Transfer Between LCM, GCD for Rings?
$begingroup$
I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.
abstract-algebra ring-theory
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
$endgroup$
add a comment |
$begingroup$
I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.
abstract-algebra ring-theory
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
$endgroup$
add a comment |
$begingroup$
I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.
abstract-algebra ring-theory
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
$endgroup$
I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
edited Mar 18 '17 at 0:33
Jacob Wakem
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
asked Mar 19 '14 at 3:44
Jacob WakemJacob Wakem
1,910621
1,910621
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus
$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
color{#c00}iff& rm a',b'mid c'\[2px]
iff & rm lcm(a',b')mid c'\
color{#c00}iff & rm cmid lcm(a',b')'\
{rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
end{align}quad $$
The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.
Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
add a comment |
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$begingroup$
A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus
$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
color{#c00}iff& rm a',b'mid c'\[2px]
iff & rm lcm(a',b')mid c'\
color{#c00}iff & rm cmid lcm(a',b')'\
{rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
end{align}quad $$
The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.
Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
add a comment |
$begingroup$
A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus
$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
color{#c00}iff& rm a',b'mid c'\[2px]
iff & rm lcm(a',b')mid c'\
color{#c00}iff & rm cmid lcm(a',b')'\
{rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
end{align}quad $$
The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.
Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
add a comment |
$begingroup$
A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus
$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
color{#c00}iff& rm a',b'mid c'\[2px]
iff & rm lcm(a',b')mid c'\
color{#c00}iff & rm cmid lcm(a',b')'\
{rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
end{align}quad $$
The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.
Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
$endgroup$
A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus
$$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
color{#c00}iff& rm a',b'mid c'\[2px]
iff & rm lcm(a',b')mid c'\
color{#c00}iff & rm cmid lcm(a',b')'\
{rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
end{align}quad $$
The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.
Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
edited Dec 6 '18 at 20:49
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
answered Mar 19 '14 at 4:06
Bill DubuqueBill Dubuque
212k29195650
212k29195650
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
add a comment |
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 16:34
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
$endgroup$
– Bill Dubuque
Mar 19 '14 at 16:48
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
$endgroup$
– Jacob Wakem
Mar 19 '14 at 17:03
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
@Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
$endgroup$
– Bill Dubuque
Mar 19 '14 at 18:25
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
$begingroup$
See also this answer.
$endgroup$
– Bill Dubuque
Sep 6 '14 at 5:22
add a comment |
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