Fundamental Theorem on Homomorphisms - Application
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
add a comment |
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
add a comment |
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
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 2 days ago
Antonios-Alexandros Robotis
9,40741640
9,40741640
asked 2 days ago
KingDingeling
505
505
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060672%2ffundamental-theorem-on-homomorphisms-application%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
add a comment |
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 $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
add a comment |
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 $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
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 $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$
edited 2 days ago
answered 2 days ago
Antonios-Alexandros Robotis
9,40741640
9,40741640
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
add a comment |
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
Thank you for your answer and for taking the time :)
– KingDingeling
yesterday
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060672%2ffundamental-theorem-on-homomorphisms-application%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown