Mixing
What's the operation on $Bbb Zleft[frac1Pright]/(xsim Px)$, and is $lvert px+prod_pin Pp^nu_p(x)rvertleqlvert xrvert$? [on hold]
Clash Royale CLAN TAG#URR8PPP
up vote
-4
down vote
favorite
This is Collatz-related although I think it stands alone as a question.
What's the operation on $X_P=Bbb Zleft[frac1Pright]/(xsim Px)$ and is $$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$?
Let $P$ be some set of primes
Then $Bbb Zleft[frac1Pright]$ is the ring generated by $leftfrac1p:pin Pright$
And let $sim$ be the transitive closure of $forall pin P:xsim px$
In other words: $xsim yiffexists z_1,z_2,ldotsinBbb Z:p_1^z_1cdot p_2^z_2cdots x=y$
Then let $Bbb Zleft[frac1Pright]/sim$ be a quotient.
I can see that it contains cosets indexed by, at most, the integers coprime to $P$.
Unless I'm mistaken, multiplicatively, it seems fairly simple that $X_P>0$ is the multiplicative group generated by the primes not in $P$. I'm not sure about the negative numbers but it seems $X_Psetminus0$ might be generated by $-1cuppnotin P$
Seen as an additive object, addition is simply addition of the unique element of $langle xrangle$ coprime to $P$, multiplicatively modulo the primes in $P$, so if $r,s$ are coprime to $P$, and if we notate $P^<omega$ as the set of all products of powers of primes in $P$, and $p,qin P^<omega$ then:
$pr+qs=(p+s)cdot P^<omega$.
I'm particularly interested in $P=2,3$. It looks like $X_2,3$ is the set $equiv-1,1pmod6$. I think the operation is simply addition, after dividing out the primes in $P$ then the result is $xP^<omega$
Let $p^nu_p(x)$ be the highest power of $p$ that divides $x$ so $2^nu_2(28)=4$
Then I have the notion that in $X_2,3$ we have the following inequality on the orders of elements of a group: $leftlvert3x+2^nu_2(x)cdot3^nu_3(x)rightrvert leqleftlvert xrightrvert$ which is equivalent to the Collatz conjecture.
The identity class would be the $3$-smooth numbers.
and possibly even the more general:
$$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$
Looks like it might be a "lift" to me but I'm still a novice in that regard. I've tagged torsion groups which is speculation more than certainty - please feel free to remove if inappropriate.
EDIT
By now we've determined that $x^2=x^3=x$ (operation-wise) and therefore $X_2,3$ can't be an additive group other than the trivial group. Furthermore, since $x^5neq x$, it's not the trivial group so it can't be an additive group at all. Therefore $lvertcdotrvert$ cannot measure an element's order in an additive group. So for the positive integers to sit in an ordered structure according to the inequality, with orders satisfying the given inequality, there would need to be some additional structure that preserves integer values such as e.g. $x^2=-x$ or the order would need to be some other algebraic ordering structure.
abstract-algebra group-theory collatz
asked Jul 26 at 19:52
Robert Frost
3,886936
put on hold as unclear what you're asking by Did, Alex Provost, Taroccoesbrocco, José Carlos Santos, stressed out Aug 8 at 22:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
 |Â
show 18 more comments
up vote
-4
down vote
favorite
This is Collatz-related although I think it stands alone as a question.
What's the operation on $X_P=Bbb Zleft[frac1Pright]/(xsim Px)$ and is $$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$?
Let $P$ be some set of primes
Then $Bbb Zleft[frac1Pright]$ is the ring generated by $leftfrac1p:pin Pright$
And let $sim$ be the transitive closure of $forall pin P:xsim px$
In other words: $xsim yiffexists z_1,z_2,ldotsinBbb Z:p_1^z_1cdot p_2^z_2cdots x=y$
Then let $Bbb Zleft[frac1Pright]/sim$ be a quotient.
I can see that it contains cosets indexed by, at most, the integers coprime to $P$.
Unless I'm mistaken, multiplicatively, it seems fairly simple that $X_P>0$ is the multiplicative group generated by the primes not in $P$. I'm not sure about the negative numbers but it seems $X_Psetminus0$ might be generated by $-1cuppnotin P$
Seen as an additive object, addition is simply addition of the unique element of $langle xrangle$ coprime to $P$, multiplicatively modulo the primes in $P$, so if $r,s$ are coprime to $P$, and if we notate $P^<omega$ as the set of all products of powers of primes in $P$, and $p,qin P^<omega$ then:
$pr+qs=(p+s)cdot P^<omega$.
