Mixing
How to recognise an algebra form a ring?
$begingroup$
Suppose $mathcal{V}$ be a variety of algebras of some signature $Sigma ={f_1 ,f_2dots f_n}$. Let $mathfrak{A}={A, f_1, f_2,dots ,f_n}$ be an algebra in $mathcal{V}$.
Sometimes we are able to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ such that $mathfrak{A} $ form a ring.
Sometimes it is possible to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ in uniform way such that every algebra of $mathcal{V}$ form a ring.
Well known example of such is the variety of Boolean algebras. An arbitrary Boolean algebra form Boolean ring with respect to meet $t_*(a,b)=awedge b$ and simmetric differene $t_+(a,b)=(awedge neg b)vee(neg a wedge b)$ operation terms.
Is it possible to recognise somehow that either the algebra from a variety or every algebra from variety from a ring (possibly non-commutative) with respect to some terms $o,iin mathfrak{A}$ (either for zero-element and unity, or without unity) and $t_{+},t_{times}:mathfrak{A}^2 to mathfrak{A} $ in signature $Sigma$?
abstract-algebra reference-request ring-theory universal-algebra
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
$endgroup$
|
show 6 more comments
$begingroup$
Suppose $mathcal{V}$ be a variety of algebras of some signature $Sigma ={f_1 ,f_2dots f_n}$. Let $mathfrak{A}={A, f_1, f_2,dots ,f_n}$ be an algebra in $mathcal{V}$.
Sometimes we are able to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ such that $mathfrak{A} $ form a ring.
Sometimes it is possible to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ in uniform way such that every algebra of $mathcal{V}$ form a ring.
Well known example of such is the variety of Boolean algebras. An arbitrary Boolean algebra form Boolean ring with respect to meet $t_*(a,b)=awedge b$ and simmetric differene $t_+(a,b)=(awedge neg b)vee(neg a wedge b)$ operation terms.
Is it possible to recognise somehow that either the algebra from a variety or every algebra from variety from a ring (possibly non-commutative) with respect to some terms $o,iin mathfrak{A}$ (either for zero-element and unity, or without unity) and $t_{+},t_{times}:mathfrak{A}^2 to mathfrak{A} $ in signature $Sigma$?
abstract-algebra reference-request ring-theory universal-algebra
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
$endgroup$
1
$begingroup$
You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
$endgroup$
– Rob Arthan
Dec 28 '18 at 21:13
1
$begingroup$
this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
$endgroup$
– Eran
Dec 29 '18 at 16:23
$begingroup$
@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
$endgroup$
– Evgeny Kuznetsov
Dec 29 '18 at 18:23
1
$begingroup$
Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
$endgroup$
– Evgeny Kuznetsov
Dec 31 '18 at 9:11
2
$begingroup$
Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
$endgroup$
– amrsa
Dec 31 '18 at 10:15
|
show 6 more comments
$begingroup$
Suppose $mathcal{V}$ be a variety of algebras of some signature $Sigma ={f_1 ,f_2dots f_n}$. Let $mathfrak{A}={A, f_1, f_2,dots ,f_n}$ be an algebra in $mathcal{V}$.
Sometimes we are able to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ such that $mathfrak{A} $ form a ring.
Sometimes it is possible to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ in uniform way such that every algebra of $mathcal{V}$ form a ring.
Well known example of such is the variety of Boolean algebras. An arbitrary Boolean algebra form Boolean ring with respect to meet $t_*(a,b)=awedge b$ and simmetric differene $t_+(a,b)=(awedge neg b)vee(neg a wedge b)$ operation terms.
Is it possible to recognise somehow that either the algebra from a variety or every algebra from variety from a ring (possibly non-commutative) with respect to some terms $o,iin mathfrak{A}$ (either for zero-element and unity, or without unity) and $t_{+},t_{times}:mathfrak{A}^2 to mathfrak{A} $ in signature $Sigma$?
abstract-algebra reference-request ring-theory universal-algebra
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
$endgroup$
Suppose $mathcal{V}$ be a variety of algebras of some signature $Sigma ={f_1 ,f_2dots f_n}$. Let $mathfrak{A}={A, f_1, f_2,dots ,f_n}$ be an algebra in $mathcal{V}$.
