Why is the direct product $\mathbb{R}\times\mathbb{R}$ undefined when $\mathbb{R}$ is viewed as a field?

Question

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:

Why can we not define $\mathbb{R}\times\mathbb{R}$ as $\mathbb{C}$, when viewing $\mathbb{R}$ as the field of real numbers?

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?

Answer 1

A product $A \times B$ also comes with a pair of projection maps $\pi_1: A \times B \rightarrow A$ and $\pi_2: A \times B \rightarrow B$. These are a fundamental part of the definition of a direct product.

We want these projections to be morphisms in the category we're considering. In this case, we want them to be ring homomorphisms, so they need to preserve the operations of the ring. Using this, we can prove that the operation of the product has to be as expected.

$$\pi_1((x_1,x_2)*_{\times}(y_1,y_2)) = \pi_1(x_1,x_2) *\pi_1(y_1,y_2) = x_1 * y_1$$

So the first entry of the result under the operation has to be $x_1 * y_1$. The same idea works for the second element, giving $(x_1,x_2)*_{\times}(y_1,y_2):=(x_1 * y_1,x_2 * y_2)$.

If you try to define the product of $\mathbb{R}$ with itself as $\mathbb{C}$, you find that no pair of projection maps can work, in part because any ring homomorphism out of a field must be injective.

Answer 2

Chessanator has answered the stated question, but one might plausibly ask whether there is any way to take the product $\mathbb{R} \times \mathbb{R}$ in the category of fields.

The answer is no. For such a product consists of a field $\mathbb{R} \times \mathbb{R}$ together with two projection maps $\pi_1, \pi_2 : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ which are field homomorphisms.

Now consider the diagonal map $\Delta : \mathbb{R} \to \mathbb{R} \times \mathbb{R}$. Note that $\pi_i \circ \Delta$ is the identity map for $i = 1, 2$. Since each $\pi_i$ is a field homomorphism and hence injective, it follows that each $\pi_i$ is a bijection with $\Delta$ its inverse. Thus, we have $\pi_1 = \pi_2$, and without loss of generality, we can take the product to be $\mathbb{R}$ with the projections the identity map.

It’s easy to conclude based on this fact that for any field $k$, there is at most one field homomorphism $k \to \mathbb{R}$.

Now consider, for instance, $k = \mathbb{Q}(\sqrt{2})$. There are two maps $f, g : k \to \mathbb{R}$ sending $\sqrt{2}$ to $\pm \sqrt{2}$. So we have arrived at our contradiction.