Show that $equiv$ is a congruence on $Mtimes S$ The 2019 Stack Overflow Developer Survey...
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Show that $equiv$ is a congruence on $Mtimes S$
The 2019 Stack Overflow Developer Survey Results Are InConstruction of the monoid of fractionsWhen is a face of a monoid contained in a minimal direct summand?Proof: Every normal subgroup has corresponding Congruence relation and vice versaProve that $(mathbb{R}setminus {0},sim):= ab>0$ is transitive.If $x + y sim x$ in a commutative monoid, does this imply $y sim 0$?Generated congruence on monoidsGroup Congruence on a Free monogenic semigroupCongruence if and only if Left and Right CongruenceProof verification: Show that the set of congruence classes form a partition of $mathbb{Z}$.Objects of a congruence category
$begingroup$
I'm sorry if a similar question has been posted before, but I was unable to find one based on my searches. This is an extra practice problem for a number theory class. I've been trying to prove this out, and I'm getting stuck on the transitivity step. This is also a check on myself to make sure that I understand what I need to demonstrate in order to show a congruence.
Here is the full question. I include the last sentence for extra context:
For a submonoid $S$ of a commutative monoid $(M,*)$, define a relation $equiv$ on the monoid $Mtimes S$ by $(a,s)equiv(b,t)$ if there is a $uin S$ such that $a*t*u=b*s*u$. Show that $equiv$ is a congruence on $Mtimes S$. $M$ is a commutative monoid $M$ and $S$ is an arbitrary submonoid of $M$.
Here's what I've written so far. Since I'm dealing with monoids, I don't think I need the part that starts "And finally" (that only applies to groups). Please let me know if I'm mistaken:
We want to see that $equiv$ is a congruence. That means we need to demonstrate that $equiv$ is an equivalence relation. Then we want to see that for any $(a,s),(a',s'),(b,t),(b',t') in M times S$, if $(a,s) sim (a',s')$ and $(b,t) sim (b',t')$, then $(a,s)*(b,t) sim (a',s')*(b',t')$. And finally, we want to see that for any $(a,s),(a',s') in M times S$, if $(a,s) sim (a',s')$, then $(a,s)^{-1} sim (a',s')^{-1}$. Note that $a*t*u=b*s*u$ implies that $u in S$ is at least right cancellable. And since $S$ is a submonoid of a commutative monoid $M$, $u$ is left cancellable, thus cancellable in general.
First, we'll demonstrate that $equiv$ is an equivalence relation.
Reflexivity
Suppose that $a*s*u=a*s*u$. Then it follows immediately that $(a,s) equiv (a,s)$.
Symmetry
Suppose that $a*t*u=b*s*u$. Then $(a,s),(b,t) in M times S$, and $(a,s) equiv (b,t)$. Since $=$ is an equality, we have $(b,t) equiv (a,s)$.
Transitivity
Suppose that $a*t*u_1=b*s*u_1$ and $b*v*u_2=c*t*u_2$ where $u_1,u_2 in S$.
At this point, I'm stuck trying to show how $(a,s) equiv (c,v)$. I've tried writing this out multiplicatively, and I thought I would have to use the fact that $u$ is cancellable, but I don't know how to proceed without violating rules.
For instance, I've written on scratch paper:
$frac{a}{t}*frac{1}{u_1}=frac{b}{s}*frac{1}{u_1}$ and $frac{b}{v}*frac{1}{u_2}=frac{c}{t}*frac{1}{u_2}$. I know that I want to be able to say $frac{a}{v}*frac{1}{u_3}=frac{c}{s}*frac{1}{u_3}$, but I'm not sure how to get there. I've also written this additively but that led to a dead end as well.
Thanks ahead of time for any help / hints / clues.
abstract-algebra number-theory localization monoid congruence-relations
asked Mar 21 at 21:13
dnem41dnem41
446
$endgroup$
add a comment |
$begingroup$
I'm sorry if a similar question has been posted before, but I was unable to find one based on my searches. This is an extra practice problem for a number theory class. I've been trying to prove this out, and I'm getting stuck on the transitivity step. This is also a check on myself to make sure that I understand what I need to demonstrate in order to show a congruence.