Sometimes we are able to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ such that $mathfrak{A} $ form a ring.
Sometimes it is possible to define operations $t_+, t_*$ by means of $f_1, f_2,dots,f_n$ in uniform way such that every algebra of $mathcal{V}$ form a ring.
Well known example of such is the variety of Boolean algebras. An arbitrary Boolean algebra form Boolean ring with respect to meet $t_*(a,b)=awedge b$ and simmetric differene $t_+(a,b)=(awedge neg b)vee(neg a wedge b)$ operation terms.
Is it possible to recognise somehow that either the algebra from a variety or every algebra from variety from a ring (possibly non-commutative) with respect to some terms $o,iin mathfrak{A}$ (either for zero-element and unity, or without unity) and $t_{+},t_{times}:mathfrak{A}^2 to mathfrak{A} $ in signature $Sigma$?
abstract-algebra reference-request ring-theory universal-algebra
abstract-algebra reference-request ring-theory universal-algebra
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
edited Dec 29 '18 at 10:29
Evgeny Kuznetsov
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
asked Dec 28 '18 at 21:00
Evgeny KuznetsovEvgeny Kuznetsov
274212
274212
1
$begingroup$
You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
$endgroup$
– Rob Arthan
Dec 28 '18 at 21:13
1
$begingroup$
this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
$endgroup$
– Eran
Dec 29 '18 at 16:23
$begingroup$
@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
$endgroup$
– Evgeny Kuznetsov
Dec 29 '18 at 18:23
1
$begingroup$
Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
$endgroup$
– Evgeny Kuznetsov
Dec 31 '18 at 9:11
2
$begingroup$
Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
$endgroup$
– amrsa
Dec 31 '18 at 10:15
|
show 6 more comments
1
$begingroup$
You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
$endgroup$
– Rob Arthan
Dec 28 '18 at 21:13
1
$begingroup$
this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
$endgroup$
– Eran
Dec 29 '18 at 16:23
$begingroup$
@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
$endgroup$
– Evgeny Kuznetsov
Dec 29 '18 at 18:23
1
$begingroup$
Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
$endgroup$
– Evgeny Kuznetsov
Dec 31 '18 at 9:11
2
$begingroup$
Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
$endgroup$
– amrsa
Dec 31 '18 at 10:15
1
1
$begingroup$
You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
$endgroup$
– Rob Arthan
Dec 28 '18 at 21:13
$begingroup$
You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
$endgroup$
– Rob Arthan
Dec 28 '18 at 21:13
1
1
$begingroup$
this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
$endgroup$
– Eran
Dec 29 '18 at 16:23
$begingroup$
this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
$endgroup$
– Eran
Dec 29 '18 at 16:23
$begingroup$
@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
$endgroup$
– Evgeny Kuznetsov
Dec 29 '18 at 18:23
$begingroup$
@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
$endgroup$
– Evgeny Kuznetsov
Dec 29 '18 at 18:23
1
1
$begingroup$
Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
$endgroup$
– Evgeny Kuznetsov
Dec 31 '18 at 9:11
$begingroup$
Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
$endgroup$
– Evgeny Kuznetsov
Dec 31 '18 at 9:11
2
2
$begingroup$
Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
$endgroup$
– amrsa
Dec 31 '18 at 10:15
$begingroup$
Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
$endgroup$
– amrsa
Dec 31 '18 at 10:15
|
show 6 more comments
1 Answer
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$begingroup$
I think this question is hopelessly broad, but I will give some partial answers that may be helpful.
Sufficient conditions for an algebra to be term-equivalent to a ring
The only way I can see to do this would be algorithmically. To expand upon my comment above, if $mathbf{A}=langle A;f_1,dots, f_nrangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.
