$\mathbb{R}^3$ isn't isomorphic to $\mathbb{R} \times \mathbb{C}$ How to show the two rings $\mathbb{R}^3 = \mathbb{R}\times\mathbb{R}\times\mathbb{R}$ and $\mathbb{R} \times \mathbb{C}$ aren't ring isomorphic to each other?
My attempt:
Since $\mathbb{R}^2$ isn't an Integral Domain as $(1,0).(0,1) = (0,0)$, It's not a field. But $\mathbb{C}$ is a field. So $\mathbb{R}^2$ isn't isomorphic to $\mathbb{C}$.
Therefore, $\mathbb{R}^3$ isn't Ring Isomorphic to $\mathbb{R} \times \mathbb{C}$
Is the logic okay? If not, then could you tell me how to show this one.
 A: It is not generally true for rings that $A \times B \cong A \times C$ implies $B \cong C$, so your proof does not work as stated.
Try looking at roots of unity.  In $\mathbb C$ there is the unit circle, which contains many, many elements $z$ with the property that $z^n = 1$ for some $n \in \mathbb N$.  Therefore in the ring $R = \mathbb R \times \mathbb C$, the same is true.
What are the roots of unity in $S = \mathbb R^3$?
A: As D_S stated in their answer, your proof does not work as stated because for rings, $R_{0}\times R_{1}\cong R_{0}\times R_{2}$ does not imply that $R_{1}\cong R_{2}.$ What follows is a very direct line of argument. I encourage you to read the next paragraph, think about how you would prove it, and only then reveal the spoilers.
In $\mathbb{R}^{3},$ there exist three distinct non-zero elements $a_{1}=(1,0,0),$ $a_{2}=(0,1,0)$ and $a_{3}=(0,0,1)$ such that the product of any two of them is zero: $a_{1}a_{2}=a_{2}a_{3}=a_{3}a_{1}=(0,0,0).$ No such triple of elements exists in $\mathbb{R}\times\mathbb{C},$ and therefore the two rings cannot be isomorphic to each other.

 To prove this, suppose, for a contradiction, that $A_{1},A_{2},A_{3}\in\mathbb{R}\times\mathbb{C}$ are distinct, non-zero, and satisfy the conditions $A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{1}=(0,0).$ Let $\pi_{1}\colon \mathbb{R}\times\mathbb{C} \to \mathbb{R}$ and $\pi_{2}\colon \mathbb{R}\times\mathbb{C}\to \mathbb{C}$ denote the coordinate functions (so for example, $\pi_{2}(1,2+3i)=2+3i$). Then for $j=1,2,$ $$\pi_{j}(A_{1})\pi_{j}(A_{2})=\pi_{j}(A_{2})\pi_{j}(A_{3})=\pi_{j}(A_{3})\pi_{j}(A_{1})=0.$$ Without loss of generality, we may assume that $\pi_{1}(A_{1})=0$ (otherwise, just relabel things). Since $A_{1}$ is non-zero, we must now have $\pi_{2}(A_{1})\neq0,$ and therefore $\pi_{2}(A_{2})=\pi_{2}(A_{3})=0.$ Now it follows that $\pi_{1}(A_{2})\neq0$ and $\pi_{1}(A_{3})\neq0$ since $A_{2}$ and $A_{3}$ are non-zero. But this means that $\pi_{1}(A_{2})\pi_{1}(A_{3})\neq0$ also, a contradiction.


 The key fact that makes this argument work is that $\mathbb{R}$ and $\mathbb{C}$ are both integral domains (so, in your attempt, you were right to pick up on the distinguishing feature that $\mathbb{R}^{2}$ is not an integral domain, whereas $\mathbb{C}$ is). This is what allowed us to deduce that $\pi_{2}(A_{2})=\pi_{2}(A_{3})=0.$ Without this, the argument falls apart. I encourage you to think about how this argument is essentially a form of the pigeonhole principle in action.

A: Hm, the existing solutions are a bit more complicated than the one that I would use.
$\mathbb R^3$ has 8 idempotents and $\mathbb R\times \mathbb C$ only has four.  An idempotent is an element $e$ such that $e^2=e$.  Naturally an isomorphism must match up idempotents to idempotents.
