Proving a ring-homomorphism using a group-homomorphism
Let f : R → R' be a group homomorphism. Show that the induced map φ : R[x] → R'[x], where φ(anxn + . . . + a0) = f(an)xn + . . . + f(a0), is a ring homomorphism.
I know that φ(0) = f(0) = 0 since f is a group homomorphism. I also know how to show the additive and multiplicative properties for the ring homomorphism, but how can I prove that φ(1) = f(1) = 1?
Also, we're only using commutative rings in my class; I forgot to specify that because I never have to in my work.
I believe it's safe to assume that R, R', R[x], and R'[x] are all rings.
group-theory ring-theory group-homomorphism ring-homomorphism
asked Nov 29 at 21:14
Jon D.
164
|
show 4 more comments
Let f : R → R' be a group homomorphism. Show that the induced map φ : R[x] → R'[x], where φ(anxn + . . . + a0) = f(an)xn + . . . + f(a0), is a ring homomorphism.
I know that φ(0) = f(0) = 0 since f is a group homomorphism. I also know how to show the additive and multiplicative properties for the ring homomorphism, but how can I prove that φ(1) = f(1) = 1?
Also, we're only using commutative rings in my class; I forgot to specify that because I never have to in my work.
I believe it's safe to assume that R, R', R[x], and R'[x] are all rings.
group-theory ring-theory group-homomorphism ring-homomorphism
asked Nov 29 at 21:14
Jon D.
164
2
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
1
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
1
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13
|
show 4 more comments
Let f : R → R' be a group homomorphism. Show that the induced map φ : R[x] → R'[x], where φ(anxn + . . . + a0) = f(an)xn + . . . + f(a0), is a ring homomorphism.
I know that φ(0) = f(0) = 0 since f is a group homomorphism. I also know how to show the additive and multiplicative properties for the ring homomorphism, but how can I prove that φ(1) = f(1) = 1?
Also, we're only using commutative rings in my class; I forgot to specify that because I never have to in my work.
I believe it's safe to assume that R, R', R[x], and R'[x] are all rings.
group-theory ring-theory group-homomorphism ring-homomorphism
asked Nov 29 at 21:14
Jon D.
164
Let f : R → R' be a group homomorphism. Show that the induced map φ : R[x] → R'[x], where φ(anxn + . . . + a0) = f(an)xn + . . . + f(a0), is a ring homomorphism.
I know that φ(0) = f(0) = 0 since f is a group homomorphism. I also know how to show the additive and multiplicative properties for the ring homomorphism, but how can I prove that φ(1) = f(1) = 1?
Also, we're only using commutative rings in my class; I forgot to specify that because I never have to in my work.
I believe it's safe to assume that R, R', R[x], and R'[x] are all rings.
group-theory ring-theory group-homomorphism ring-homomorphism
group-theory ring-theory group-homomorphism ring-homomorphism
asked Nov 29 at 21:14
Jon D.
164
asked Nov 29 at 21:14
Jon D.
164
edited Nov 29 at 21:39
asked Nov 29 at 21:14
Jon D.
164
asked Nov 29 at 21:14
Jon D.
164
asked Nov 29 at 21:14
Jon D.
164
164
2
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
1
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
1
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13
|
show 4 more comments
2
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
1
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
1
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13
2
2
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
1
1
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
1
1
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13
|
show 4 more comments
2 Answers
2
active
oldest
votes
To prove that $varphi: R[x]rightarrow R'[x]$, is a ring homomorphism you have to show that: $$ varphi(f+g)=varphi(f)+varphi(g), qquad varphi(fg)=varphi(f)varphi(g) qquad text{and} qquad varphi(1)=1.$$ Note that you have suppose that $f: Rrightarrow R'$ is a ring homomorphis.
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
add a comment |
It should be: "$f:Rto R'$ is a ring homomorphism". Otherwise this is not true. Indeed, if $f$ is not a ring homomorphism then $f(ab)neq f(a)f(b)$ for some $a,bin R$. It is clear that $varphi(ab)neqvarphi(a)varphi(b)$ as well where $a,b$ are now treated as polynomials of degree $0$. Note that for polynomial $r$ of degree $0$ we have $varphi(r)=f(r)$.
As an example of such group homomorphism that is not a ring homomorphism but even satisfies $f(1)=1$ consider this: let $R=R'=mathbb{Z}^2$ (with pointwise multiplication) and let $f(x,y)=(x,2x-y)$. I leave it as an exercise that $f$ is a group homomorphism. But it is not a ring homomorphism because
$$f((2,1)cdot (2,1))=f(4,1)=(4,7)$$
$$f(2,1)cdot f(2,1)=(2,3)cdot (2,3)=(4,9)$$
BTW: this example shows that your I also know how to show the additive and multiplicative properties for the ring homomorphism statement cannot be correct (more precisely I'm refering to the "multiplicative" part).
