Mixing
If $V$ is free, show that $f$ is surjective.
$begingroup$
Let $R$ be a commutative ring with $1$. Let $V$ and $W$ be $R$-modules.
a) Exhibit a canonical $R$-linear map $f: V^* otimes V to R $
b) If $V$ is free, show that $f$ is surjective.
Now for the first part $f:V^* otimes V to R$ would be $f(beta otimes v)=beta(v)$ $forall beta otimes v in V^* otimes V$
How can I prove the second part?
modules manifolds homological-algebra tensor-products free-modules
asked Jan 26 at 4:59
GimgimGimgim
29813
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring with $1$. Let $V$ and $W$ be $R$-modules.
a) Exhibit a canonical $R$-linear map $f: V^* otimes V to R $
b) If $V$ is free, show that $f$ is surjective.
Now for the first part $f:V^* otimes V to R$ would be $f(beta otimes v)=beta(v)$ $forall beta otimes v in V^* otimes V$
How can I prove the second part?
modules manifolds homological-algebra tensor-products free-modules
asked Jan 26 at 4:59
GimgimGimgim
29813
$endgroup$
add a comment |
$begingroup$
Let $R$ be a commutative ring with $1$. Let $V$ and $W$ be $R$-modules.
a) Exhibit a canonical $R$-linear map $f: V^* otimes V to R $
b) If $V$ is free, show that $f$ is surjective.
Now for the first part $f:V^* otimes V to R$ would be $f(beta otimes v)=beta(v)$ $forall beta otimes v in V^* otimes V$
How can I prove the second part?
modules manifolds homological-algebra tensor-products free-modules
asked Jan 26 at 4:59
GimgimGimgim
29813
$endgroup$
Let $R$ be a commutative ring with $1$. Let $V$ and $W$ be $R$-modules.
a) Exhibit a canonical $R$-linear map $f: V^* otimes V to R $
b) If $V$ is free, show that $f$ is surjective.
Now for the first part $f:V^* otimes V to R$ would be $f(beta otimes v)=beta(v)$ $forall beta otimes v in V^* otimes V$
How can I prove the second part?
modules manifolds homological-algebra tensor-products free-modules
modules manifolds homological-algebra tensor-products free-modules
asked Jan 26 at 4:59
GimgimGimgim
29813
asked Jan 26 at 4:59
GimgimGimgim
29813
asked Jan 26 at 4:59
GimgimGimgim
29813
asked Jan 26 at 4:59
GimgimGimgim
29813
asked Jan 26 at 4:59
GimgimGimgim
29813
29813
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is true if $V=Rcoprod V_1$,consider the $R$-homomorphism $alpha:Vrightarrow R$ which maps $(a,b)$ to $a$.then $f(alphaotimes (1,0))=1$.
Hence your question is a special case.
answered Jan 26 at 7:17
SkySky
1,243312
$endgroup$
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
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%2f3087916%2fif-v-is-free-show-that-f-is-surjective%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
$begingroup$
This is true if $V=Rcoprod V_1$,consider the $R$-homomorphism $alpha:Vrightarrow R$ which maps $(a,b)$ to $a$.then $f(alphaotimes (1,0))=1$.
Hence your question is a special case.
answered Jan 26 at 7:17
SkySky
1,243312
$endgroup$
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
add a comment |
$begingroup$
This is true if $V=Rcoprod V_1$,consider the $R$-homomorphism $alpha:Vrightarrow R$ which maps $(a,b)$ to $a$.then $f(alphaotimes (1,0))=1$.
Hence your question is a special case.
answered Jan 26 at 7:17
SkySky
1,243312
$endgroup$
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
add a comment |
$begingroup$
This is true if $V=Rcoprod V_1$,consider the $R$-homomorphism $alpha:Vrightarrow R$ which maps $(a,b)$ to $a$.then $f(alphaotimes (1,0))=1$.
Hence your question is a special case.
answered Jan 26 at 7:17
SkySky
1,243312
$endgroup$
This is true if $V=Rcoprod V_1$,consider the $R$-homomorphism $alpha:Vrightarrow R$ which maps $(a,b)$ to $a$.then $f(alphaotimes (1,0))=1$.
Hence your question is a special case.
answered Jan 26 at 7:17
SkySky
1,243312
answered Jan 26 at 7:17
SkySky
1,243312
answered Jan 26 at 7:17
SkySky
1,243312
answered Jan 26 at 7:17
SkySky
1,243312
1,243312
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
add a comment |
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 8:19
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
And what is the notation $V=Rcoprod V_1$? Is it direct product?
$endgroup$
– Gimgim
Jan 26 at 8:21
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
@Gimgim yes,that is direct product.
$endgroup$
– Sky
Jan 26 at 10:16
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
Otherwise, what is a counter example?
$endgroup$
– Gimgim
Jan 26 at 11:20
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
$begingroup$
@Gimgim Let $R=mathbb{Z}$ and $V=mathbb{Z}/nmathbb{Z}$. Then $V^*=Hom(mathbb{Z}/nmathbb{Z},mathbb{Z})=0$ so $V^*otimes Vto mathbb{Z}$ is not surjective.
$endgroup$
– Roland
Jan 26 at 12:46
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%2f3087916%2fif-v-is-free-show-that-f-is-surjective%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