Here is the full question. I include the last sentence for extra context:
For a submonoid $S$ of a commutative monoid $(M,*)$, define a relation $equiv$ on the monoid $Mtimes S$ by $(a,s)equiv(b,t)$ if there is a $uin S$ such that $a*t*u=b*s*u$. Show that $equiv$ is a congruence on $Mtimes S$. $M$ is a commutative monoid $M$ and $S$ is an arbitrary submonoid of $M$.
Here's what I've written so far. Since I'm dealing with monoids, I don't think I need the part that starts "And finally" (that only applies to groups). Please let me know if I'm mistaken:
We want to see that $equiv$ is a congruence. That means we need to demonstrate that $equiv$ is an equivalence relation. Then we want to see that for any $(a,s),(a',s'),(b,t),(b',t') in M times S$, if $(a,s) sim (a',s')$ and $(b,t) sim (b',t')$, then $(a,s)*(b,t) sim (a',s')*(b',t')$. And finally, we want to see that for any $(a,s),(a',s') in M times S$, if $(a,s) sim (a',s')$, then $(a,s)^{-1} sim (a',s')^{-1}$. Note that $a*t*u=b*s*u$ implies that $u in S$ is at least right cancellable. And since $S$ is a submonoid of a commutative monoid $M$, $u$ is left cancellable, thus cancellable in general.
First, we'll demonstrate that $equiv$ is an equivalence relation.
Reflexivity
Suppose that $a*s*u=a*s*u$. Then it follows immediately that $(a,s) equiv (a,s)$.
Symmetry
Suppose that $a*t*u=b*s*u$. Then $(a,s),(b,t) in M times S$, and $(a,s) equiv (b,t)$. Since $=$ is an equality, we have $(b,t) equiv (a,s)$.
Transitivity
Suppose that $a*t*u_1=b*s*u_1$ and $b*v*u_2=c*t*u_2$ where $u_1,u_2 in S$.
At this point, I'm stuck trying to show how $(a,s) equiv (c,v)$. I've tried writing this out multiplicatively, and I thought I would have to use the fact that $u$ is cancellable, but I don't know how to proceed without violating rules.
For instance, I've written on scratch paper:
$frac{a}{t}*frac{1}{u_1}=frac{b}{s}*frac{1}{u_1}$ and $frac{b}{v}*frac{1}{u_2}=frac{c}{t}*frac{1}{u_2}$. I know that I want to be able to say $frac{a}{v}*frac{1}{u_3}=frac{c}{s}*frac{1}{u_3}$, but I'm not sure how to get there. I've also written this additively but that led to a dead end as well.
Thanks ahead of time for any help / hints / clues.
abstract-algebra number-theory localization monoid congruence-relations
asked Mar 21 at 21:13
dnem41dnem41
446
$endgroup$
$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
$endgroup$
– chhro
Mar 21 at 21:29
$begingroup$
You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
$endgroup$
– dnem41
Mar 21 at 21:40
add a comment |
$begingroup$
I'm sorry if a similar question has been posted before, but I was unable to find one based on my searches. This is an extra practice problem for a number theory class. I've been trying to prove this out, and I'm getting stuck on the transitivity step. This is also a check on myself to make sure that I understand what I need to demonstrate in order to show a congruence.
Here is the full question. I include the last sentence for extra context:
For a submonoid $S$ of a commutative monoid $(M,*)$, define a relation $equiv$ on the monoid $Mtimes S$ by $(a,s)equiv(b,t)$ if there is a $uin S$ such that $a*t*u=b*s*u$. Show that $equiv$ is a congruence on $Mtimes S$. $M$ is a commutative monoid $M$ and $S$ is an arbitrary submonoid of $M$.
