Mixing
$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because Choose the correct option
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$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because
Choose the correct option
$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$
$b)$GCD$(9,12) = 3neq 1$
$c)$ $9$ doesnot divide $12$
I thinks option $c)$ will be correct by Lagrange Theorem
abstract-algebra
asked 2 days ago
jasmine
1,304416
|
show 4 more comments
up vote
0
down vote
favorite
$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because
Choose the correct option
$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$
$b)$GCD$(9,12) = 3neq 1$
$c)$ $9$ doesnot divide $12$
I thinks option $c)$ will be correct by Lagrange Theorem
abstract-algebra
asked 2 days ago
jasmine
1,304416
2
Can you explain why you think that?
– fleablood
2 days ago
1
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
1
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
1
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
1
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because
Choose the correct option
$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$
$b)$GCD$(9,12) = 3neq 1$
$c)$ $9$ doesnot divide $12$
I thinks option $c)$ will be correct by Lagrange Theorem
abstract-algebra
asked 2 days ago
jasmine
1,304416
$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because
Choose the correct option
$a)$ $mathbb{Z_9}$ is not a subset of $mathbb{Z_{12}}$
$b)$GCD$(9,12) = 3neq 1$
$c)$ $9$ doesnot divide $12$
I thinks option $c)$ will be correct by Lagrange Theorem
abstract-algebra
abstract-algebra
asked 2 days ago
jasmine
1,304416
asked 2 days ago
jasmine
1,304416
edited 2 days ago
asked 2 days ago
jasmine
1,304416
asked 2 days ago
jasmine
1,304416
asked 2 days ago
jasmine
1,304416
1,304416
2
Can you explain why you think that?
– fleablood
2 days ago
1
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
1
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
1
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
1
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago
|
show 4 more comments
2
Can you explain why you think that?
– fleablood
2 days ago
1
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
1
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
1
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
1
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago
2
2
Can you explain why you think that?
– fleablood
2 days ago
Can you explain why you think that?
– fleablood
2 days ago
1
1
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
1
1
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
1
1
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
1
1
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
up vote
1
down vote
accepted
Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.
answered 2 days ago
Omnomnomnom
124k788176
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.
answered 2 days ago
Omnomnomnom
124k788176
Your choice and justification are fine. In particular: in order for $Bbb Z_9$ to be a subring of $Bbb Z_{12}$, the group $(Bbb Z_9,+)$ would have to be a subgroup of $(Bbb Z_{12},+)$. However, Lagrange's theorem says that since $9$ does not divide $12$, this cannot be the case.
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
answered 2 days ago
Omnomnomnom
124k788176
124k788176
add a comment |
add a comment |
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2
Can you explain why you think that?
– fleablood
2 days ago
1
So Legrange Theorem says the order of every subgroup of a finite group divides the order of a group. And as rings are groups (under their primary operation (in this case addition)) that would apply to rings as well. And the order of $mathbb Z_{12}$ is $12$ and the order of $mathbb Z_9$ is $9$ and $9not mid 12$. Yep.... seems like you've figured it out.
– fleablood
2 days ago
1
.... and a) is obviously false as $Z_9 subset Z_{12}$ and b) is not a requirement of subrings. In fact by legrange the gcd of (non trivial) subrings must never be $1$.
– fleablood
2 days ago
1
I don't really have any more to add to i707107's hint. Need to show it is closed undo addition and multiplication, and that the identitys of C[x] are in C[x^5] and that every element has an additive inverse.
– fleablood
2 days ago
1
@fleablood (a) isn't obviously false as you have suggested. It depends on what definition of $Bbb{Z}_{12}$ and $Bbb{Z}_{9}$ is one using. If you consider $Bbb{Z}_{9}={[0]_9, [1]_9, ldots, [8]_9}$ and $Bbb{Z}_{12}={[0]_{12}, [1]_{12}, ldots, [11]_{12}}$ as quotients of $Bbb{Z}$ by the relevant congruence relation, then $Bbb{Z}_{9}$ isn't a subset of $Bbb{Z}_{12}$.
– Anurag A
2 days ago