I'm particularly interested in $P=2,3$. It looks like $X_2,3$ is the set $equiv-1,1pmod6$. I think the operation is simply addition, after dividing out the primes in $P$ then the result is $xP^<omega$
Let $p^nu_p(x)$ be the highest power of $p$ that divides $x$ so $2^nu_2(28)=4$
Then I have the notion that in $X_2,3$ we have the following inequality on the orders of elements of a group: $leftlvert3x+2^nu_2(x)cdot3^nu_3(x)rightrvert leqleftlvert xrightrvert$ which is equivalent to the Collatz conjecture.
The identity class would be the $3$-smooth numbers.
and possibly even the more general:
$$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$
Looks like it might be a "lift" to me but I'm still a novice in that regard. I've tagged torsion groups which is speculation more than certainty - please feel free to remove if inappropriate.
EDIT
By now we've determined that $x^2=x^3=x$ (operation-wise) and therefore $X_2,3$ can't be an additive group other than the trivial group. Furthermore, since $x^5neq x$, it's not the trivial group so it can't be an additive group at all. Therefore $lvertcdotrvert$ cannot measure an element's order in an additive group. So for the positive integers to sit in an ordered structure according to the inequality, with orders satisfying the given inequality, there would need to be some additional structure that preserves integer values such as e.g. $x^2=-x$ or the order would need to be some other algebraic ordering structure.
abstract-algebra group-theory collatz
asked Jul 26 at 19:52
Robert Frost
3,886936
put on hold as unclear what you're asking by Did, Alex Provost, Taroccoesbrocco, José Carlos Santos, stressed out Aug 8 at 22:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
2
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
1
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
1
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
7
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23
 |Â
show 18 more comments
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
This is Collatz-related although I think it stands alone as a question.
What's the operation on $X_P=Bbb Zleft[frac1Pright]/(xsim Px)$ and is $$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$?
Let $P$ be some set of primes
Then $Bbb Zleft[frac1Pright]$ is the ring generated by $leftfrac1p:pin Pright$
And let $sim$ be the transitive closure of $forall pin P:xsim px$
In other words: $xsim yiffexists z_1,z_2,ldotsinBbb Z:p_1^z_1cdot p_2^z_2cdots x=y$
Then let $Bbb Zleft[frac1Pright]/sim$ be a quotient.
I can see that it contains cosets indexed by, at most, the integers coprime to $P$.
Unless I'm mistaken, multiplicatively, it seems fairly simple that $X_P>0$ is the multiplicative group generated by the primes not in $P$. I'm not sure about the negative numbers but it seems $X_Psetminus0$ might be generated by $-1cuppnotin P$
Seen as an additive object, addition is simply addition of the unique element of $langle xrangle$ coprime to $P$, multiplicatively modulo the primes in $P$, so if $r,s$ are coprime to $P$, and if we notate $P^<omega$ as the set of all products of powers of primes in $P$, and $p,qin P^<omega$ then:
$pr+qs=(p+s)cdot P^<omega$.
I'm particularly interested in $P=2,3$. It looks like $X_2,3$ is the set $equiv-1,1pmod6$. I think the operation is simply addition, after dividing out the primes in $P$ then the result is $xP^<omega$
Let $p^nu_p(x)$ be the highest power of $p$ that divides $x$ so $2^nu_2(28)=4$
Then I have the notion that in $X_2,3$ we have the following inequality on the orders of elements of a group: $leftlvert3x+2^nu_2(x)cdot3^nu_3(x)rightrvert leqleftlvert xrightrvert$ which is equivalent to the Collatz conjecture.
The identity class would be the $3$-smooth numbers.
and possibly even the more general:
$$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$
Looks like it might be a "lift" to me but I'm still a novice in that regard. I've tagged torsion groups which is speculation more than certainty - please feel free to remove if inappropriate.
EDIT
By now we've determined that $x^2=x^3=x$ (operation-wise) and therefore $X_2,3$ can't be an additive group other than the trivial group. Furthermore, since $x^5neq x$, it's not the trivial group so it can't be an additive group at all. Therefore $lvertcdotrvert$ cannot measure an element's order in an additive group. So for the positive integers to sit in an ordered structure according to the inequality, with orders satisfying the given inequality, there would need to be some additional structure that preserves integer values such as e.g. $x^2=-x$ or the order would need to be some other algebraic ordering structure.
abstract-algebra group-theory collatz
asked Jul 26 at 19:52
Robert Frost
3,886936
This is Collatz-related although I think it stands alone as a question.
What's the operation on $X_P=Bbb Zleft[frac1Pright]/(xsim Px)$ and is $$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$?
Let $P$ be some set of primes
Then $Bbb Zleft[frac1Pright]$ is the ring generated by $leftfrac1p:pin Pright$
And let $sim$ be the transitive closure of $forall pin P:xsim px$
In other words: $xsim yiffexists z_1,z_2,ldotsinBbb Z:p_1^z_1cdot p_2^z_2cdots x=y$
Then let $Bbb Zleft[frac1Pright]/sim$ be a quotient.
