Problem reagarding the annihilator of a $mathbb Z$-module The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Annihilator of an ideal in an R-moduleShow that $mathbbQ^+/mathbbZ^+$ cannot be decomposed into the direct sum of cyclic groups.number of subgroups of $mathbbZ_5timesmathbbZ_5$Questions of a completely reducible moduleDescribe all the one dimensional complex representation of the cyclic group $C_n$ .Annihilator - Product of cyclic groupspossible cyclic group from fundamental theorem of finite abelianSimple $F[T]$ module whose annihilator ideal is generated by an irreducible polynomialDescribe the following group: $mathbbZ_60^x$, as a direct product of cyclic groups of prime power orderIs Ring $mathbb C[x]/(x^2+1) cong mathbb C$
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Problem reagarding the annihilator of a $mathbb Z$-module
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Annihilator of an ideal in an R-moduleShow that $mathbbQ^+/mathbbZ^+$ cannot be decomposed into the direct sum of cyclic groups.number of subgroups of $mathbbZ_5timesmathbbZ_5$Questions of a completely reducible moduleDescribe all the one dimensional complex representation of the cyclic group $C_n$ .Annihilator - Product of cyclic groupspossible cyclic group from fundamental theorem of finite abelianSimple $F[T]$ module whose annihilator ideal is generated by an irreducible polynomialDescribe the following group: $mathbbZ_60^x$, as a direct product of cyclic groups of prime power orderIs Ring $mathbb C[x]/(x^2+1) cong mathbb C$
$begingroup$
Let $M=mathbb Z/24mathbb Ztimes mathbb Z/15mathbb Ztimes mathbb Z/50mathbb Z$.
a) Find the annihilator of $M$ in $mathbb Z$.
b) Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
Annihilator of $M$ is $rin mathbb Z$.
My attempt:
a) Annihilator of $M=dmathbb Z$ where $d=gcd(24,50,15)$.
b) Actually I do not get this question. I would be thankful if someone help me. Please give me hint.
Any help will be appreciated.
abstract-algebra ring-theory modules
edited Apr 8 at 21:58
user26857
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
$endgroup$
add a comment |
$begingroup$
Let $M=mathbb Z/24mathbb Ztimes mathbb Z/15mathbb Ztimes mathbb Z/50mathbb Z$.
a) Find the annihilator of $M$ in $mathbb Z$.
b) Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
Annihilator of $M$ is $rin mathbb Z$.
My attempt:
a) Annihilator of $M=dmathbb Z$ where $d=gcd(24,50,15)$.
b) Actually I do not get this question. I would be thankful if someone help me. Please give me hint.
Any help will be appreciated.
abstract-algebra ring-theory modules
edited Apr 8 at 21:58
user26857
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
$endgroup$
add a comment |
$begingroup$
Let $M=mathbb Z/24mathbb Ztimes mathbb Z/15mathbb Ztimes mathbb Z/50mathbb Z$.
a) Find the annihilator of $M$ in $mathbb Z$.
b) Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
Annihilator of $M$ is $rin mathbb Z$.
My attempt:
a) Annihilator of $M=dmathbb Z$ where $d=gcd(24,50,15)$.
b) Actually I do not get this question. I would be thankful if someone help me. Please give me hint.
Any help will be appreciated.
abstract-algebra ring-theory modules
edited Apr 8 at 21:58
user26857
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
$endgroup$
Let $M=mathbb Z/24mathbb Ztimes mathbb Z/15mathbb Ztimes mathbb Z/50mathbb Z$.
a) Find the annihilator of $M$ in $mathbb Z$.
b) Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
Annihilator of $M$ is $rin mathbb Z$.
My attempt:
a) Annihilator of $M=dmathbb Z$ where $d=gcd(24,50,15)$.
b) Actually I do not get this question. I would be thankful if someone help me. Please give me hint.
Any help will be appreciated.
abstract-algebra ring-theory modules
abstract-algebra ring-theory modules
edited Apr 8 at 21:58
user26857
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
edited Apr 8 at 21:58
user26857
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
edited Apr 8 at 21:58
user26857
39.5k124284
edited Apr 8 at 21:58
user26857
39.5k124284
edited Apr 8 at 21:58
user26857
39.5k124284
39.5k124284
asked Apr 8 at 13:49
MathLoverMathLover
58711
asked Apr 8 at 13:49
MathLoverMathLover
58711
asked Apr 8 at 13:49
MathLoverMathLover
58711
58711
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
a) By your work, $gcd(24,50,15)=1$ and the annihilator is $1mathbb Z=mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.
b)
Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
By definition, the annihilator of $I$ in $M$ is $min Mmid Im=0$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.
Good luck.
edited Apr 8 at 21:59
user26857
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
add a comment |
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$begingroup$
a) By your work, $gcd(24,50,15)=1$ and the annihilator is $1mathbb Z=mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.
b)
Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
By definition, the annihilator of $I$ in $M$ is $min Mmid Im=0$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.
Good luck.
edited Apr 8 at 21:59
user26857
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
add a comment |
$begingroup$
a) By your work, $gcd(24,50,15)=1$ and the annihilator is $1mathbb Z=mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.
b)
Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
By definition, the annihilator of $I$ in $M$ is $min Mmid Im=0$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.
Good luck.
edited Apr 8 at 21:59
user26857
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
$endgroup$
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
add a comment |
$begingroup$
a) By your work, $gcd(24,50,15)=1$ and the annihilator is $1mathbb Z=mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.
b)
Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
By definition, the annihilator of $I$ in $M$ is $min Mmid Im=0$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.
Good luck.
edited Apr 8 at 21:59
user26857
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
$endgroup$
a) By your work, $gcd(24,50,15)=1$ and the annihilator is $1mathbb Z=mathbb Z$. Does that sound right? Perhaps instead of fishing for an answer that sounds plausible, you should think of at least one element you know can annihilate $M$, and then reconsider what you've written.
b)
Let $I=2mathbb Z$. Describe the annihilator of $I$ in $M$ as direct product of cyclic groups.
By definition, the annihilator of $I$ in $M$ is $min Mmid Im=0$. This is what you should be looking for. Locate these elements and see if you can decompose them as a product of cyclic groups. Knowing it is a subgroup of $M$ should help you a lot.
Good luck.
edited Apr 8 at 21:59
user26857
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
edited Apr 8 at 21:59
user26857
39.5k124284
edited Apr 8 at 21:59
user26857
39.5k124284
edited Apr 8 at 21:59
user26857
39.5k124284
39.5k124284
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
answered Apr 8 at 15:18
rschwiebrschwieb
108k12104253
108k12104253
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
add a comment |
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
Thanks a lot Sir. Actually I wanted to say LCM instead of gcd . and for second $12,0times 0times 25,0$ which is isomorphic to $Z_2times 0times Z_2$. is I am correct?
$endgroup$
– MathLover
Apr 9 at 11:56
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
$begingroup$
@MathLover Yes, clearly anything in $(lcm(...))$ annihilates $M$, and hopefully you've taken steps to show the reverse containment too. I agree with your solution to the second part as well. Good work! Sorry if I sounded curt earlier... rereading it makes me cringe a little.
$endgroup$
– rschwieb
Apr 9 at 13:53
add a comment |
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