euclid.rmjm/1181072068

Many of these concepts of irreducible/indivisible diverge with zero-divisors present. It may be a bit confusing because what you have called indivisible is what is called irreducible in this paper. Your definition of irreducible is very strongly irreducible. In fact there are two other notions that are in between these two concepts. $\pi$ is m-irreducible if it is maximal among principal ideals. $\pi$ is strongly irreducible if $\pi=bc$ implies $\pi \approx b$ or $\pi \approx c$ where $a\approx b$ means there is a unit $\lambda$ such that $a=\lambda b$.

Then for non-zero elements you have very strongly irreducible $\Rightarrow$ m-irreducible $\Rightarrow$ strongly irreducible $\Rightarrow$ irreducible. Lastly, prime $\Rightarrow$ irreducible (your indivisible) but none of the others.

If you look at example 2.3 in the article I linked, they provide the example $F[X,Y,Z]/(X-XYZ)$ where $F$ is a field. Then $x,y,z$ are the image of $X,Y,Z$, respectively. Then $x$ is prime, and hence irreducible (indivisible in your terms) but not even strongly irreducible and hence neither m-irreducible nor very strongly irreducible. Other examples showing these are distinct in rings with zero-divisors are provided in this article following Theorem 2.12.

Regarding the last few questions I would point you towards (ON INTEGRAL DOMAINS WITH NO ATOMS) which are cleverly called "anti-matter domains" https://www.ndsu.edu/pubweb/~coykenda/paper13.pdf

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