Do there exist finite commutative rings with identity that are not Bézout rings?

A similar question has been asked before: Example of finite ring which is not a Bézout ring, but has not been answered.

There also seems to be a dearth of resources online regarding this question. Some finite commutative rings that come to mind are $$\mathbb{Z}/n\mathbb{Z}, \quad\mathbb{Z}_2\times\mathbb{Z}_2,$$ but all of them are Bézout rings. I was wondering if such rings are even possible, and what such an example of a ring might be.

To be clear, by Bézout ring, I mean a ring where Bézout's identity holds. Danke.

• For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition? – Bernard Mar 26 at 23:50
• I haven't covered ideals yet in my studies, so I am honestly not sure. – magikarrrp Mar 27 at 0:00
• Be careful, the ring $M_n(\mathbb{F}_q)$ is not commutative if $n$ is at least 2. – Captain Lama Mar 27 at 0:22
• @CaptainLama: good point! I have updated the description – magikarrrp Mar 27 at 0:25
• The DaRT search for this request yields this hit, a commutative local ring of 8 elements. – rschwieb Mar 27 at 14:04

I will work with the definition of Bézout ring provided by Bernard in the comments. Since every ideal of a finite ring is manifestly finitely generated, this amounts to asking whether there are finite rings which are not principal ideal rings (i.e., rings in which every ideal is principal).

Indeed, there are many examples of such rings. Here is one construction: let $$F$$ be any finite field you like, let $$R = F[X, Y], \mathfrak{m} = \langle X, Y\rangle$$, and put $$A = R/\mathfrak{m}^{2}$$. Then $$A$$ is a finite ring; indeed, it is an $$F$$-vector space of dimension $$3$$, with basis $$\overline{1}, \overline{X}, \overline{Y}$$, and so has $$|F|^{3}$$ elements. However, the ideal $$I := \mathfrak{m}/\mathfrak{m}^{2}$$ of $$A$$ is not principal. There are a number of ways to see this, but the point is that $$I$$ is $$I$$-torsion as an $$A$$-module, so the $$A$$-module structure on $$I$$ coincides with the induced $$A/I \cong R/\mathfrak{m} \cong F$$-module structure on $$I$$. Clearly, $$I$$ is free of rank two as an $$F$$-module on the classes $$\overline{X}, \overline{Y}$$, so $$I$$ requires two generators as an $$A$$-module.

Incidentally, $$A$$ is also a local ring with unique maximal ideal $$I$$, so this gives an answer to one of the questions in the (unanswered) linked question in your post.

• Since the OP states they are not familiar with ideals, maybe it is useful to give a more ad hoc description of $A$: you can see $A$ as $F^3$ as an additive group, with the product $(x,y,z)\cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$. – Captain Lama Mar 27 at 0:29
• @CaptainLama I think the question makes only sense if you accept the definition of an ideal (which is not too long or complcated). Actually, the OP himself wrote $\Bbb Z/n\Bbb Z$ with an ideal $n\Bbb Z$. – Dietrich Burde Mar 27 at 9:55
• I'll admit this answer has gone over my head as I was only looking for the property of Bezout's identity, as described here: planetmath.org/bezoutdomain. However, I would like to acknowledge the effort that has gone into this response and I presume it's correct based on other's endorsements. – magikarrrp Mar 27 at 14:39

Based on your comment to Bernard, I'm fairly sure that this answer will not be helpful to you, since you say that you aren't yet familiar with ideals. However, I have no idea how to approach this question without such notions.

I'm assuming your definition of Bezout ring is the same as that given by Bernard in the comments, that a ring $$R$$ is a Bezout ring if its finitely generated ideals are principal. Since $$R$$ is finite, $$R$$ is a Bezout ring if and only if all of its ideals are principal (since every ideal is finite, and thus finitely generated).

The answer is no.

Let $$k$$ be a finite field. $$V$$ a finite dimensional vector space over the field.

Define $$R=k\oplus V$$ to be the ring with multiplication $$(c,v)\cdot (d,w)=(cd,cw+dv)$$.

The proper ideals of $$R$$ are the vector subspaces of $$V$$, and the proper ideals generated by a single element are the zero and one-dimensional subspaces of $$V$$. Thus if $$V$$ is two dimensional, the ideal $$V$$ is not principal.

Attempting to translate this into more elementary language:

Let $$\Bbb{F}_p=\Bbb{Z}/p\Bbb{Z}$$ for some prime $$p$$.

Define $$R=\Bbb{F}_p^3$$, with pointwise addition and multiplication given by $$(a,b,c)(d,e,f) = (ad,ae+db,af+dc)$$. Then the ideal $$(0,*,*)$$ (I'm using $$*$$ to denote allowing that element of the tuple to be anything in the field) is not principal, since the ideal generated by a single element $$(a,b,c)$$ is either $$\{(0,0,0)\}$$ if $$a=b=c=0$$, or $$R$$ if $$a\ne 0$$ (since $$(a^{-1},-a^{-2}b,-a^{-2}c)(a,b,c)=(1,0,0),$$ which is the unit of $$R$$), or $$\{ (0,tb,tc) : t\in\Bbb{F}_p \},$$ when $$a=0$$, since $$(0,b,c)(t,x,y)=(0,tb,tc).$$

This means that the elements $$e_1=(0,1,0)$$ and $$e_2=(0,0,1)$$ do not satisfy a Bezout type identity, (though to be honest, it's not entirely what that identity should be when we are not working in a domain).

• I had to go do work after composing most of this, and after posting, I see that there is another answer with a similar strategy. Oh well. – jgon Mar 27 at 2:53
• Nice answer! I don't really see any problem with the similarity between our two answers, since yours is (as you say) translated into more elementary language that might appeal to the OP. (I elected not to, since, as you point out, I'm not sure what a Bezout identity should be in non-domains.) – Alex Wertheim Mar 27 at 3:36