Do there exist finite commutative rings with identity that are not Bézout rings?Example of finite ring which is not a Bézout ringWhen does a finite ring become a finite field?Does there exist an ordered ring, with $mathbbZ$ as an ordered subring, such that some ring of p-adic integers can be formed as a quotient ring?Characteristic collection of rings?Examples of Commutative Rings with $1$ that are not integral domains besides $mathbb Z/nmathbb Z$?Is there a theory of “rings” with partially defined multiplication?Characterize all finite unital rings with only zero divisorsThere are $10$ commutative rings of order $8$Enumerating finite local commutative rings effectivelyIs there an elementary way to prove that the algebraic integers are a Bézout domain?Does there exist a homomorphism of commutative rings with unit from $mathbbZ[x]/(x^2+3)$ to $mathbbZ[x]/(x^2-x+1)$

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


Example of finite ring which is not a Bézout ringWhen does a finite ring become a finite field?Does there exist an ordered ring, with $mathbbZ$ as an ordered subring, such that some ring of p-adic integers can be formed as a quotient ring?Characteristic collection of rings?Examples of Commutative Rings with $1$ that are not integral domains besides $mathbb Z/nmathbb Z$?Is there a theory of “rings” with partially defined multiplication?Characterize all finite unital rings with only zero divisorsThere are $10$ commutative rings of order $8$Enumerating finite local commutative rings effectivelyIs there an elementary way to prove that the algebraic integers are a Bézout domain?Does there exist a homomorphism of commutative rings with unit from $mathbbZ[x]/(x^2+3)$ to $mathbbZ[x]/(x^2-x+1)$













4












$begingroup$


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 $$mathbbZ/nmathbbZ, quadmathbbZ_2timesmathbbZ_2,$$ 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.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
    $endgroup$
    – Bernard
    Mar 26 at 23:50










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    Mar 27 at 0:00






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:22










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    Mar 27 at 0:25






  • 1




    $begingroup$
    The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
    $endgroup$
    – rschwieb
    Mar 27 at 14:04
















4












$begingroup$


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 $$mathbbZ/nmathbbZ, quadmathbbZ_2timesmathbbZ_2,$$ 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.










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
    $endgroup$
    – Bernard
    Mar 26 at 23:50










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    Mar 27 at 0:00






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:22










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    Mar 27 at 0:25






  • 1




    $begingroup$
    The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
    $endgroup$
    – rschwieb
    Mar 27 at 14:04














4












4








4


1



$begingroup$


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 $$mathbbZ/nmathbbZ, quadmathbbZ_2timesmathbbZ_2,$$ 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.










share|cite|improve this question











$endgroup$




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 $$mathbbZ/nmathbbZ, quadmathbbZ_2timesmathbbZ_2,$$ 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.







abstract-algebra ring-theory finite-fields finite-rings






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 27 at 0:31









Captain Lama

10.1k1030




10.1k1030










asked Mar 26 at 23:39









magikarrrpmagikarrrp

264




264







  • 2




    $begingroup$
    For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
    $endgroup$
    – Bernard
    Mar 26 at 23:50










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    Mar 27 at 0:00






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:22










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    Mar 27 at 0:25






  • 1




    $begingroup$
    The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
    $endgroup$
    – rschwieb
    Mar 27 at 14:04













  • 2




    $begingroup$
    For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
    $endgroup$
    – Bernard
    Mar 26 at 23:50










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    Mar 27 at 0:00






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:22










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    Mar 27 at 0:25






  • 1




    $begingroup$
    The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
    $endgroup$
    – rschwieb
    Mar 27 at 14:04








2




2




$begingroup$
For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
$endgroup$
– Bernard
Mar 26 at 23:50




$begingroup$
For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
$endgroup$
– Bernard
Mar 26 at 23:50












$begingroup$
I haven't covered ideals yet in my studies, so I am honestly not sure.
$endgroup$
– magikarrrp
Mar 27 at 0:00




$begingroup$
I haven't covered ideals yet in my studies, so I am honestly not sure.
$endgroup$
– magikarrrp
Mar 27 at 0:00




1




1




$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
Mar 27 at 0:22




$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
Mar 27 at 0:22












$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
Mar 27 at 0:25




$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
Mar 27 at 0:25




1




1




$begingroup$
The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
$endgroup$
– rschwieb
Mar 27 at 14:04





$begingroup$
The DaRT search for this request yields this hit, a commutative local ring of 8 elements.
$endgroup$
– rschwieb
Mar 27 at 14:04











