Proving a ring-homomorphism using a group-homomorphism

Question

Let f : R → R' be a group homomorphism. Show that the induced map φ : R[x] → R'[x], where φ(a_nxⁿ + . . . + a₀) = f(a_n)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.

Answer 1

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 $f: R\rightarrow R'$ is a ring homomorphis.

Answer 2

It should be: "$f:R\to R'$ 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 $a,b\in R$. It is clear that $\varphi(ab)\neq\varphi(a)\varphi(b)$ 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.