Proving a ring-homomorphism using a group-homomorphism












0














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







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  • 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
















0














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







share|cite|improve this question




















  • 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














0












0








0


0





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 21:39

























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














  • 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










2 Answers
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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.






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    0














    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.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      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.






      share|cite|improve this answer


























        0














        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.






        share|cite|improve this answer
























          0












          0








          0






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 at 14:01









          Domenico Vuono

          2,3201523




          2,3201523























              0














              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.






              share|cite|improve this answer




























                0














                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.






                share|cite|improve this answer


























                  0












                  0








                  0






                  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.






                  share|cite|improve this answer














                  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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 at 14:29

























                  answered Nov 30 at 14:17









                  freakish

                  11.2k1629




                  11.2k1629






























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