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Homomorphic image of modular lattice is modular?
Clash Royale CLAN TAG#URR8PPP
up vote
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Let L be modular lattice, M be lattice and $f:Lto M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..
We already know that homomorphic image of lattice is lattice.
So we only want to show that if $f(a)leq f(b)$ then $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for $a,b,x in L$
Since L is modular $aleq b$ implies $a vee(xwedge b)= (avee x)wedge b$
My problem is:
I must begin with the assumption $f(a)leq f(b)$ then show $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for which I need to use $a vee(xwedge b)= (avee x)wedge b$
But $f(a)leq f(b)$ need not necessarily imply $aleq b$ since it is only homomorphism and not isomorphism.
discrete-mathematics
asked 4 hours ago
So Lo
59118
add a comment |Â
up vote
4
down vote
favorite
Let L be modular lattice, M be lattice and $f:Lto M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..
We already know that homomorphic image of lattice is lattice.
So we only want to show that if $f(a)leq f(b)$ then $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for $a,b,x in L$
Since L is modular $aleq b$ implies $a vee(xwedge b)= (avee x)wedge b$
My problem is:
I must begin with the assumption $f(a)leq f(b)$ then show $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for which I need to use $a vee(xwedge b)= (avee x)wedge b$
But $f(a)leq f(b)$ need not necessarily imply $aleq b$ since it is only homomorphism and not isomorphism.
discrete-mathematics
asked 4 hours ago
So Lo
59118
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let L be modular lattice, M be lattice and $f:Lto M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..
We already know that homomorphic image of lattice is lattice.
So we only want to show that if $f(a)leq f(b)$ then $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for $a,b,x in L$
Since L is modular $aleq b$ implies $a vee(xwedge b)= (avee x)wedge b$
My problem is:
I must begin with the assumption $f(a)leq f(b)$ then show $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for which I need to use $a vee(xwedge b)= (avee x)wedge b$
But $f(a)leq f(b)$ need not necessarily imply $aleq b$ since it is only homomorphism and not isomorphism.
discrete-mathematics
asked 4 hours ago
So Lo
59118
Let L be modular lattice, M be lattice and $f:Lto M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..
We already know that homomorphic image of lattice is lattice.
So we only want to show that if $f(a)leq f(b)$ then $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for $a,b,x in L$
Since L is modular $aleq b$ implies $a vee(xwedge b)= (avee x)wedge b$
My problem is:
I must begin with the assumption $f(a)leq f(b)$ then show $f(a) vee (f(x)wedge f(b))= (f(a)vee f(x))wedge f(b) $ for which I need to use $a vee(xwedge b)= (avee x)wedge b$
But $f(a)leq f(b)$ need not necessarily imply $aleq b$ since it is only homomorphism and not isomorphism.
discrete-mathematics
discrete-mathematics
asked 4 hours ago
So Lo
59118
asked 4 hours ago
So Lo
59118
asked 4 hours ago
So Lo
59118
asked 4 hours ago
So Lo
59118
asked 4 hours ago
So Lo
59118
59118
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
answered 2 hours ago
bof
48.1k449114
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
add a comment |Â
up vote
0
down vote
You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.
answered 3 hours ago
Matt Samuel
35.9k63462
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
answered 2 hours ago
bof
48.1k449114
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
answered 2 hours ago
bof
48.1k449114
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
answered 2 hours ago
bof
48.1k449114
A lattice is modular if and only if it satisfies the modular law:
$$ale bimplies alor(xland b)=(alor x)land b.tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(cland b)lor(xland b)=[(cland b)lor x)land b.tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $ale b$ if and only if $a=cland b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,xin L$ with $f(a)le f(b)$, we want to show that
$$f(a)lor(f(x)land f(b))=(f(a)lor f(x))land f(b).tag3$$
We don't know if $ale b$, so let $a_0=aland b$. Then $a_0le b$, so by the modular law we have
$$a_0lor(xland b)=(a_0lor x)land b.tag4$$
Moreover, $f(a_0)=f(aland b)=f(a)land f(b)=f(b)$ since $f(a)le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
answered 2 hours ago
bof
48.1k449114
edited 2 hours ago
answered 2 hours ago
bof
48.1k449114
answered 2 hours ago
bof
48.1k449114
answered 2 hours ago
bof
48.1k449114
48.1k449114
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
add a comment |Â
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
thanks. But my textbook gave only the first definition. So is it possible to show it using that? If not, it is fine too.
– So Lo
2 hours ago
1
1
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
Yes, of course. Since your textbook gives only the first definition, you need to include the proofs of (1) implies (2) and (2) implies (1) as part of your solution.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
@SoLo I edited my answer to include a simple direct argument, which does not mention any alternative definition of modular lattices.
– bof
2 hours ago
add a comment |Â
up vote
0
down vote
You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.
answered 3 hours ago
Matt Samuel
35.9k63462
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
add a comment |Â
up vote
0
down vote
You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.
answered 3 hours ago
Matt Samuel
35.9k63462
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.
answered 3 hours ago
Matt Samuel
35.9k63462
You can instead replace $b$ with $avee b$, since $f(avee b) =f(a) vee f(b) =f(b) $.
answered 3 hours ago
Matt Samuel
35.9k63462
answered 3 hours ago
Matt Samuel
35.9k63462
answered 3 hours ago
Matt Samuel
35.9k63462
answered 3 hours ago
Matt Samuel
35.9k63462
35.9k63462
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
add a comment |Â
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
I don't understand this. Can you explain.
– So Lo
3 hours ago
I don't understand this. Can you explain.
– So Lo
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
@So Your issue was that it may not be the case that $aleq b$. But $aleq avee b$ and $f(avee b)=f(b) $.
– Matt Samuel
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
Yes but I don't get to use it anywhere. $f(a) vee (f(x)wedge f(b)) = f(a)vee(f(xwedge b)) = f(avee(xwedge b))$ need not imply $f((avee x)wedge b)$
– So Lo
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
@So Everywhere you see $b$, replace it with $avee b$. As far as $f$ is concerned, they're identical, so if you need $aleq b$, replace $b$ with $avee b$.
– Matt Samuel
3 hours ago
add a comment |Â
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