My question is that why converse need not be true? [closed]












0












$begingroup$


The homomorphic image of ring with unity is a ring with unity but converse need not be true .



My question is that why converse need not be true ?



I mean



If $R '$ is a homomorphics image of a ring $R$ where $R'$ is a ring with unity then$ R$ need not have unity



As i know that The homomorphic image of ring with unity is a ring with unity take $f : mathbb{Z_2} rightarrow mathbb{Z_2}$



Any hints/solution



thanks u






abstract-algebra







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Jyrki Lahtonen, KReiser, Arnaud D., José Carlos Santos, Cesareo Jan 11 at 11:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    The converse is what?
    $endgroup$
    – Wuestenfux
    Jan 9 at 10:53






  • 1




    $begingroup$
    Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
    $endgroup$
    – Tobias Kildetoft
    Jan 9 at 10:53






  • 1




    $begingroup$
    There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
    $endgroup$
    – Arthur
    Jan 9 at 10:57












  • $begingroup$
    @TobiasKildetoft i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58










  • $begingroup$
    @Wuestenfux i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58
















0












$begingroup$


The homomorphic image of ring with unity is a ring with unity but converse need not be true .



My question is that why converse need not be true ?



I mean



If $R '$ is a homomorphics image of a ring $R$ where $R'$ is a ring with unity then$ R$ need not have unity



As i know that The homomorphic image of ring with unity is a ring with unity take $f : mathbb{Z_2} rightarrow mathbb{Z_2}$



Any hints/solution



thanks u






abstract-algebra







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Jyrki Lahtonen, KReiser, Arnaud D., José Carlos Santos, Cesareo Jan 11 at 11:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    The converse is what?
    $endgroup$
    – Wuestenfux
    Jan 9 at 10:53






  • 1




    $begingroup$
    Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
    $endgroup$
    – Tobias Kildetoft
    Jan 9 at 10:53






  • 1




    $begingroup$
    There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
    $endgroup$
    – Arthur
    Jan 9 at 10:57












  • $begingroup$
    @TobiasKildetoft i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58










  • $begingroup$
    @Wuestenfux i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58














0












0








0





$begingroup$


The homomorphic image of ring with unity is a ring with unity but converse need not be true .



My question is that why converse need not be true ?



I mean



If $R '$ is a homomorphics image of a ring $R$ where $R'$ is a ring with unity then$ R$ need not have unity



As i know that The homomorphic image of ring with unity is a ring with unity take $f : mathbb{Z_2} rightarrow mathbb{Z_2}$



Any hints/solution



thanks u






abstract-algebra







share|cite|improve this question











$endgroup$




The homomorphic image of ring with unity is a ring with unity but converse need not be true .



My question is that why converse need not be true ?



I mean



If $R '$ is a homomorphics image of a ring $R$ where $R'$ is a ring with unity then$ R$ need not have unity



As i know that The homomorphic image of ring with unity is a ring with unity take $f : mathbb{Z_2} rightarrow mathbb{Z_2}$



Any hints/solution



thanks u






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 Jan 9 at 10:57







jasmine

















asked Jan 9 at 10:46









jasminejasmine

1,974420




1,974420




closed as unclear what you're asking by Jyrki Lahtonen, KReiser, Arnaud D., José Carlos Santos, Cesareo Jan 11 at 11:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Jyrki Lahtonen, KReiser, Arnaud D., José Carlos Santos, Cesareo Jan 11 at 11:27


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $begingroup$
    The converse is what?
    $endgroup$
    – Wuestenfux
    Jan 9 at 10:53






  • 1




    $begingroup$
    Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
    $endgroup$
    – Tobias Kildetoft
    Jan 9 at 10:53






  • 1




    $begingroup$
    There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
    $endgroup$
    – Arthur
    Jan 9 at 10:57












  • $begingroup$
    @TobiasKildetoft i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58










  • $begingroup$
    @Wuestenfux i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58














  • 1




    $begingroup$
    The converse is what?
    $endgroup$
    – Wuestenfux
    Jan 9 at 10:53






  • 1




    $begingroup$
    Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
    $endgroup$
    – Tobias Kildetoft
    Jan 9 at 10:53






