isomorphism between a field and non-field rings

Question

I think that once you have a field and a ring which is not a field, you can conclude that there is no isomorphism between these two.

Is it right? if not, is there an example? if true, can someone give a rigorous proof?

Answer 1

This is correct, and the intuition should be that "being a field" is a property that can be stated entirely in terms of ring-theoretic statements, and therefore must be preserved by isomorphism. Here's the proof in full detail.

Suppose $F$ is a field and $R$ is a ring and $\varphi : F \to R$ is a ring isomorphism.

First, we will show that $R$ is commutative. Let $a,b \in R$ be arbitrary. Since $F$ is surjective, $a = \varphi(x)$ and $b = \varphi(y)$ for some $x,y \in F$. Then $$ab = \varphi(x) \varphi(y) = \varphi(xy) = \varphi(yx) = \varphi(y) \varphi(x) = ba.$$ Since $a$ and $b$ are arbitrary, $R$ is commutative.

Next, since $0 \neq 1$ in $F$, $\varphi$ is injective, and $\varphi(0) = 0, \varphi(1) = 1$, $0 \neq 1$ in $R$.

Finally, we will show that nonzero element of $R$ are invertible. Let $r \in R \setminus \{0\}$ be arbitrary. Since $F$ is surjective, there is some $x \in F$ such that $\varphi(x) = r$. Since $r \neq 0$ and $\varphi(0) = 0$, $x \neq 0$. Since $F$ is a field, there is some $y \in F$ such that $xy = yx = 1$. Now $$1 = \varphi(1) = \varphi(xy) = \varphi(x) \varphi(y) = r \varphi(y)$$ and $$1 = \varphi(1) = \varphi(yx) = \varphi(y) \varphi(x) = \varphi(y) r,$$ so $r$ is invertible in $R$.

This completes the proof that $R$ is a field.

Answer 2

Let $A$ be a ring which is not a field and $K$ a field, and suppose there is an isomorphism $f:K\to A$. Obviously, we have $A\neq 0$, so by definition, there is a $a\in A\setminus \{0\}$ which is not invertible.

However, $f^{-1}(a)\neq 0$ has an inverse $b$ in $K$. Now, $a\cdot f(b)=f(f^{-1}(a)\cdot b)=f(1)=1$, a contradiction.