Question about ring homomorphism.

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I have a ring homomorphism $f : mathbbZ[x] to mathbbZ$ such that $f$ sends $x$ to $3$. I want to find $f(x^3+2)$.
My attempt is, since $f$ is a homomorphism
$$f(x^3+2)=f^3(x)+f(2)=27+f(2)$$ but what is $f(2)$?






abstract-algebra







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    A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
    – quasi
    4 hours ago















up vote
4
down vote

favorite












I have a ring homomorphism $f : mathbbZ[x] to mathbbZ$ such that $f$ sends $x$ to $3$. I want to find $f(x^3+2)$.
My attempt is, since $f$ is a homomorphism
$$f(x^3+2)=f^3(x)+f(2)=27+f(2)$$ but what is $f(2)$?






abstract-algebra







share|cite|improve this question

















  • 3




    A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
    – quasi
    4 hours ago













up vote
4
down vote

favorite









up vote
4
down vote

favorite











I have a ring homomorphism $f : mathbbZ[x] to mathbbZ$ such that $f$ sends $x$ to $3$. I want to find $f(x^3+2)$.
My attempt is, since $f$ is a homomorphism
$$f(x^3+2)=f^3(x)+f(2)=27+f(2)$$ but what is $f(2)$?






abstract-algebra







share|cite|improve this question













I have a ring homomorphism $f : mathbbZ[x] to mathbbZ$ such that $f$ sends $x$ to $3$. I want to find $f(x^3+2)$.
My attempt is, since $f$ is a homomorphism
$$f(x^3+2)=f^3(x)+f(2)=27+f(2)$$ but what is $f(2)$?






abstract-algebra




abstract-algebra






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asked 4 hours ago









paradox

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444







  • 3




    A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
    – quasi
    4 hours ago













  • 3




    A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
    – quasi
    4 hours ago








3




3




A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
– quasi
4 hours ago





A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$
– quasi
4 hours ago











1 Answer
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As mentioned in the comments, $f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$, so $f(x^3 + 2) = 29$.






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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    As mentioned in the comments, $f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$, so $f(x^3 + 2) = 29$.






    share|cite|improve this answer
























      up vote
      5
      down vote



      accepted










      As mentioned in the comments, $f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$, so $f(x^3 + 2) = 29$.






      share|cite|improve this answer






















        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        As mentioned in the comments, $f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$, so $f(x^3 + 2) = 29$.






        share|cite|improve this answer












        As mentioned in the comments, $f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$, so $f(x^3 + 2) = 29$.







        share|cite|improve this answer












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        answered 4 hours ago









        Matheus Andrade

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