  • 1




    $begingroup$
    There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
    $endgroup$
    – Arthur
    Jan 9 at 10:57












  • $begingroup$
    @TobiasKildetoft i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58










  • $begingroup$
    @Wuestenfux i have edit its
    $endgroup$
    – jasmine
    Jan 9 at 10:58








1




1




$begingroup$
The converse is what?
$endgroup$
– Wuestenfux
Jan 9 at 10:53




$begingroup$
The converse is what?
$endgroup$
– Wuestenfux
Jan 9 at 10:53




1




1




$begingroup$
Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
$endgroup$
– Tobias Kildetoft
Jan 9 at 10:53




$begingroup$
Could you clarify what you mean by the converse? There are a couple of different ways to state this as an implication, but these statements do not have the same converses.
$endgroup$
– Tobias Kildetoft
Jan 9 at 10:53




1




1




$begingroup$
There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
$endgroup$
– Arthur
Jan 9 at 10:57






$begingroup$
There are many things to convert here. For instance, do you mean "given two rings with unity and a function between them, then even if the image is a ring the function isn't necessarily a homomorphism"?
$endgroup$
– Arthur
Jan 9 at 10:57














$begingroup$
@TobiasKildetoft i have edit its
$endgroup$
– jasmine
Jan 9 at 10:58




$begingroup$
@TobiasKildetoft i have edit its
$endgroup$
– jasmine
Jan 9 at 10:58












$begingroup$
@Wuestenfux i have edit its
$endgroup$
– jasmine
Jan 9 at 10:58




$begingroup$
@Wuestenfux i have edit its
$endgroup$
– jasmine
Jan 9 at 10:58










1 Answer
1






active

oldest

votes


















2












$begingroup$

Consider the homomorphism $f:Bbb Zto Bbb Ztimes2Bbb Z$ given by $f(n)=(n,0)$. The image of $f$ is a unital ring (being isomorphic to $Bbb Z$), but $Bbb Ztimes 2Bbb Z$ is not unital.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    superb answer i gots its
    $endgroup$
    – jasmine
    Jan 9 at 11:02


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Consider the homomorphism $f:Bbb Zto Bbb Ztimes2Bbb Z$ given by $f(n)=(n,0)$. The image of $f$ is a unital ring (being isomorphic to $Bbb Z$), but $Bbb Ztimes 2Bbb Z$ is not unital.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    superb answer i gots its
    $endgroup$
    – jasmine
    Jan 9 at 11:02
















2












$begingroup$

Consider the homomorphism $f:Bbb Zto Bbb Ztimes2Bbb Z$ given by $f(n)=(n,0)$. The image of $f$ is a unital ring (being isomorphic to $Bbb Z$), but $Bbb Ztimes 2Bbb Z$ is not unital.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    superb answer i gots its
    $endgroup$
    – jasmine
    Jan 9 at 11:02














2












2








2





$begingroup$

Consider the homomorphism $f:Bbb Zto Bbb Ztimes2Bbb Z$ given by $f(n)=(n,0)$. The image of $f$ is a unital ring (being isomorphic to $Bbb Z$), but $Bbb Ztimes 2Bbb Z$ is not unital.






share|cite|improve this answer









$endgroup$



Consider the homomorphism $f:Bbb Zto Bbb Ztimes2Bbb Z$ given by $f(n)=(n,0)$. The image of $f$ is a unital ring (being isomorphic to $Bbb Z$), but $Bbb Ztimes 2Bbb Z$ is not unital.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 9 at 11:00









ArthurArthur

123k7122211




123k7122211












  • $begingroup$
    superb answer i gots its
    $endgroup$
    – jasmine
    Jan 9 at 11:02


















  • $begingroup$
    superb answer i gots its
    $endgroup$
    – jasmine
    Jan 9 at 11:02
















$begingroup$
superb answer i gots its
$endgroup$
– jasmine
Jan 9 at 11:02




$begingroup$
superb answer i gots its
$endgroup$
– jasmine
Jan 9 at 11:02



-

Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Image
0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
Read more

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
Read more