Reading through the Wiki article on direct products (<https://en.wikipedia.org/wiki/Direct_product>), an example is given using the set of real numbers, $\mathbb{R}$, viewed as a set endowed with different algebraic structures. In all cases the underlying set of $\mathbb{R}\times\mathbb{R}$ is (of course) $\mathbb{R}^2$. However, when considering $\mathbb{R}$ as a field, the article states that the direct product is undefined, since the set $\mathbb{R}^2$ endowed with the naïve addition $(a,b)+(c,d)=(a+c,b+d)$ and multiplication $(a,b)\cdot(c,d)=(ac,bd)$ is clearly not a field.

My problem is that the set $\mathbb{R}^2$ **can** be endowed with a multiplicative structure that makes it into a field (since $\mathbb{R}^2=\mathbb{C}$ as a set), so my question is the following:

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> Why can we **not** define $\mathbb{R}\times\mathbb{R}$ as $\mathbb{C}$, when viewing $\mathbb{R}$ as the field of real numbers?
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Is there an implicit assumption in the definition of direct product that if $X$ has some algebraic operation $\*$, then $X\times X$ must have the corresponding operation $\*\_{\times}$ defined as $(x\_1,x\_2)\*\_{\times}(y\_1,y\_2):=(x\_1 \* y\_1,x\_2 \* y\_2)$? If so, is this because of the way the direct product is defined in category theory?