Here's what I've written so far. Since I'm dealing with monoids, I don't think I need the part that starts "And finally" (that only applies to groups). Please let me know if I'm mistaken:
We want to see that $equiv$ is a congruence. That means we need to demonstrate that $equiv$ is an equivalence relation. Then we want to see that for any $(a,s),(a',s'),(b,t),(b',t') in M times S$, if $(a,s) sim (a',s')$ and $(b,t) sim (b',t')$, then $(a,s)*(b,t) sim (a',s')*(b',t')$. And finally, we want to see that for any $(a,s),(a',s') in M times S$, if $(a,s) sim (a',s')$, then $(a,s)^{-1} sim (a',s')^{-1}$. Note that $a*t*u=b*s*u$ implies that $u in S$ is at least right cancellable. And since $S$ is a submonoid of a commutative monoid $M$, $u$ is left cancellable, thus cancellable in general.
First, we'll demonstrate that $equiv$ is an equivalence relation.
Reflexivity
Suppose that $a*s*u=a*s*u$. Then it follows immediately that $(a,s) equiv (a,s)$.
Symmetry
Suppose that $a*t*u=b*s*u$. Then $(a,s),(b,t) in M times S$, and $(a,s) equiv (b,t)$. Since $=$ is an equality, we have $(b,t) equiv (a,s)$.
Transitivity
Suppose that $a*t*u_1=b*s*u_1$ and $b*v*u_2=c*t*u_2$ where $u_1,u_2 in S$.
At this point, I'm stuck trying to show how $(a,s) equiv (c,v)$. I've tried writing this out multiplicatively, and I thought I would have to use the fact that $u$ is cancellable, but I don't know how to proceed without violating rules.
For instance, I've written on scratch paper:
$frac{a}{t}*frac{1}{u_1}=frac{b}{s}*frac{1}{u_1}$ and $frac{b}{v}*frac{1}{u_2}=frac{c}{t}*frac{1}{u_2}$. I know that I want to be able to say $frac{a}{v}*frac{1}{u_3}=frac{c}{s}*frac{1}{u_3}$, but I'm not sure how to get there. I've also written this additively but that led to a dead end as well.
Thanks ahead of time for any help / hints / clues.
abstract-algebra number-theory localization monoid congruence-relations
asked Mar 21 at 21:13
dnem41dnem41
446
$endgroup$
I'm sorry if a similar question has been posted before, but I was unable to find one based on my searches. This is an extra practice problem for a number theory class. I've been trying to prove this out, and I'm getting stuck on the transitivity step. This is also a check on myself to make sure that I understand what I need to demonstrate in order to show a congruence.
Here is the full question. I include the last sentence for extra context:
For a submonoid $S$ of a commutative monoid $(M,*)$, define a relation $equiv$ on the monoid $Mtimes S$ by $(a,s)equiv(b,t)$ if there is a $uin S$ such that $a*t*u=b*s*u$. Show that $equiv$ is a congruence on $Mtimes S$. $M$ is a commutative monoid $M$ and $S$ is an arbitrary submonoid of $M$.
Here's what I've written so far. Since I'm dealing with monoids, I don't think I need the part that starts "And finally" (that only applies to groups). Please let me know if I'm mistaken:
We want to see that $equiv$ is a congruence. That means we need to demonstrate that $equiv$ is an equivalence relation. Then we want to see that for any $(a,s),(a',s'),(b,t),(b',t') in M times S$, if $(a,s) sim (a',s')$ and $(b,t) sim (b',t')$, then $(a,s)*(b,t) sim (a',s')*(b',t')$. And finally, we want to see that for any $(a,s),(a',s') in M times S$, if $(a,s) sim (a',s')$, then $(a,s)^{-1} sim (a',s')^{-1}$. Note that $a*t*u=b*s*u$ implies that $u in S$ is at least right cancellable. And since $S$ is a submonoid of a commutative monoid $M$, $u$ is left cancellable, thus cancellable in general.
First, we'll demonstrate that $equiv$ is an equivalence relation.
Reflexivity
Suppose that $a*s*u=a*s*u$. Then it follows immediately that $(a,s) equiv (a,s)$.
Symmetry
Suppose that $a*t*u=b*s*u$. Then $(a,s),(b,t) in M times S$, and $(a,s) equiv (b,t)$. Since $=$ is an equality, we have $(b,t) equiv (a,s)$.
Transitivity
Suppose that $a*t*u_1=b*s*u_1$ and $b*v*u_2=c*t*u_2$ where $u_1,u_2 in S$.
At this point, I'm stuck trying to show how $(a,s) equiv (c,v)$. I've tried writing this out multiplicatively, and I thought I would have to use the fact that $u$ is cancellable, but I don't know how to proceed without violating rules.
For instance, I've written on scratch paper:
$frac{a}{t}*frac{1}{u_1}=frac{b}{s}*frac{1}{u_1}$ and $frac{b}{v}*frac{1}{u_2}=frac{c}{t}*frac{1}{u_2}$. I know that I want to be able to say $frac{a}{v}*frac{1}{u_3}=frac{c}{s}*frac{1}{u_3}$, but I'm not sure how to get there. I've also written this additively but that led to a dead end as well.
Thanks ahead of time for any help / hints / clues.
abstract-algebra number-theory localization monoid congruence-relations
abstract-algebra number-theory localization monoid congruence-relations
asked Mar 21 at 21:13
dnem41dnem41
446
asked Mar 21 at 21:13
dnem41dnem41
446
edited Mar 21 at 21:40
dnem41
asked Mar 21 at 21:13
dnem41dnem41
446
asked Mar 21 at 21:13
dnem41dnem41
446
asked Mar 21 at 21:13
dnem41dnem41
446
446
$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
$endgroup$
– chhro
Mar 21 at 21:29
$begingroup$
You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
$endgroup$
– dnem41
Mar 21 at 21:40
add a comment |
$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
$endgroup$
– chhro
Mar 21 at 21:29
$begingroup$
You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
$endgroup$
– dnem41
Mar 21 at 21:40
$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
$endgroup$
– chhro
Mar 21 at 21:29
$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
$endgroup$
– chhro
Mar 21 at 21:29
$begingroup$
You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
$endgroup$
– dnem41
Mar 21 at 21:40
$begingroup$
You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
$endgroup$
– dnem41
Mar 21 at 21:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $atu_1=bsu_1$ and $bvu_2=ctu_2$ for some $a,bin M$, $s,t,u_1,u_2in S$.
Multiplying the first equation by $u_2v$, and noting that multiplication is associative and commutative in this setting, we get
$$(av)(u_1u_2t)=atu_1u_2v=bsu_1u_2v=(bvu_2)u_1s=(ctu_2)u_1s=(cs)(u_1u_2t).$$
Observe that $u_1u_2tin S$ since $S$ is a submonoid.
answered Mar 21 at 21:33
chhrochhro
1,442311
$endgroup$
$begingroup$
Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
$endgroup$
– dnem41
Mar 21 at 21:42
$begingroup$
That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
$endgroup$
– chhro
Mar 21 at 21:43
$begingroup$
Ah that makes sense. Thanks again.
$endgroup$
– dnem41
Mar 21 at 21:44
1
$begingroup$
Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
$endgroup$
– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Let $atu_1=bsu_1$ and $bvu_2=ctu_2$ for some $a,bin M$, $s,t,u_1,u_2in S$.
Multiplying the first equation by $u_2v$, and noting that multiplication is associative and commutative in this setting, we get
$$(av)(u_1u_2t)=atu_1u_2v=bsu_1u_2v=(bvu_2)u_1s=(ctu_2)u_1s=(cs)(u_1u_2t).$$
Observe that $u_1u_2tin S$ since $S$ is a submonoid.
answered Mar 21 at 21:33
chhrochhro
1,442311
$endgroup$
$begingroup$
Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
$endgroup$
– dnem41
Mar 21 at 21:42
$begingroup$
That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
$endgroup$
– chhro
Mar 21 at 21:43
$begingroup$
Ah that makes sense. Thanks again.
$endgroup$
– dnem41
Mar 21 at 21:44
1
$begingroup$
Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
$endgroup$
– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
|
show 3 more comments
$begingroup$
Let $atu_1=bsu_1$ and $bvu_2=ctu_2$ for some $a,bin M$, $s,t,u_1,u_2in S$.
