I am totally aware of the fact that If $R$ and $S$ are rings then $R \times S$ is never a field. I have seen this solution a lot of places. But if I take the multiplication differently, suppose $(r_1,r_2)\in R$ and $(s_1,s_2)\in S$, then I can define the multiplication structure as complex number structure, i.e., $(r_1,r_2)\cdot(s_1,s_2)=(r_1s_1-r_2s_2,r_1s_2+r_2s_1)$. I think in this scenario, $R \times S$ produces a field.
If I am wrong, please point out the flaws in my approach. Thanks in advance.
Best Answer
First of all, you should be aware that this doesn't define a product at all for $R \neq S$, as in this case, $r_1 s_1$ (etc.) is not defined.
For the case $R = S$, this can indeed form a field; clearly, it works for $R = \mathbb{R}$. However, this is not the direct product of $R$ with itself, since this by definition (and also necessitated by the universal property of products) carries the pointwise ring structure.
In any case, you do get a ring structure for $R^2$ like this (in fact, this essentially amounts to $R[i]$, for $i^2 + 1 = 0$). From multiplication of complex numbers, we know that $(a, b)^{-1} = (da, -db)$ (if defined) for $d(a^2 + b^2)=1$, thus, $(a,b)$ is a unit in $R^2$ iff. $a^2 + b^2$ is a unit in $R$. For $\mathbb{Z}$, this gives us some easy counterexamples of when this does not form a ring.