If $fcolonmathbb{Z}to F$ is an onto morphism and $F$ is a field, then $mathbb{Z}_pcong F$ where $p$ is a...
up vote
0
down vote
favorite
Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$
Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.
On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.
Correct?
Thanks!
abstract-algebra proof-verification proof-explanation
asked Nov 20 at 9:04
Jack J.
6031319
|
show 2 more comments
up vote
0
down vote
favorite
Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$
Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.
On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.
Correct?
Thanks!
abstract-algebra proof-verification proof-explanation
asked Nov 20 at 9:04
Jack J.
6031319
1
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
1
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$
Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.
On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.
Correct?
Thanks!
abstract-algebra proof-verification proof-explanation
asked Nov 20 at 9:04
Jack J.
6031319
Let $F$ a field ed $fcolonmathbb{Z}to F$ an onto morphism. We know that $ker f$ is an ideal of $mathbb{Z}$, then $ker f=(n)=nmathbb{Z}$ for same $ninmathbb{Z}$. For the first isomorphism theorem we have that $$mathbb{Z}_n:=mathbb{Z}/nmathbb{Z}=mathbb{Z}/ker fcong F.$$
Now, if $n=0$, the canonical projection is $picolonmathbb{Z}to mathbb{Z}$, but the only non-zero morphism from $mathbb{Z}$ to $mathbb{Z}$ is $id_{mathbb{Z}}$ which is, in particular, injective. Since $ker f=nmathbb{Z}$, then $tilde{f}colonmathbb{Z}to F$ is injective, but for the first isomorphism theorem $f=tilde{f}circpi$, then $f$ is injective, moreover, for hypotesis, $f$ is onto, then $f$ is an isomorphism. But this is absurd, because $mathbb{Z}$ is not a field. Therefore $n$ can not be $0$.
On the other hand $mathbb{Z}_ncong F$, then $mathbb{Z}_n$ is a field, then $n$ must be prime.
Correct?
Thanks!
abstract-algebra proof-verification proof-explanation
abstract-algebra proof-verification proof-explanation
asked Nov 20 at 9:04
Jack J.
6031319
asked Nov 20 at 9:04
Jack J.
6031319
edited Nov 20 at 9:21
asked Nov 20 at 9:04
Jack J.
6031319
asked Nov 20 at 9:04
Jack J.
6031319
asked Nov 20 at 9:04
Jack J.
6031319
6031319
1
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
1
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35
|
show 2 more comments
1
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
1
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35
1
1
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
1
1
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.
Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:
Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
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',
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%2f3006094%2fif-f-colon-mathbbz-to-f-is-an-onto-morphism-and-f-is-a-field-then-mathb%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
up vote
1
down vote
accepted
We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.
Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:
Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
add a comment |
up vote
1
down vote
accepted
We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.
Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:
Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.
Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:
Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
We can assume $ker(f)neq (0)$, since this would imply that $Bbb Z$ is a field.
Since $Bbb Z$ is a PID, we can assume $ker(f)=(n)$, where $ninBbb Z^+$. You are ofcourse right that $Bbb Z_n$ is a field $iff n$ prime. One way to show this:
Suppose $n=ab$ so that $ker(f)=(ab)$, then $f(ab)=f(a)f(b)=0$ in $F$. Since $F$ is a field, and so an integral domain we need to have $f(a)=0$ or $f(b)=0$, which forces either $a,b$ to be a unit. This means by definition that $n$ is prime.
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
answered Nov 20 at 9:55
cansomeonehelpmeout
6,5933834
6,5933834
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
add a comment |
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
@cansomeonehelpmeoutThanks for yourn answer. So my reasoning to show that $n$ can not be zero is correct?
– Jack J.
Nov 20 at 10:14
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%2f3006094%2fif-f-colon-mathbbz-to-f-is-an-onto-morphism-and-f-is-a-field-then-mathb%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
1
I don't understand your claim. $mathbb{Z}$ is not isomorphic to a field because it is not a field: $2$ has no multiplicative inverse. And there is an onto homomorphism from $mathbb{Z}$ to a field, take the quotient map $mathbb{Z}tomathbb{Z}/pmathbb{Z}$ for prime $p$. But obviously not every field is a homomorphic image of $mathbb{Z}$ (e.g. reals).
– freakish
Nov 20 at 9:11
Sorry but I have not explained well, now I correct my question.
– Jack J.
Nov 20 at 9:18
The kernel cannot be zero, since this would imply that $2$, for exemple, has multiplicative inverse (since its image in F does). Now, by the isomorphism theorem you have $Z/nZ cong F$. Thus, $Z/nZ$ is a field. Now prove that $n$ must be prime.
– math.h
Nov 20 at 10:17
@math.h I would like to know if the way I have shown that $n$ can not be zero is correct
– Jack J.
Nov 20 at 10:23
1
Yes, it is. But I think you are "talking too much" (no offense, really) for just a simple thing. The morphism $Zrightarrow F$ cannot be injective (otherwise the morphism would be bijective), since $F$ is a field. You seem to know that $Z$ can't be isomorphic to a field, so you gotta accept this argument. As you said, the kernel should be of the form: $nZ$. As it is not zero (the kernel), then $n$ cannot be zero.
– math.h
Nov 20 at 10:35