Necessary conditions
If you have a good understanding of the congruences of the algebra, then here are some things you could check:
Congruence-permutability: if $mathbf{A}$ is term-equivalent to a ring and $theta,psiinmathrm{Con}(mathbf{A})$, then $thetacircpsi:={(x,z)in A^2:exists yin A, (x,y)intheta,(y,z)inpsi}$ is equal to $psicirctheta$. This also implies $thetaveepsi=thetacircpsi$.
Congruence modularity: if $mathbf{A}$ is term-equivalent to a ring, then $mathrm{Con}(mathbf{A})$ is a modular lattice.
Congruence-uniformity: if $mathbf{A}$ is term-equivalent to a ring, $a,bin A$, and $thetainmathrm{Con}(mathbf{A})$, then there is a bijection between $a/theta$ and $b/theta$. In other words, each of the congruence classes of a congruence are the same size.
Congruence-regularity: if $mathbf{A}$ is term-equivalent to a ring, $ain A$, and $theta,psiinmathrm{Con}(mathbf{A})$, then $a/theta=a/psi$ implies $theta=psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.
This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:
- If $mathbf{A}$ and $mathbf{B}$ are both term-equivalent to rings with identity, then $mathrm{Con}(mathbf{A}timesmathbf{B})$ is isomorphic to $mathrm{Con}(mathbf{A})timesmathrm{Con}(mathbf{B})$. This is definitely not true for groups.
answered Jan 2 at 17:53
EranEran
1,260818
$endgroup$
$begingroup$
Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
$endgroup$
– Evgeny Kuznetsov
Jan 3 at 12:49
add a comment |
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$begingroup$
I think this question is hopelessly broad, but I will give some partial answers that may be helpful.
Sufficient conditions for an algebra to be term-equivalent to a ring
The only way I can see to do this would be algorithmically. To expand upon my comment above, if $mathbf{A}=langle A;f_1,dots, f_nrangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.
Necessary conditions
If you have a good understanding of the congruences of the algebra, then here are some things you could check:
Congruence-permutability: if $mathbf{A}$ is term-equivalent to a ring and $theta,psiinmathrm{Con}(mathbf{A})$, then $thetacircpsi:={(x,z)in A^2:exists yin A, (x,y)intheta,(y,z)inpsi}$ is equal to $psicirctheta$. This also implies $thetaveepsi=thetacircpsi$.
Congruence modularity: if $mathbf{A}$ is term-equivalent to a ring, then $mathrm{Con}(mathbf{A})$ is a modular lattice.
Congruence-uniformity: if $mathbf{A}$ is term-equivalent to a ring, $a,bin A$, and $thetainmathrm{Con}(mathbf{A})$, then there is a bijection between $a/theta$ and $b/theta$. In other words, each of the congruence classes of a congruence are the same size.
Congruence-regularity: if $mathbf{A}$ is term-equivalent to a ring, $ain A$, and $theta,psiinmathrm{Con}(mathbf{A})$, then $a/theta=a/psi$ implies $theta=psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.
This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:
- If $mathbf{A}$ and $mathbf{B}$ are both term-equivalent to rings with identity, then $mathrm{Con}(mathbf{A}timesmathbf{B})$ is isomorphic to $mathrm{Con}(mathbf{A})timesmathrm{Con}(mathbf{B})$. This is definitely not true for groups.
answered Jan 2 at 17:53
EranEran
1,260818
$endgroup$
$begingroup$
Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
$endgroup$
– Evgeny Kuznetsov
Jan 3 at 12:49
add a comment |
$begingroup$
I think this question is hopelessly broad, but I will give some partial answers that may be helpful.
Sufficient conditions for an algebra to be term-equivalent to a ring
The only way I can see to do this would be algorithmically. To expand upon my comment above, if $mathbf{A}=langle A;f_1,dots, f_nrangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.
