Mixing
The variety of rings which contain a fixed ring as a subring
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
From Barnes and Mack's "An Algebraic Introduction to Mathematical Logic"
S is a commutative ring with 1. Show that the class of commutative rings R with 1_R = 1_S and which contain S as a subring is a variety (in the sense of universal algebra).
I figure the sets will need to be +, $cdot$, -, 0, 1, S algebras, where everything in S has arity-0. Then, add on all the laws for a commutative ring, plus a bunch of laws of the form $(s + s', (s +_S s'))$. However, I can't seem to see how to make S actually inject into any algebra in the variety. For example, how do I avoid the case where $s = 0$ for all $s in S$?
universal-algebra
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
 |Â
show 1 more comment
up vote
2
down vote
favorite
From Barnes and Mack's "An Algebraic Introduction to Mathematical Logic"
S is a commutative ring with 1. Show that the class of commutative rings R with 1_R = 1_S and which contain S as a subring is a variety (in the sense of universal algebra).
I figure the sets will need to be +, $cdot$, -, 0, 1, S algebras, where everything in S has arity-0. Then, add on all the laws for a commutative ring, plus a bunch of laws of the form $(s + s', (s +_S s'))$. However, I can't seem to see how to make S actually inject into any algebra in the variety. For example, how do I avoid the case where $s = 0$ for all $s in S$?
universal-algebra
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
1
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
From Barnes and Mack's "An Algebraic Introduction to Mathematical Logic"
S is a commutative ring with 1. Show that the class of commutative rings R with 1_R = 1_S and which contain S as a subring is a variety (in the sense of universal algebra).
I figure the sets will need to be +, $cdot$, -, 0, 1, S algebras, where everything in S has arity-0. Then, add on all the laws for a commutative ring, plus a bunch of laws of the form $(s + s', (s +_S s'))$. However, I can't seem to see how to make S actually inject into any algebra in the variety. For example, how do I avoid the case where $s = 0$ for all $s in S$?
universal-algebra
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
From Barnes and Mack's "An Algebraic Introduction to Mathematical Logic"
S is a commutative ring with 1. Show that the class of commutative rings R with 1_R = 1_S and which contain S as a subring is a variety (in the sense of universal algebra).
I figure the sets will need to be +, $cdot$, -, 0, 1, S algebras, where everything in S has arity-0. Then, add on all the laws for a commutative ring, plus a bunch of laws of the form $(s + s', (s +_S s'))$. However, I can't seem to see how to make S actually inject into any algebra in the variety. For example, how do I avoid the case where $s = 0$ for all $s in S$?
universal-algebra
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
edited Mar 20 '17 at 22:40
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
asked Mar 20 '17 at 22:35
Duncan Ramage
3,9491026
3,9491026
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
1
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38
 |Â
show 1 more comment
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
1
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
1
1
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38
 |Â
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2195772%2fthe-variety-of-rings-which-contain-a-fixed-ring-as-a-subring%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Your thinking appears tp be correct: the class of rings that have a subring isomorphic to $BbbZ_2$ is not a variety. No equational axiom can exclude the possibility that $1 = 0$. However, the phrasing of the question from the book (as given in your quotation) seems to be very confused: are you sure you have quoted it correctly?
– Rob Arthan
Mar 20 '17 at 22:49
Aside from the parenthetical at the end, it is a direct quotation. i.imgur.com/zF1LFIa.png
– Duncan Ramage
Mar 20 '17 at 22:57
Alright, thanks for the help. I figure it was just a poorly written question; it should probably be "contains an image of S under some homomorphism".
– Duncan Ramage
Mar 21 '17 at 2:18
Your last comment seems to be correct. As it is originally (according to the image you posted) it is wrong. The reason is that, in the sense of Universal Algebra, every variety must have one element algebras, since this is the case for the quotient of any algebra $mathbfA / theta$ by the maximal congruence $theta = (a,b) : a, b in A $.
– amrsa
Mar 21 '17 at 9:58
1
I think the uttermost misleading requirement is that $1_R=1_S$. That class is not even closed under isomorphism!
– Pedro Sánchez Terraf
Mar 21 '17 at 13:38