$A\times B\simeq C\times D \implies A\simeq C\text{ or } A\simeq D$

Question

Let $A,B,C,D$ be integral domains such that $A\times B$ is isomorphic to $C\times D$. Prove that $A$ is isomorphic to either $C$ or $D$.

Normally I include my thoughts about the problem, but I don't have any thoughts on this one... Any help is appreciated.

Answer 1

Let $\varphi: A \times B \to C \times D$ be the isomorphism.

Then, $\varphi(A \times \{0\})$ is an ideal of $C \times D$.

By the structure theorem for ideal of product, $\varphi(A)$ can be expressed in the form $\mathfrak c \times \mathfrak d$, where $\mathfrak c$ and $\mathfrak d$ are ideals of $C$ and $D$ respectively.

If they are both not zero, then letting $\varphi(a_c,0) \in C \times \{0\}$ and $\varphi(a_d,0) \in \{0\} \times D$ be the non-trivial elements, we see that $\varphi(a_c a_d, 0) = 0$, so $a_c = 0$ or $a_d = 0$, contradiction.

Therefore, either $\mathfrak c = 0$ or $\mathfrak d = 0$. In the first case, $A \cong D$; in the second case, $A \cong C$.

Answer 2

Hint. $A \times B$ has exactly two ideals that are maximal among those consisting solely of zero divisors.