Necessary conditions
If you have a good understanding of the congruences of the algebra, then here are some things you could check:
Congruence-permutability: if $mathbf{A}$ is term-equivalent to a ring and $theta,psiinmathrm{Con}(mathbf{A})$, then $thetacircpsi:={(x,z)in A^2:exists yin A, (x,y)intheta,(y,z)inpsi}$ is equal to $psicirctheta$. This also implies $thetaveepsi=thetacircpsi$.
Congruence modularity: if $mathbf{A}$ is term-equivalent to a ring, then $mathrm{Con}(mathbf{A})$ is a modular lattice.
Congruence-uniformity: if $mathbf{A}$ is term-equivalent to a ring, $a,bin A$, and $thetainmathrm{Con}(mathbf{A})$, then there is a bijection between $a/theta$ and $b/theta$. In other words, each of the congruence classes of a congruence are the same size.
Congruence-regularity: if $mathbf{A}$ is term-equivalent to a ring, $ain A$, and $theta,psiinmathrm{Con}(mathbf{A})$, then $a/theta=a/psi$ implies $theta=psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.
This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:
- If $mathbf{A}$ and $mathbf{B}$ are both term-equivalent to rings with identity, then $mathrm{Con}(mathbf{A}timesmathbf{B})$ is isomorphic to $mathrm{Con}(mathbf{A})timesmathrm{Con}(mathbf{B})$. This is definitely not true for groups.
answered Jan 2 at 17:53
EranEran
1,260818
$endgroup$
$begingroup$
Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
$endgroup$
– Evgeny Kuznetsov
Jan 3 at 12:49
add a comment |
$begingroup$
I think this question is hopelessly broad, but I will give some partial answers that may be helpful.
Sufficient conditions for an algebra to be term-equivalent to a ring
The only way I can see to do this would be algorithmically. To expand upon my comment above, if $mathbf{A}=langle A;f_1,dots, f_nrangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.
Necessary conditions
If you have a good understanding of the congruences of the algebra, then here are some things you could check:
Congruence-permutability: if $mathbf{A}$ is term-equivalent to a ring and $theta,psiinmathrm{Con}(mathbf{A})$, then $thetacircpsi:={(x,z)in A^2:exists yin A, (x,y)intheta,(y,z)inpsi}$ is equal to $psicirctheta$. This also implies $thetaveepsi=thetacircpsi$.
Congruence modularity: if $mathbf{A}$ is term-equivalent to a ring, then $mathrm{Con}(mathbf{A})$ is a modular lattice.
Congruence-uniformity: if $mathbf{A}$ is term-equivalent to a ring, $a,bin A$, and $thetainmathrm{Con}(mathbf{A})$, then there is a bijection between $a/theta$ and $b/theta$. In other words, each of the congruence classes of a congruence are the same size.
Congruence-regularity: if $mathbf{A}$ is term-equivalent to a ring, $ain A$, and $theta,psiinmathrm{Con}(mathbf{A})$, then $a/theta=a/psi$ implies $theta=psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.
This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:
- If $mathbf{A}$ and $mathbf{B}$ are both term-equivalent to rings with identity, then $mathrm{Con}(mathbf{A}timesmathbf{B})$ is isomorphic to $mathrm{Con}(mathbf{A})timesmathrm{Con}(mathbf{B})$. This is definitely not true for groups.
answered Jan 2 at 17:53
EranEran
1,260818
$endgroup$
I think this question is hopelessly broad, but I will give some partial answers that may be helpful.
Sufficient conditions for an algebra to be term-equivalent to a ring
The only way I can see to do this would be algorithmically. To expand upon my comment above, if $mathbf{A}=langle A;f_1,dots, f_nrangle$ is a finite algebra with finite signature, you can list out all the binary term operations of the algebra, and then check to see if any two of them correspond to addition and multiplication. The Universal Algebra Calculator can list out the binary term operations, but you would have to figure out whether any two of them could be addition/multiplication yourself. Finally, if you did find addition and multiplication, you'd then have to check if $f_1,dots,f_n$ could be obtained from addition, multiplication, (unary) negation, and (constants) 0 and 1.
