$mathbb{Z_9}$ is not a subring of $mathbb{Z_{12}}$ because Choose the correct option












0














$mathbb{Z_9}$ 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







share|cite|improve this question




















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 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
    Nov 19 at 23:39








  • 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
    Nov 19 at 23:43






  • 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
    Nov 19 at 23:56






  • 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
    Nov 20 at 0:00


















0














$mathbb{Z_9}$ 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







share|cite|improve this question




















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 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
    Nov 19 at 23:39








  • 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
    Nov 19 at 23:43






  • 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
    Nov 19 at 23:56






  • 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
    Nov 20 at 0:00
















0












0








0







$mathbb{Z_9}$ 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







share|cite|improve this question















$mathbb{Z_9}$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 23:35

























asked Nov 19 at 23:29









jasmine

1,522416




1,522416








  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 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
    Nov 19 at 23:39








  • 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
    Nov 19 at 23:43






  • 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
    Nov 19 at 23:56






  • 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
    Nov 20 at 0:00
















  • 2




    Can you explain why you think that?
    – fleablood
    Nov 19 at 23:34






  • 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
    Nov 19 at 23:39








  • 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
    Nov 19 at 23:43






  • 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
    Nov 19 at 23:56






  • 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
    Nov 20 at 0:00










2




2




Can you explain why you think that?
– fleablood
Nov 19 at 23:34




Can you explain why you think that?
– fleablood
Nov 19 at 23:34




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
Nov 19 at 23:39






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
Nov 19 at 23:39






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
Nov 19 at 23:43




.... 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
Nov 19 at 23:43




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
Nov 19 at 23:56




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
Nov 19 at 23:56




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
Nov 20 at 0:00






@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
Nov 20 at 0:00












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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.






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    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.






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      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.






      share|cite|improve this answer
























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        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 23:37









        Omnomnomnom

        126k788176




        126k788176






























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