Fundamental Theorem on Homomorphisms - Application
$begingroup$
I want to show
1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.
I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?
2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.
I believe that one can solve this easily if 1.) is solved.
abstract-algebra ring-theory modules ideals
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
$endgroup$
add a comment |
$begingroup$
I want to show
1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.
I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?
2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.
I believe that one can solve this easily if 1.) is solved.
abstract-algebra ring-theory modules ideals
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
$endgroup$
add a comment |
$begingroup$
I want to show
1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.
I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?
2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.
I believe that one can solve this easily if 1.) is solved.
abstract-algebra ring-theory modules ideals
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
$endgroup$
I want to show
1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.
I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?
2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.
I believe that one can solve this easily if 1.) is solved.
abstract-algebra ring-theory modules ideals
abstract-algebra ring-theory modules ideals
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
10.6k41741
10.6k41741
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
asked Jan 3 at 15:25
KingDingelingKingDingeling
1857
1857
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
$begingroup$
If $f : Gto G'$ be a homomorphic and onto then $h : dfrac{G}{K} to G'$ is an isomorphism and onto where $K = ker f$.
edited 2 days ago
Max
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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2 Answers
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$begingroup$
Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
$begingroup$
Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
$begingroup$
Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
$$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
edited Jan 3 at 15:40
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
$begingroup$
If $f : Gto G'$ be a homomorphic and onto then $h : dfrac{G}{K} to G'$ is an isomorphism and onto where $K = ker f$.
edited 2 days ago
Max
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $f : Gto G'$ be a homomorphic and onto then $h : dfrac{G}{K} to G'$ is an isomorphism and onto where $K = ker f$.
edited 2 days ago
Max
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If $f : Gto G'$ be a homomorphic and onto then $h : dfrac{G}{K} to G'$ is an isomorphism and onto where $K = ker f$.
edited 2 days ago
Max
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If $f : Gto G'$ be a homomorphic and onto then $h : dfrac{G}{K} to G'$ is an isomorphism and onto where $K = ker f$.
edited 2 days ago
Max
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Max
9771319
edited 2 days ago
Max
9771319
edited 2 days ago
Max
9771319
9771319
answered 2 days ago
Yamini Singh Yamini Singh
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Yamini Singh Yamini Singh
1
answered 2 days ago
Yamini Singh Yamini Singh
1
1
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yamini Singh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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