Why is $f(x) = 5x$ not a homomorphism?
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
add a comment |
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
0
0
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
edited Dec 13 '18 at 21:52
Henrik
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
edited Dec 13 '18 at 21:52
Henrik
6,04592030
edited Dec 13 '18 at 21:52
Henrik
6,04592030
edited Dec 13 '18 at 21:52
Henrik
6,04592030
6,04592030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
asked Dec 13 '18 at 21:48
imariptideimariptide
72
asked Dec 13 '18 at 21:48
imariptideimariptide
72
72
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
1
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
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
});
}
});
draft saved
draft discarded
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%2f3038607%2fwhy-is-fx-5x-not-a-homomorphism%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
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
3
3
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
$endgroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
answered Dec 13 '18 at 21:52
user289143user289143
1,037313
1,037313
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
1
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3038607%2fwhy-is-fx-5x-not-a-homomorphism%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
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17