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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 $mathbbZ[x]/(x^2-x+1)$
$begingroup$
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.
abstract-algebra ring-theory finite-fields finite-rings
New contributor
magikarrrp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
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.
abstract-algebra ring-theory finite-fields finite-rings
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
4 hours ago
1
$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
4 hours ago
$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
3 hours ago
add a comment |
$begingroup$
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.
abstract-algebra ring-theory finite-fields finite-rings
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
abstract-algebra ring-theory finite-fields finite-rings
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.
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.
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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
4 hours ago
1
$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
4 hours ago
$begingroup$
@CaptainLama: good point! I have updated the description
$endgroup$
– magikarrrp
3 hours ago
add a comment |
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
4 hours ago
1
$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
4 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
4 hours ago
$begingroup$
I haven't covered ideals yet in my studies, so I am honestly not sure.
$endgroup$
– magikarrrp
4 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
4 hours ago
$begingroup$
Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
$endgroup$
– Captain Lama
4 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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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).
$endgroup$
$begingroup$
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
46 mins ago
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 4 hours ago
answered 4 hours ago
![](https://i.stack.imgur.com/iFJFd.jpg?s=32&g=1)
![](https://i.stack.imgur.com/iFJFd.jpg?s=32&g=1)
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
add a comment |
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
add a comment |
$begingroup$
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).
$endgroup$
$begingroup$
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
46 mins ago
add a comment |
$begingroup$
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).
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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.
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– jgon
1 hour ago
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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.)
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– Alex Wertheim
46 mins ago
add a comment |
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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).
$endgroup$
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).
answered 1 hour ago
jgonjgon
15.8k32143
15.8k32143
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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.
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– jgon
1 hour ago
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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.)
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– Alex Wertheim
46 mins ago
add a comment |
$begingroup$
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
46 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
46 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
46 mins ago
add a comment |
magikarrrp is a new contributor. Be nice, and check out our Code of Conduct.
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For Bourbaki, a Bézout ring is a unital ring in which finitely generated ideals are principal. Is it equivalent to your definition?
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– Bernard
4 hours ago
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I haven't covered ideals yet in my studies, so I am honestly not sure.
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– magikarrrp
4 hours ago
1
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Be careful, the ring $M_n(mathbbF_q)$ is not commutative if $n$ is at least 2.
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– Captain Lama
4 hours ago
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@CaptainLama: good point! I have updated the description
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– magikarrrp
3 hours ago