2 Answers
2






active

oldest

votes


















7












$begingroup$

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 $R = F[X, Y], mathfrakm = langle X, Yrangle$, and put $A = R/mathfrakm^2$. Then $A$ is a finite ring; indeed, it is an $F$-vector space of dimension $3$, with basis $overline1, overlineX, overlineY$, and so has $|F|^3$ elements. However, the ideal $I := mathfrakm/mathfrakm^2$ 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 $A/I cong R/mathfrakm cong F$-module structure on $I$. Clearly, $I$ is free of rank two as an $F$-module on the classes $overlineX, overlineY$, 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.






share|cite|improve this answer











$endgroup$








  • 3




    $begingroup$
    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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:29






  • 2




    $begingroup$
    @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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
    $endgroup$
    – Dietrich Burde
    Mar 27 at 9:55











  • $begingroup$
    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.
    $endgroup$
    – magikarrrp
    Mar 27 at 14:39


















5












$begingroup$

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 $R=koplus V$ 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 $BbbF_p=BbbZ/pBbbZ$ for some prime $p$.



Define $R=BbbF_p^3$, 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 $ane 0$ (since $$(a^-1,-a^-2b,-a^-2c)(a,b,c)=(1,0,0),$$ which is the unit of $R$), or
$$ (0,tb,tc) : tinBbbF_p ,$$
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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – jgon
    Mar 27 at 2:53










  • $begingroup$
    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.)
    $endgroup$
    – Alex Wertheim
    Mar 27 at 3:36











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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 $R = F[X, Y], mathfrakm = langle X, Yrangle$, and put $A = R/mathfrakm^2$. Then $A$ is a finite ring; indeed, it is an $F$-vector space of dimension $3$, with basis $overline1, overlineX, overlineY$, and so has $|F|^3$ elements. However, the ideal $I := mathfrakm/mathfrakm^2$ 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 $A/I cong R/mathfrakm cong F$-module structure on $I$. Clearly, $I$ is free of rank two as an $F$-module on the classes $overlineX, overlineY$, 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.






share|cite|improve this answer











$endgroup$








  • 3




    $begingroup$
    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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:29






  • 2




    $begingroup$
    @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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
    $endgroup$
    – Dietrich Burde
    Mar 27 at 9:55











  • $begingroup$
    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.
    $endgroup$
    – magikarrrp
    Mar 27 at 14:39















7












$begingroup$

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 $R = F[X, Y], mathfrakm = langle X, Yrangle$, and put $A = R/mathfrakm^2$. Then $A$ is a finite ring; indeed, it is an $F$-vector space of dimension $3$, with basis $overline1, overlineX, overlineY$, and so has $|F|^3$ elements. However, the ideal $I := mathfrakm/mathfrakm^2$ 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 $A/I cong R/mathfrakm cong F$-module structure on $I$. Clearly, $I$ is free of rank two as an $F$-module on the classes $overlineX, overlineY$, 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.






share|cite|improve this answer











$endgroup$








  • 3




    $begingroup$
    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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:29






  • 2




    $begingroup$
    @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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
    $endgroup$
    – Dietrich Burde
    Mar 27 at 9:55











  • $begingroup$
    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.
    $endgroup$
    – magikarrrp
    Mar 27 at 14:39













7












7








7





$begingroup$

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 $R = F[X, Y], mathfrakm = langle X, Yrangle$, and put $A = R/mathfrakm^2$. Then $A$ is a finite ring; indeed, it is an $F$-vector space of dimension $3$, with basis $overline1, overlineX, overlineY$, and so has $|F|^3$ elements. However, the ideal $I := mathfrakm/mathfrakm^2$ 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 $A/I cong R/mathfrakm cong F$-module structure on $I$. Clearly, $I$ is free of rank two as an $F$-module on the classes $overlineX, overlineY$, 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.






share|cite|improve this answer











$endgroup$



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 $R = F[X, Y], mathfrakm = langle X, Yrangle$, and put $A = R/mathfrakm^2$. Then $A$ is a finite ring; indeed, it is an $F$-vector space of dimension $3$, with basis $overline1, overlineX, overlineY$, and so has $|F|^3$ elements. However, the ideal $I := mathfrakm/mathfrakm^2$ 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 $A/I cong R/mathfrakm cong F$-module structure on $I$. Clearly, $I$ is free of rank two as an $F$-module on the classes $overlineX, overlineY$, 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 27 at 0:16

























answered Mar 27 at 0:10









Alex WertheimAlex Wertheim

16.2k22848




16.2k22848







  • 3




    $begingroup$
    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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:29






