Is $\mathbb{R}^2$ a Ring?

Question

From what I know, $\mathbb{R}^2$ is a group under addition, defined as $(a, b) + (c, d) = (a+b,c+d)$. However, this answer on another question seems to suggest that $\mathbb{R}^2$ is actually a ring with multiplication defined as $(a, b)\cdot (c, d) = (ac, bd)$. I thought that we usually only define multiplication over the group $\mathbb{R}^2$ as $(a, b)\cdot (c,d) = (ac - bd, ad+bc)$, and in result end up making it a field called $\mathbb{C}$?

Answer 1

As the answer you have linked indicates, the key here is that the exact meaning of $\Bbb R^2$ (or equivalently $\Bbb R \times \Bbb R$) depends in the context that we are working in.

In settings where $\Bbb R$ and/or $\Bbb C$ are the only rings/fields being discussed (typically in problems of an area whose name includes the word "analysis"), $\Bbb R^2$ typically refers to the abelian group/vector space over the set $\Bbb R^2$. In other words, no multiplication between elements is defined or considered.

However, in settings where $\Bbb R$ and/or $\Bbb C$ are considered to be one ring among many (typically in problems of an area whose name includes the word "algebra(ic)"), $\Bbb R^2$ refers to the multiplication associated with the product $\Bbb R \times \Bbb R$ of rings. That is, $\Bbb R^2$ a ring with multiplication defined by $(a,b)\cdot(c,d) = (ac,bd)$.

The symbol $\Bbb R^2$ is never used to refer to $\Bbb R^2$ with the complex-number multiplication $(a,b)\cdot (c,d) = (ac - bd,ad + bc)$, except perhaps for pedagogical reasons. Where the set $\Bbb R^2$ is given this multiplication rule, the symbol $\Bbb C$ is used instead.

Answer 2

You have that from the same abelian group (in this case $\mathbb{R}^2$) you can obtain different non-isomorphic rings if you consider it with different multiplications.
In this case $\mathbb{C}$ and $\mathbb{R}^2$ are both rings but with different multiplications.

In the first case, as Randall commented, you have that the group you are considering became a field. In the second case it's not a field (for example $(1,0)$ doesn't admit any inverse).

So, even if the abelian group is the same, you can construct different rings on it.