How do we show that the set of module homomorphisms is a commutative group?
$begingroup$
Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.
From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.
abstract-algebra abelian-groups
edited Jan 15 at 20:51
user26857
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
$endgroup$
add a comment |
$begingroup$
Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.
From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.
abstract-algebra abelian-groups
edited Jan 15 at 20:51
user26857
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
$endgroup$
2
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31
add a comment |
$begingroup$
Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.
From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.
abstract-algebra abelian-groups
edited Jan 15 at 20:51
user26857
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
$endgroup$
Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.
From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.
abstract-algebra abelian-groups
abstract-algebra abelian-groups
edited Jan 15 at 20:51
user26857
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
edited Jan 15 at 20:51
user26857
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
edited Jan 15 at 20:51
user26857
39.4k124183
edited Jan 15 at 20:51
user26857
39.4k124183
edited Jan 15 at 20:51
user26857
39.4k124183
39.4k124183
asked Jan 15 at 7:24
NawajNawaj
283
asked Jan 15 at 7:24
NawajNawaj
283
asked Jan 15 at 7:24
NawajNawaj
283
283
2
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31
add a comment |
2
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31
2
2
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So you only have problems with inverses? Then here you go:
first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
$endgroup$
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%2f3074158%2fhow-do-we-show-that-the-set-of-module-homomorphisms-is-a-commutative-group%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
$begingroup$
So you only have problems with inverses? Then here you go:
first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
$endgroup$
add a comment |
$begingroup$
So you only have problems with inverses? Then here you go:
first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
$endgroup$
add a comment |
$begingroup$
So you only have problems with inverses? Then here you go:
first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
$endgroup$
So you only have problems with inverses? Then here you go:
first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:
neutral element:
$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.
additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.
additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$
closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.
Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
answered Jan 15 at 8:22
EnkiduEnkidu
1,32619
1,32619
add a comment |
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.
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%2f3074158%2fhow-do-we-show-that-the-set-of-module-homomorphisms-is-a-commutative-group%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
2
$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31