I can see that it contains cosets indexed by, at most, the integers coprime to $P$.
Unless I'm mistaken, multiplicatively, it seems fairly simple that $X_P>0$ is the multiplicative group generated by the primes not in $P$. I'm not sure about the negative numbers but it seems $X_Psetminus0$ might be generated by $-1cuppnotin P$
Seen as an additive object, addition is simply addition of the unique element of $langle xrangle$ coprime to $P$, multiplicatively modulo the primes in $P$, so if $r,s$ are coprime to $P$, and if we notate $P^<omega$ as the set of all products of powers of primes in $P$, and $p,qin P^<omega$ then:
$pr+qs=(p+s)cdot P^<omega$.
I'm particularly interested in $P=2,3$. It looks like $X_2,3$ is the set $equiv-1,1pmod6$. I think the operation is simply addition, after dividing out the primes in $P$ then the result is $xP^<omega$
Let $p^nu_p(x)$ be the highest power of $p$ that divides $x$ so $2^nu_2(28)=4$
Then I have the notion that in $X_2,3$ we have the following inequality on the orders of elements of a group: $leftlvert3x+2^nu_2(x)cdot3^nu_3(x)rightrvert leqleftlvert xrightrvert$ which is equivalent to the Collatz conjecture.
The identity class would be the $3$-smooth numbers.
and possibly even the more general:
$$leftlvert px+prod_pin Pp^nu_p(x)rightrvertleq leftlvert xrightrvert$$
Looks like it might be a "lift" to me but I'm still a novice in that regard. I've tagged torsion groups which is speculation more than certainty - please feel free to remove if inappropriate.
EDIT
By now we've determined that $x^2=x^3=x$ (operation-wise) and therefore $X_2,3$ can't be an additive group other than the trivial group. Furthermore, since $x^5neq x$, it's not the trivial group so it can't be an additive group at all. Therefore $lvertcdotrvert$ cannot measure an element's order in an additive group. So for the positive integers to sit in an ordered structure according to the inequality, with orders satisfying the given inequality, there would need to be some additional structure that preserves integer values such as e.g. $x^2=-x$ or the order would need to be some other algebraic ordering structure.
abstract-algebra group-theory collatz
asked Jul 26 at 19:52
Robert Frost
3,886936
edited Jul 29 at 9:34
asked Jul 26 at 19:52
Robert Frost
3,886936
asked Jul 26 at 19:52
Robert Frost
3,886936
asked Jul 26 at 19:52
Robert Frost
3,886936
3,886936
put on hold as unclear what you're asking by Did, Alex Provost, Taroccoesbrocco, José Carlos Santos, stressed out Aug 8 at 22:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Did, Alex Provost, Taroccoesbrocco, José Carlos Santos, stressed out Aug 8 at 22:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
2
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
1
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
1
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
7
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23
 |Â
show 18 more comments
1
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
2
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
1
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
1
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
7
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23
1
1
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
2
2
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
1
1
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
1
1
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
7
7
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23
 |Â
show 18 more comments
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
FYI: $BbbQ$ has no subfields. It is a so called prime field. Did you mean the smallest ring containing $1/p$, $pin P$?
– Jyrki Lahtonen
Jul 26 at 20:06
2
@RobertFrost That's a well-defined but highly unnatural operation on equivalence classes (in general, the point of equivalence classes is that any element is as good as any other), but I'm afraid you still don't have a group (or you have a trivial one); $langle1rangle+langle1rangle=2inlangle1rangle$, so either $langle 1rangle$ is your identity (in which case I'm pretty sure it's the only element in your group) or what you've defined is not a group operation.
– Steven Stadnicki
Jul 27 at 3:51
1
You still just don't have a group operation. $langle 5rangle+langle1rangle=langle(5+1)rangle=langle1rangle$. $langle5rangle+langle7rangle=langle1rangle$, but so is $langle5rangle+langle19rangle$, etc. And you don't have associativity either; examples are left as an exercise. I don't see any sensible way that you can make these equivalence classes into a group under addition.
– Steven Stadnicki
Jul 27 at 4:37
1
And to nitpick: you probably want $BbbZ[1/P]$ to be the smalles subring containing those $1/p$, i.e. the subring generated by the reciprocals of the chosen primes. You know, you would really benefit from actually studying a basic text on structures of abstract algebra.
– Jyrki Lahtonen
Jul 27 at 4:56
7
If you can't arsed to open a book on abstract algebra, then there is no hope of you making any progress this way.
– Jyrki Lahtonen
Jul 27 at 8:23