Suppose R and S are commutative rings with unity. Let $\varphi : R \rightarrow S$ be a ring isomorphism. Prove that R is a field iff S is a field.

Question

We had a similar proof (R is an ID IFF S is an ID) where we proved by contradiction, but I'm a little confused on how to go about proving this one.

What I have so far ($\Rightarrow$) Let $r\in R$ be nonzero. $\forall b \in R$ where $b\neq0$, $rb = 1_R$. Observe then, $\varphi(1_R) = \varphi(rb) = \varphi(r)\cdot \varphi(b) = 1_S.$

Is it safe to assume that if r is a unit, then it would be true for all $b\neq0$ that $rb = 1$?

any help would be appreciated.

Answer 1

Definition: A field $F$ is a commutative ring such that $1_F\neq 0_F$ and every nonzero $x\in F$ has a multiplicative inverse in $F$.

Now, observe that if $\varphi:R\to S$ is a ring isomorphism, then $\varphi^{-1}:S\to R$ is a ring isomorphism, so you only need to prove the statement in one direction.

Suppose $R$ is a field. We can show that $S$ is a field in two steps by verifying the definition of a field holds for $S$:

1. We have $1_R\neq 0_R$. Also, $\varphi(1_R)=1_S$ and $\varphi(0_R)=0_S$. What can we conclude from the fact that $\varphi$ is injective?

2. Suppose $s\in S$ is nonzero. We want to show that $s$ has a multiplicative inverse. Well, $\varphi$ is surjective, so there exists $r\in R$ such that $\varphi(r)=s$. Since $s\neq 0_S$ and $\varphi$ is injective, $r\neq 0_R$ (why?). It follows that $r^{-1}\in R$ exists. Can you guess a candidate for $s^{-1}$?