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I have a ring homomorphism $f : \mathbb{Z}[x] \to \mathbb{Z}$ 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 $f(2)$?

paradox
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    A ring homomorphism sends the identity to the identity, hence $f(1)=1$, so $f(2)=f(1+1)=,...$ – quasi Sep 16 '18 at 05:46
  • @quasi A ring-with-identity homomorphism sends the identity to the identity; that can't be literally true for a ring homomorphism, since a ring need not have an identity. (And in fact if $R$ is a ring with identity and $f:R\to R$ iis defined by $f(x)=0$ then $f$ is a ring homomorphism.) In this example it's not hard to prove that $f(1)=1$, in spite of that (and without assuming that we really havve a ring-with-identity homomorphism): $3=f(x)=f(1x)=3f(1)$. – David C. Ullrich Sep 16 '18 at 16:46
  • @David C. Ullrich: Many authors require, as part of the definition of ring, the existence of a multiplicative identity $1$, and for such rings, they require, as part of the definition of homomorphism, that a ring homomorphism sends $1$ to $1$. But I agree, if rings are not required to have a multiplicative identity, then your argument nicely avoids the issue. – quasi Sep 16 '18 at 22:35

1 Answers1

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

Matheus Andrade
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