  • 2




    $begingroup$
    @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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
    $endgroup$
    – Dietrich Burde
    Mar 27 at 9:55











  • $begingroup$
    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.
    $endgroup$
    – magikarrrp
    Mar 27 at 14:39












  • 3




    $begingroup$
    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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
    $endgroup$
    – Captain Lama
    Mar 27 at 0:29






  • 2




    $begingroup$
    @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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
    $endgroup$
    – Dietrich Burde
    Mar 27 at 9:55











  • $begingroup$
    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.
    $endgroup$
    – magikarrrp
    Mar 27 at 14:39







3




3




$begingroup$
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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
$endgroup$
– Captain Lama
Mar 27 at 0:29




$begingroup$
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 $(x,y,z)cdot (x',y',z') = (xx',xy'+x'y,xz'+x'z)$.
$endgroup$
– Captain Lama
Mar 27 at 0:29




2




2




$begingroup$
@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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
$endgroup$
– Dietrich Burde
Mar 27 at 9:55





$begingroup$
@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 $Bbb Z/nBbb Z$ with an ideal $nBbb Z$.
$endgroup$
– Dietrich Burde
Mar 27 at 9:55













$begingroup$
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.
$endgroup$
– magikarrrp
Mar 27 at 14:39




$begingroup$
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.
$endgroup$
– magikarrrp
Mar 27 at 14:39











5












$begingroup$

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 $R=koplus V$ 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 $BbbF_p=BbbZ/pBbbZ$ for some prime $p$.



Define $R=BbbF_p^3$, 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 $ane 0$ (since $$(a^-1,-a^-2b,-a^-2c)(a,b,c)=(1,0,0),$$ which is the unit of $R$), or
$$ (0,tb,tc) : tinBbbF_p ,$$
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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – jgon
    Mar 27 at 2:53










  • $begingroup$
    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.)
    $endgroup$
    – Alex Wertheim
    Mar 27 at 3:36















5












$begingroup$

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 $R=koplus V$ 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 $BbbF_p=BbbZ/pBbbZ$ for some prime $p$.



Define $R=BbbF_p^3$, 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 $ane 0$ (since $$(a^-1,-a^-2b,-a^-2c)(a,b,c)=(1,0,0),$$ which is the unit of $R$), or
$$ (0,tb,tc) : tinBbbF_p ,$$
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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – jgon
    Mar 27 at 2:53










  • $begingroup$
    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.)
    $endgroup$
    – Alex Wertheim
    Mar 27 at 3:36













5












5








5





$begingroup$

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 $R=koplus V$ 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 $BbbF_p=BbbZ/pBbbZ$ for some prime $p$.



Define $R=BbbF_p^3$, 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 $ane 0$ (since $$(a^-1,-a^-2b,-a^-2c)(a,b,c)=(1,0,0),$$ which is the unit of $R$), or
$$ (0,tb,tc) : tinBbbF_p ,$$
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).






share|cite|improve this answer









$endgroup$



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 $R=koplus V$ 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 $BbbF_p=BbbZ/pBbbZ$ for some prime $p$.



Define $R=BbbF_p^3$, 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 $ane 0$ (since $$(a^-1,-a^-2b,-a^-2c)(a,b,c)=(1,0,0),$$ which is the unit of $R$), or
$$ (0,tb,tc) : tinBbbF_p ,$$
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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 27 at 2:52









jgonjgon

16.4k32143




16.4k32143











  • $begingroup$
    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.
    $endgroup$
    – jgon
    Mar 27 at 2:53










  • $begingroup$
    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.)
    $endgroup$
    – Alex Wertheim
    Mar 27 at 3:36
















  • $begingroup$
    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.
    $endgroup$
    – jgon
    Mar 27 at 2:53










  • $begingroup$
    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.)
    $endgroup$
    – Alex Wertheim
    Mar 27 at 3:36















$begingroup$
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.
$endgroup$
– jgon
Mar 27 at 2:53




$begingroup$
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.
$endgroup$
– jgon
Mar 27 at 2:53












$begingroup$
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.)
$endgroup$
– Alex Wertheim
Mar 27 at 3:36




$begingroup$
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.)
$endgroup$
– Alex Wertheim
Mar 27 at 3:36

















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