Mixing
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
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
$endgroup$
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
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
$endgroup$
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
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
$endgroup$
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
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
9,75241640
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
9,75241640
edited Jan 3 at 15:39
Antonios-Alexandros Robotis
9,75241640
9,75241640
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
asked Jan 3 at 15:25
KingDingelingKingDingeling
1366
1366
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
9,75241640
$endgroup$
$begingroup$
Thank you for your answer and for taking the time :)
$endgroup$
– KingDingeling
Jan 3 at 16:02
add a comment |
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1 Answer
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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
9,75241640
$endgroup$
$begingroup$
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
9,75241640
$endgroup$
$begingroup$
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
9,75241640
$endgroup$
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
9,75241640
edited Jan 3 at 15:40
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
answered Jan 3 at 15:34
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,75241640
9,75241640
$begingroup$
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 |
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