Necessary conditions
If you have a good understanding of the congruences of the algebra, then here are some things you could check:
Congruence-permutability: if $mathbf{A}$ is term-equivalent to a ring and $theta,psiinmathrm{Con}(mathbf{A})$, then $thetacircpsi:={(x,z)in A^2:exists yin A, (x,y)intheta,(y,z)inpsi}$ is equal to $psicirctheta$. This also implies $thetaveepsi=thetacircpsi$.
Congruence modularity: if $mathbf{A}$ is term-equivalent to a ring, then $mathrm{Con}(mathbf{A})$ is a modular lattice.
Congruence-uniformity: if $mathbf{A}$ is term-equivalent to a ring, $a,bin A$, and $thetainmathrm{Con}(mathbf{A})$, then there is a bijection between $a/theta$ and $b/theta$. In other words, each of the congruence classes of a congruence are the same size.
Congruence-regularity: if $mathbf{A}$ is term-equivalent to a ring, $ain A$, and $theta,psiinmathrm{Con}(mathbf{A})$, then $a/theta=a/psi$ implies $theta=psi$. In other words, if two congruences have a congruence class in common, then they are the same congruence.
This list may seem helpful, but these four conditions can't even differentiate between groups and rings. If you are further willing to restrict to rings with identity, then I can add this to the list:
- If $mathbf{A}$ and $mathbf{B}$ are both term-equivalent to rings with identity, then $mathrm{Con}(mathbf{A}timesmathbf{B})$ is isomorphic to $mathrm{Con}(mathbf{A})timesmathrm{Con}(mathbf{B})$. This is definitely not true for groups.
answered Jan 2 at 17:53
EranEran
1,260818
answered Jan 2 at 17:53
EranEran
1,260818
answered Jan 2 at 17:53
EranEran
1,260818
answered Jan 2 at 17:53
EranEran
1,260818
1,260818
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Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
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– Evgeny Kuznetsov
Jan 3 at 12:49
add a comment |
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Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
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– Evgeny Kuznetsov
Jan 3 at 12:49
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Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
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– Evgeny Kuznetsov
Jan 3 at 12:49
$begingroup$
Thank you @Eran. I agree that this question is broad. Some partial answers, like yours are exactly what I was expected when posted the question, the answer that show the big picture. Thank you for the Universal Algebra Calculator link.
$endgroup$
– Evgeny Kuznetsov
Jan 3 at 12:49
add a comment |
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You need to supply some more context and detail: are $t_+$ and $t_*$ intended to be amongst the $f_i$ or definable from them? In what sense are we "given" the algebra?
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– Rob Arthan
Dec 28 '18 at 21:13
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this seems to be decidable when $A$ is finite: just list all binary term operations and check to see if any can be addition/multiplication. Is that what you're looking for?
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– Eran
Dec 29 '18 at 16:23
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@Eran Partially yes. The finite case is far enough good business, but it may be better in case of free algebras which may be infinite. :) Thank you for a feedback.
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– Evgeny Kuznetsov
Dec 29 '18 at 18:23
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Alex Kruckmans reading is the one I had in mind when I was writing the question. I had in mind algebraic properties of an algebra/ a variety which is at least sufficient to be a ring/a variety of rings. More relevant for me is case of not locally finite varieties. In spite of this the methods for finite algebras and locally finite varieties are appreciated as well. I am interested in the case when algebras are given as generators and relations. And it would be great to know an algorithmic side of the problem as well. I would be thankful for any information related to the question.
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– Evgeny Kuznetsov
Dec 31 '18 at 9:11
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Although it doesn't answer the question, this is similar to a result due to Gumm, Hagemann and Herrmann, for modules instead of rings. If $mathbf A$ is an algebra such that the variety it generates, $V(mathbf A)$, is congruence-permutable, then it gives three equivalent conditions, one of which is that the algebra is polynomially equivalent to a left module over some ring $mathbf R$. See, for example, Theorem II.13.4, in Burris and Sankappanavar.
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– amrsa
Dec 31 '18 at 10:15