So the assumption "$f$ is a group homomorphism" is a mistake (it is not strong enough) and it should be "$f$ is a ring homomorphism".
Also note that the identity of $R[X]$ is $1$ (treated as a polynomial of degree $0$). Therefore $varphi(1)=1$ if and only if $f(1)=1$. It's quite trivial. More difficult is to show that $varphi$ preserves multiplication if $f$ does.
answered Nov 30 at 14:17
freakish
11.2k1629
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%2f3019229%2fproving-a-ring-homomorphism-using-a-group-homomorphism%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
To prove that $varphi: R[x]rightarrow R'[x]$, is a ring homomorphism you have to show that: $$ varphi(f+g)=varphi(f)+varphi(g), qquad varphi(fg)=varphi(f)varphi(g) qquad text{and} qquad varphi(1)=1.$$ Note that you have suppose that $f: Rrightarrow R'$ is a ring homomorphis.
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
add a comment |
To prove that $varphi: R[x]rightarrow R'[x]$, is a ring homomorphism you have to show that: $$ varphi(f+g)=varphi(f)+varphi(g), qquad varphi(fg)=varphi(f)varphi(g) qquad text{and} qquad varphi(1)=1.$$ Note that you have suppose that $f: Rrightarrow R'$ is a ring homomorphis.
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
add a comment |
To prove that $varphi: R[x]rightarrow R'[x]$, is a ring homomorphism you have to show that: $$ varphi(f+g)=varphi(f)+varphi(g), qquad varphi(fg)=varphi(f)varphi(g) qquad text{and} qquad varphi(1)=1.$$ Note that you have suppose that $f: Rrightarrow R'$ is a ring homomorphis.
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
To prove that $varphi: R[x]rightarrow R'[x]$, is a ring homomorphism you have to show that: $$ varphi(f+g)=varphi(f)+varphi(g), qquad varphi(fg)=varphi(f)varphi(g) qquad text{and} qquad varphi(1)=1.$$ Note that you have suppose that $f: Rrightarrow R'$ is a ring homomorphis.
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
answered Nov 30 at 14:01
Domenico Vuono
2,3201523
2,3201523
add a comment |
add a comment |
It should be: "$f:Rto R'$ is a ring homomorphism". Otherwise this is not true. Indeed, if $f$ is not a ring homomorphism then $f(ab)neq f(a)f(b)$ for some $a,bin R$. It is clear that $varphi(ab)neqvarphi(a)varphi(b)$ as well where $a,b$ are now treated as polynomials of degree $0$. Note that for polynomial $r$ of degree $0$ we have $varphi(r)=f(r)$.
As an example of such group homomorphism that is not a ring homomorphism but even satisfies $f(1)=1$ consider this: let $R=R'=mathbb{Z}^2$ (with pointwise multiplication) and let $f(x,y)=(x,2x-y)$. I leave it as an exercise that $f$ is a group homomorphism. But it is not a ring homomorphism because
$$f((2,1)cdot (2,1))=f(4,1)=(4,7)$$
$$f(2,1)cdot f(2,1)=(2,3)cdot (2,3)=(4,9)$$
BTW: this example shows that your I also know how to show the additive and multiplicative properties for the ring homomorphism statement cannot be correct (more precisely I'm refering to the "multiplicative" part).
So the assumption "$f$ is a group homomorphism" is a mistake (it is not strong enough) and it should be "$f$ is a ring homomorphism".
Also note that the identity of $R[X]$ is $1$ (treated as a polynomial of degree $0$). Therefore $varphi(1)=1$ if and only if $f(1)=1$. It's quite trivial. More difficult is to show that $varphi$ preserves multiplication if $f$ does.
answered Nov 30 at 14:17
freakish
11.2k1629
add a comment |
It should be: "$f:Rto R'$ is a ring homomorphism". Otherwise this is not true. Indeed, if $f$ is not a ring homomorphism then $f(ab)neq f(a)f(b)$ for some $a,bin R$. It is clear that $varphi(ab)neqvarphi(a)varphi(b)$ as well where $a,b$ are now treated as polynomials of degree $0$. Note that for polynomial $r$ of degree $0$ we have $varphi(r)=f(r)$.
As an example of such group homomorphism that is not a ring homomorphism but even satisfies $f(1)=1$ consider this: let $R=R'=mathbb{Z}^2$ (with pointwise multiplication) and let $f(x,y)=(x,2x-y)$. I leave it as an exercise that $f$ is a group homomorphism. But it is not a ring homomorphism because
$$f((2,1)cdot (2,1))=f(4,1)=(4,7)$$
$$f(2,1)cdot f(2,1)=(2,3)cdot (2,3)=(4,9)$$
BTW: this example shows that your I also know how to show the additive and multiplicative properties for the ring homomorphism statement cannot be correct (more precisely I'm refering to the "multiplicative" part).
