Why are the rings $A$ and $B$ not isomorphic?

Question

When the ring $A$ has no zero divisors, but the ring $B$ has, why are the rings $A$ and $B$ not isomorphic??

For example when $A=\mathbb{Z}[\sqrt{2}]$ and $B=\begin{Bmatrix} \begin{pmatrix} a & 2b\\ 2b & a \end{pmatrix}:a,b \in \mathbb{Z} \end{Bmatrix}$

Answer 1

Simply because an isomorphism preserves all ring-theoretic structure.

Here's a short proof: Suppose there is an isomorphism $f:A \to B$. $B$ has a zero-divisor, i.e. a pair of elements $a,b$, neither being zero, but such that $ab=0$. Since $f$ is surjective (it is an isomorphism), there must be elements $a^\prime,b^\prime$ in $A$ such that $f(a^\prime)=a$ and $f(b^\prime)=b$. But then $f(a^\prime b^\prime)=f(a^\prime)f(b^\prime)=ab=0$. This implies (since $f$ is injective) that $a^\prime b^\prime=0$. But $f$ has no zero-divisors, so either $a^\prime$ or $b^\prime$ must be zero. But they are not. (what would happen if they were?) Contradiction.

Answer 2

Assume they are isomorphic i.e. $\phi B\to A $ be ismorphism.

Let $b,c \in B$ such that $bc=0$ but $b,c$ are nonzero elements of $B$.

$$0=\phi(0)=\phi(bc)=\phi(b)\phi(c)$$ Since A has no zero divisiar, $\phi(b)$ or $\phi(c)$ must be zero whic is contradicting the fact that $\phi$ is a isomorphism.

Answer 3

If $\phi\colon B\to A$ is an isomorphism and $u,v\in B$ with $uv=0$, $u,v\ne0$. Then $\phi(u)\phi(v)=\phi(uv)=\phi(0)=0$ in $A$, hence $\phi(u)=0$ or $\phi(v)=0$ becasue $A$ has no zero divisors. On the other hand $\phi(u)\ne \phi(0)=0$, $\phi(v)\ne\phi(0)=0$ becasue $\phi$ is injective.