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 $φ(1) = 1$.

Source at zimmer.csufresno.edu

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 $φ(1)=1$?

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

You are ordering the equalities wrong, so it seems to assume the result when it isn't:

$$ \varphi(1)\cdot \varphi(1) = \varphi(1 \cdot 1) = \varphi(1) = 1 \cdot \varphi(1)$$ 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 $$ \varphi(1)(\varphi(1) - 1)= 0$$

Hope it helps.

  • $\begingroup$ Oh right, since D is an integral domain (which means D is a commutative ring), D contains a multiplicative identity element 1 for each φ(1) ∈ D. We are able to say φ(1) = 1·φ(1) =φ(1)·1. $\endgroup$ – user263362 Aug 20 '15 at 20:42
  • $\begingroup$ @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. $\endgroup$ – user2520938 Aug 20 '15 at 20:49
  • $\begingroup$ @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. $\endgroup$ – user2520938 Aug 20 '15 at 20:52

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