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 $mathbbZ[x]/(x^2-x+1)$

How do I keep an essay about "feeling flat" from feeling flat?

Was Spock the First Vulcan in Starfleet?

How can I replace every global instance of "x[2]" with "x_2"

Hide Select Output from T-SQL

Your magic is very sketchy

Where in the Bible does the greeting ("Dominus Vobiscum") used at Mass come from?

If you attempt to grapple an opponent that you are hidden from, do they roll at disadvantage?

How was Earth single-handedly capable of creating 3 of the 4 gods of chaos?

Using parameter substitution on a Bash array

Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past

What's a natural way to say that someone works somewhere (for a job)?

Student evaluations of teaching assistants

The plural of 'stomach"

How will losing mobility of one hand affect my career as a programmer?

Why are on-board computers allowed to change controls without notifying the pilots?

apt-get update is failing in debian

What defines a dissertation?

Everything Bob says is false. How does he get people to trust him?

Finding all intervals that match predicate in vector

Understanding "audieritis" in Psalm 94

Can a monster with multiattack use this ability if they are missing a limb?

How does residential electricity work?

Is expanding the research of a group into machine learning as a PhD student risky?

Are there any comparative studies done between Ashtavakra Gita and Buddhim?



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 $mathbbZ[x]/(x^2-x+1)$













2












$begingroup$


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









New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 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
    4 hours ago










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    3 hours ago






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    3 hours ago










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    3 hours ago















2












$begingroup$


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









New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 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
    4 hours ago










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    3 hours ago






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    3 hours ago










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    3 hours ago













2












2








2


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









New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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









New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Captain Lama

10.1k1030




10.1k1030






New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









magikarrrpmagikarrrp

112




112




New contributor




magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 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
    4 hours ago










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    3 hours ago






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    3 hours ago










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    3 hours ago












  • 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
    4 hours ago










  • $begingroup$
    I haven't covered ideals yet in my studies, so I am honestly not sure.
    $endgroup$
    – magikarrrp
    3 hours ago






  • 1




    $begingroup$
    Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
    $endgroup$
    – Captain Lama
    3 hours ago










  • $begingroup$
    @CaptainLama: good point! I have updated the description
    $endgroup$
    – magikarrrp
    3 hours ago







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
4 hours ago




$begingroup$
For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
$endgroup$
– Bernard
4 hours ago












$begingroup$
I haven't covered ideals yet in my studies, so I am honestly not sure.
$endgroup$
– magikarrrp
3 hours ago




$begingroup$
I haven't covered ideals yet in my studies, so I am honestly not sure.
$endgroup$
– magikarrrp
3 hours ago




1




1




$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
3 hours ago




$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
3 hours ago












$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
3 hours ago




$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
3 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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











$endgroup$








  • 2




    $begingroup$
    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
    3 hours ago


















1












$begingroup$

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
    1 hour ago










  • $begingroup$
    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
    21 mins ago










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);






magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.









draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163889%2fdo-there-exist-finite-commutative-rings-with-identity-that-are-not-b%25c3%25a9zout-rings%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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











$endgroup$








  • 2




    $begingroup$
    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
    3 hours ago















4












$begingroup$

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











$endgroup$








  • 2




    $begingroup$
    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
    3 hours ago













4












4








4





$begingroup$

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 3 hours ago

























answered 3 hours ago









Alex WertheimAlex Wertheim

16.1k22848




16.1k22848







  • 2




    $begingroup$
    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
    3 hours ago












  • 2




    $begingroup$
    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
    3 hours ago







2




2




$begingroup$
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
3 hours ago




$begingroup$
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
3 hours ago











1












$begingroup$

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
    1 hour ago










  • $begingroup$
    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
    21 mins ago















1












$begingroup$

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
    1 hour ago










  • $begingroup$
    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
    21 mins ago













1












1








1





$begingroup$

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 1 hour ago









jgonjgon

15.8k32143




15.8k32143











  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    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
    21 mins ago
















  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    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
    21 mins ago















$begingroup$
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
1 hour ago




$begingroup$
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
1 hour ago












$begingroup$
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
21 mins ago




$begingroup$
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
21 mins ago










magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.









draft saved

draft discarded


















magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.












magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.











magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.














Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163889%2fdo-there-exist-finite-commutative-rings-with-identity-that-are-not-b%25c3%25a9zout-rings%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Францішак Багушэвіч Змест Сям'я | Біяграфія | Творчасць | Мова Багушэвіча | Ацэнкі дзейнасці | Цікавыя факты | Спадчына | Выбраная бібліяграфія | Ушанаванне памяці | У філатэліі | Зноскі | Літаратура | Спасылкі | НавігацыяЛяхоўскі У. Рупіўся дзеля Бога і людзей: Жыццёвы шлях Лявона Вітан-Дубейкаўскага // Вольскі і Памідораў з песняй пра немца Адвакат, паэт, народны заступнік Ашмянскі веснікВ Минске появится площадь Богушевича и улица Сырокомли, Белорусская деловая газета, 19 июля 2001 г.Айцец беларускай нацыянальнай ідэі паўстаў у бронзе Сяргей Аляксандравіч Адашкевіч (1918, Мінск). 80-я гады. Бюст «Францішак Багушэвіч».Яўген Мікалаевіч Ціхановіч. «Партрэт Францішка Багушэвіча»Мікола Мікалаевіч Купава. «Партрэт зачынальніка новай беларускай літаратуры Францішка Багушэвіча»Уладзімір Іванавіч Мелехаў. На помніку «Змагарам за родную мову» Барэльеф «Францішак Багушэвіч»Памяць пра Багушэвіча на Віленшчыне Страчаная сталіца. Беларускія шыльды на вуліцах Вільні«Krynica». Ideologia i przywódcy białoruskiego katolicyzmuФранцішак БагушэвічТворы на knihi.comТворы Францішка Багушэвіча на bellib.byСодаль Уладзімір. Францішак Багушэвіч на Лідчыне;Луцкевіч Антон. Жыцьцё і творчасьць Фр. Багушэвіча ў успамінах ягоных сучасьнікаў // Запісы Беларускага Навуковага таварыства. Вільня, 1938. Сшытак 1. С. 16-34.Большая российская1188761710000 0000 5537 633Xn9209310021619551927869394п