Multiplying the first equation by $u_2v$, and noting that multiplication is associative and commutative in this setting, we get
$$(av)(u_1u_2t)=atu_1u_2v=bsu_1u_2v=(bvu_2)u_1s=(ctu_2)u_1s=(cs)(u_1u_2t).$$
Observe that $u_1u_2tin S$ since $S$ is a submonoid.
answered Mar 21 at 21:33
chhrochhro
1,442311
$endgroup$
$begingroup$
Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
$endgroup$
– dnem41
Mar 21 at 21:42
$begingroup$
That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
$endgroup$
– chhro
Mar 21 at 21:43
$begingroup$
Ah that makes sense. Thanks again.
$endgroup$
– dnem41
Mar 21 at 21:44
1
$begingroup$
Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
$endgroup$
– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
|
show 3 more comments
$begingroup$
Let $atu_1=bsu_1$ and $bvu_2=ctu_2$ for some $a,bin M$, $s,t,u_1,u_2in S$.
Multiplying the first equation by $u_2v$, and noting that multiplication is associative and commutative in this setting, we get
$$(av)(u_1u_2t)=atu_1u_2v=bsu_1u_2v=(bvu_2)u_1s=(ctu_2)u_1s=(cs)(u_1u_2t).$$
Observe that $u_1u_2tin S$ since $S$ is a submonoid.
answered Mar 21 at 21:33
chhrochhro
1,442311
$endgroup$
Let $atu_1=bsu_1$ and $bvu_2=ctu_2$ for some $a,bin M$, $s,t,u_1,u_2in S$.
Multiplying the first equation by $u_2v$, and noting that multiplication is associative and commutative in this setting, we get
$$(av)(u_1u_2t)=atu_1u_2v=bsu_1u_2v=(bvu_2)u_1s=(ctu_2)u_1s=(cs)(u_1u_2t).$$
Observe that $u_1u_2tin S$ since $S$ is a submonoid.
answered Mar 21 at 21:33
chhrochhro
1,442311
answered Mar 21 at 21:33
chhrochhro
1,442311
answered Mar 21 at 21:33
chhrochhro
1,442311
answered Mar 21 at 21:33
chhrochhro
1,442311
1,442311
$begingroup$
Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
$endgroup$
– dnem41
Mar 21 at 21:42
$begingroup$
That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
$endgroup$
– chhro
Mar 21 at 21:43
$begingroup$
Ah that makes sense. Thanks again.
$endgroup$
– dnem41
Mar 21 at 21:44
1
$begingroup$
Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
$endgroup$
– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
|
show 3 more comments
$begingroup$
Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
$endgroup$
– dnem41
Mar 21 at 21:42
$begingroup$
That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
$endgroup$
– chhro
Mar 21 at 21:43
$begingroup$
Ah that makes sense. Thanks again.
$endgroup$
– dnem41
Mar 21 at 21:44
1
$begingroup$
Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
$endgroup$
– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
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– dnem41
Mar 21 at 21:47
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Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
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– dnem41
Mar 21 at 21:42
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Thanks. I think that makes sense. But why is s,t in S? I would have thought it would be in M, if it matters.
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– dnem41
Mar 21 at 21:42
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That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
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– chhro
Mar 21 at 21:43
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That's part of the assumption since $(a,s),(b,t),(c,v)in Mtimes S$.
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– chhro
Mar 21 at 21:43
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Ah that makes sense. Thanks again.
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– dnem41
Mar 21 at 21:44
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Ah that makes sense. Thanks again.
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– dnem41
Mar 21 at 21:44
1
1
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Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
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– chhro
Mar 21 at 21:46
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Happy to help. Hope it made sense as I omitted the asterisks for easy typesetting.
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– chhro
Mar 21 at 21:46
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
$begingroup$
If anything it makes more sense. I think the asterisks were throwing me off (long day).
$endgroup$
– dnem41
Mar 21 at 21:47
|
show 3 more comments
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$begingroup$
You have a typo for the transitivity relation for $(b,t)equiv (c,v)$. Should be $bvu_2=ctu_2$. And you're right, you skip the and finally part as you're dealing with monoids where there are no guarantee of inverses to exist.
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– chhro
Mar 21 at 21:29
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You're right. I had written it correctly on my scratch paper but typed it up wrong. Fixed.
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– dnem41
Mar 21 at 21:40