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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.

Anik Bhowmick
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    For the case that $R = S = \mathbb{R}$, this indeed produces a field. However, this is not the direct product of $R, S$ anymore: The direct product is the cartesian product of sets $R \times S$ with the pointwise multiplication specifically, not just any multiplication on the underlying sets. – tolUene Jul 18 '21 at 22:09
  • @tolUene not only for $\mathbb R$, but we can also do it for any commutative ring with the identity I guess... – Anik Bhowmick Jul 18 '21 at 22:14
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    We can't! Take $R = S = \mathbb{Z}$. Try to find an inverse for $(2,1)$. You can't. By the way we get inverses of complex numbers, we know that $(a,b)^{-1} = (da, -db)$ with $d(a^2 + b^2) = 1$, thus we only have inverses when $a^2 + b^2$ is a unit. – tolUene Jul 18 '21 at 22:18
  • Let $R$ and $S$ rings of square matrices over different fields and of different sizes. How do you define $r_1s_1$? – CyclotomicField Jul 18 '21 at 23:15
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    The direct product is defined to have pointwise product, so if you choose a different product then it is no longer a direct product w.r.t. to that product. – Bill Dubuque Jul 19 '21 at 00:50
  • @tolUene That is really a solution to this post, and should be in the solutions section rather than the comments. Would you consider transferring it? – rschwieb Jul 20 '21 at 13:45
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    @rschwieb Done :) – tolUene Jul 20 '21 at 21:38

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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.

tolUene
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