Беларусь Змест Назва Гісторыя Геаграфія Сімволіка Дзяржаўны лад Палітычныя партыі Міжнароднае становішча і знешняя палітыка Адміністрацыйны падзел Насельніцтва Эканоміка Культура і грамадства Сацыяльная сфера Узброеныя сілы Заўвагі Літаратура Спасылкі НавігацыяHGЯOiТоп-2011 г. (па версіі ej.by)Топ-2013 г. (па версіі ej.by)Топ-2016 г. (па версіі ej.by)Топ-2017 г. (па версіі ej.by)Нацыянальны статыстычны камітэт Рэспублікі БеларусьШчыльнасць насельніцтва па краінахhttp://naviny.by/rubrics/society/2011/09/16/ic_articles_116_175144/А. Калечыц, У. Ксяндзоў. Спробы засялення краю неандэртальскім чалавекам.І ў Менску былі мамантыА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіГ. Штыхаў. Балты і славяне ў VI—VIII стст.М. Клімаў. Полацкае княства ў IX—XI стст.Г. Штыхаў, В. Ляўко. Палітычная гісторыя Полацкай зямліГ. Штыхаў. Дзяржаўны лад у землях-княствахГ. Штыхаў. Дзяржаўны лад у землях-княствахБеларускія землі ў складзе Вялікага Княства ЛітоўскагаЛюблінская унія 1569 г."The Early Stages of Independence"Zapomniane prawdy25 гадоў таму было аб'яўлена, што Язэп Пілсудскі — беларус (фота)Наша вадаДакументы ЧАЭС: Забруджванне тэрыторыі Беларусі « ЧАЭС Зона адчужэнняСведения о политических партиях, зарегистрированных в Республике Беларусь // Министерство юстиции Республики БеларусьСтатыстычны бюлетэнь „Полаўзроставая структура насельніцтва Рэспублікі Беларусь на 1 студзеня 2012 года і сярэднегадовая колькасць насельніцтва за 2011 год“Индекс человеческого развития Беларуси — не было бы нижеБеларусь занимает первое место в СНГ по индексу развития с учетом гендерного факцёраНацыянальны статыстычны камітэт Рэспублікі БеларусьКанстытуцыя РБ. Артыкул 17Трансфармацыйныя задачы БеларусіВыйсце з крызісу — далейшае рэфармаванне Беларускі рубель — сусветны лідар па дэвальвацыяхПра змену коштаў у кастрычніку 2011 г.Бядней за беларусаў у СНД толькі таджыкіСярэдні заробак у верасні дасягнуў 2,26 мільёна рублёўЭканомікаГаласуем за ТОП-100 беларускай прозыСучасныя беларускія мастакіАрхитектура Беларуси BELARUS.BYА. Каханоўскі. Культура Беларусі ўсярэдзіне XVII—XVIII ст.Анталогія беларускай народнай песні, гуказапісы спеваўБеларускія Музычныя IнструментыБеларускі рок, які мы страцілі. Топ-10 гуртоў«Мясцовы час» — нязгаслая легенда беларускай рок-музыкіСЯРГЕЙ БУДКІН. МЫ НЯ ЗНАЕМ СВАЁЙ МУЗЫКІМ. А. Каладзінскі. НАРОДНЫ ТЭАТРМагнацкія культурныя цэнтрыПублічная дыскусія «Беларуская новая пьеса: без беларускай мовы ці беларуская?»Беларускія драматургі па-ранейшаму лепш ставяцца за мяжой, чым на радзіме«Працэс незалежнага кіно пайшоў, і дзяржаву турбуе яго непадкантрольнасць»Беларускія філосафы ў пошуках прасторыВсе идём в библиотекуАрхіваванаАб Нацыянальнай праграме даследавання і выкарыстання касмічнай прасторы ў мірных мэтах на 2008—2012 гадыУ космас — разам.У суседнім з Барысаўскім раёне пабудуюць Камандна-вымяральны пунктСвяты і абрады беларусаў«Мірныя бульбашы з малой краіны» — 5 непраўдзівых стэрэатыпаў пра БеларусьМ. Раманюк. Беларускае народнае адзеннеУ Беларусі скарачаецца колькасць злачынстваўЛукашэнка незадаволены мінскімі ўладамі Крадзяжы складаюць у Мінску каля 70% злачынстваў Узровень злачыннасці ў Мінскай вобласці — адзін з самых высокіх у краіне Генпракуратура аналізуе стан са злачыннасцю ў Беларусі па каэфіцыенце злачыннасці У Беларусі стабілізавалася крымінагеннае становішча, лічыць генпракурорЗамежнікі сталі здзяйсняць у Беларусі больш злачынстваўМУС Беларусі турбуе рост рэцыдыўнай злачыннасціЯ з ЖЭСа. Дазволіце вас абкрасці! Рэйтынг усіх службаў і падраздзяленняў ГУУС Мінгарвыканкама вырасАб КДБ РБГісторыя Аператыўна-аналітычнага цэнтра РБГісторыя ДКФРТаможняagentura.ruБеларусьBelarus.by — Афіцыйны сайт Рэспублікі БеларусьСайт урада БеларусіRadzima.org — Збор архітэктурных помнікаў, гісторыя Беларусі«Глобус Беларуси»Гербы и флаги БеларусиАсаблівасці каменнага веку на БеларусіА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіУ. Ксяндзоў. Сярэдні каменны век (мезаліт). Засяленне краю плямёнамі паляўнічых, рыбакоў і збіральнікаўА. Калечыц, М. Чарняўскі. Плямёны на тэрыторыі Беларусі ў новым каменным веку (неаліце)А. Калечыц, У. Ксяндзоў, М. Чарняўскі. Гаспадарчыя заняткі ў каменным векуЭ. Зайкоўскі. Духоўная культура ў каменным векуАсаблівасці бронзавага веку на БеларусіФарміраванне супольнасцей ранняга перыяду бронзавага векуФотографии БеларусиРоля беларускіх зямель ва ўтварэнні і ўмацаванні ВКЛВ. Фадзеева. З гісторыі развіцця беларускай народнай вышыўкіDMOZGran catalanaБольшая российскаяBritannica (анлайн)Швейцарскі гістарычны15325917611952699xDA123282154079143-90000 0001 2171 2080n9112870100577502ge128882171858027501086026362074122714179пппппп

ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result