So the assumption "$f$ is a group homomorphism" is a mistake (it is not strong enough) and it should be "$f$ is a ring homomorphism".
Also note that the identity of $R[X]$ is $1$ (treated as a polynomial of degree $0$). Therefore $varphi(1)=1$ if and only if $f(1)=1$. It's quite trivial. More difficult is to show that $varphi$ preserves multiplication if $f$ does.
answered Nov 30 at 14:17
freakish
11.2k1629
add a comment |
It should be: "$f:Rto R'$ is a ring homomorphism". Otherwise this is not true. Indeed, if $f$ is not a ring homomorphism then $f(ab)neq f(a)f(b)$ for some $a,bin R$. It is clear that $varphi(ab)neqvarphi(a)varphi(b)$ as well where $a,b$ are now treated as polynomials of degree $0$. Note that for polynomial $r$ of degree $0$ we have $varphi(r)=f(r)$.
As an example of such group homomorphism that is not a ring homomorphism but even satisfies $f(1)=1$ consider this: let $R=R'=mathbb{Z}^2$ (with pointwise multiplication) and let $f(x,y)=(x,2x-y)$. I leave it as an exercise that $f$ is a group homomorphism. But it is not a ring homomorphism because
$$f((2,1)cdot (2,1))=f(4,1)=(4,7)$$
$$f(2,1)cdot f(2,1)=(2,3)cdot (2,3)=(4,9)$$
BTW: this example shows that your I also know how to show the additive and multiplicative properties for the ring homomorphism statement cannot be correct (more precisely I'm refering to the "multiplicative" part).
So the assumption "$f$ is a group homomorphism" is a mistake (it is not strong enough) and it should be "$f$ is a ring homomorphism".
Also note that the identity of $R[X]$ is $1$ (treated as a polynomial of degree $0$). Therefore $varphi(1)=1$ if and only if $f(1)=1$. It's quite trivial. More difficult is to show that $varphi$ preserves multiplication if $f$ does.
answered Nov 30 at 14:17
freakish
11.2k1629
It should be: "$f:Rto R'$ is a ring homomorphism". Otherwise this is not true. Indeed, if $f$ is not a ring homomorphism then $f(ab)neq f(a)f(b)$ for some $a,bin R$. It is clear that $varphi(ab)neqvarphi(a)varphi(b)$ as well where $a,b$ are now treated as polynomials of degree $0$. Note that for polynomial $r$ of degree $0$ we have $varphi(r)=f(r)$.
As an example of such group homomorphism that is not a ring homomorphism but even satisfies $f(1)=1$ consider this: let $R=R'=mathbb{Z}^2$ (with pointwise multiplication) and let $f(x,y)=(x,2x-y)$. I leave it as an exercise that $f$ is a group homomorphism. But it is not a ring homomorphism because
$$f((2,1)cdot (2,1))=f(4,1)=(4,7)$$
$$f(2,1)cdot f(2,1)=(2,3)cdot (2,3)=(4,9)$$
BTW: this example shows that your I also know how to show the additive and multiplicative properties for the ring homomorphism statement cannot be correct (more precisely I'm refering to the "multiplicative" part).
So the assumption "$f$ is a group homomorphism" is a mistake (it is not strong enough) and it should be "$f$ is a ring homomorphism".
Also note that the identity of $R[X]$ is $1$ (treated as a polynomial of degree $0$). Therefore $varphi(1)=1$ if and only if $f(1)=1$. It's quite trivial. More difficult is to show that $varphi$ preserves multiplication if $f$ does.
answered Nov 30 at 14:17
freakish
11.2k1629
edited Nov 30 at 14:29
answered Nov 30 at 14:17
freakish
11.2k1629
answered Nov 30 at 14:17
freakish
11.2k1629
answered Nov 30 at 14:17
freakish
11.2k1629
11.2k1629
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.
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%2f3019229%2fproving-a-ring-homomorphism-using-a-group-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
2
$varphi(1) neq 1$ is possible; consider $f: mathbb{Z} rightarrow mathbb{Z}$ given by $f(r)=2r$.
– Matt B
Nov 29 at 21:19
What are $R$ and $R'$? If they are (unital) rings but $f$ is just supposed to be a homomorphism between their additive groups, I see indeed no reason why $f(1_R)$ should be $1_{R'}$. (Edit: And @MattB gives a counterexample.)
– Torsten Schoeneberg
Nov 29 at 21:19
In the definition I learned for ring homomorphisms, the multiplicative identity has to be mapped to the multiplicative identity, so φ(1) = 1
– Jon D.
Nov 29 at 21:24
1
Could you double-check the statement of the question? I would guess it is supposed to say "Let $f: R to R'$ be a ring homomorphism."
– arkeet
Nov 29 at 22:34
1
Sure, but it is probably a mistake in what is written (so check with your professor?) because it is false as is.
– arkeet
Nov 29 at 23:13