# Ring homomorphism

Problem: Let $R$ be a commutative ring, and let $D$ be an integral domain. Let $φ : R → D$ be a nonzero function such that $φ(a+b) = φ(a) + φ(b)$ and $φ(ab) = φ(a)φ(b)$ for all $a,b \in R$. Show that $φ$ is a ring homomorphism.

Proof: Since $φ$ preserves both operations, but the definition of a ring homomorphism we only have to show that $φ(1) = 1$. We have $φ(1) = φ(1·1) = φ(1)φ(1)$, thus $φ(1)(φ(1)−1) = 0$. Since $D$ is an integral domain, either $φ(1) = 0 or φ(1) = 1$.

However, we will show that $φ(1) = 0$. Indeed, if $φ(1) = 0$, then for any $x ∈ R$, $$φ(x) = φ(x · 1) = φ(x)φ(1) = φ(x) · 0 = 0$$ so $φ$ is the zero function, but it was given that $φ$ is nonzero. Therefore $φ(1) = 1$.

When $φ(1) = φ(1·1) = φ(1)φ(1)$, why does $φ(1)(φ(1)−1) = 0$ follow? Because this assumes $φ(1) = φ(1·1) = φ(1)φ(1) = φ(1)·1$, but we do not know that $φ(1)=1$ because we are trying to prove that $φ(1)=1$?

• Notice that the proof should say "we will show that $\phi(1)\ne0$" – user84413 Aug 20 '15 at 21:06

$$\varphi(1)\cdot \varphi(1) = \varphi(1 \cdot 1) = \varphi(1) = 1 \cdot \varphi(1)$$ The last equality is just the fact that for every $a \in D$ , $1 \cdot a = a$. Using $\varphi(1)\cdot \varphi(1) = 1 \cdot \varphi(1)$ we can arrange this as $$\varphi(1)\cdot \varphi(1) - 1 \cdot \varphi(1) = 0$$ and taking $\varphi(1)$ as a common factor we get $$\varphi(1)(\varphi(1) - 1)= 0$$
• @user263362 Once you establish $\varphi(1)=\varphi(1)\dot\varphi(1)$ you are done, since integral domains always have the cancellation property, and the function is non zero. This doesn't change the proof in any substantial way, it just saves you having to factor the final expression explicitly. – user2520938 Aug 20 '15 at 20:49
• @user263362 Also note that even in a non-commutative ring with identity we have that $a=a\cdot 1 = 1\cdot a$; $1$ always commutes with every element of the ring. – user2520938 Aug 20 '15 at 20:52