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In order to 'turn $\mathbb{R}^2$ into a field', we just need a way to add 2-vectors and a way to multiply 2-vectors, that satisfy the field axioms.

Adding 2-vectors is easy, we can add element wise:

$$(a,b)+(x,y):=(a+x,b+y)$$

But multiplication is tricky. As complex numbers tell us, this strange product does indeed produce a field:

$$(a,b)\cdot (x,y):=(ax-by,ay+bx).$$

My question is this:

Are there any other products that produce a field? Can we prove this is the only one?

The "obvious" product:

$$(a,b)\cdot (x,y):=(ax,by),$$

doesn't work, as for example, there is no inverse element for $(1,0)$.

How about in $\mathbb{R}^n$?

It actually looks like my question has a pretty good answer here: Can we turn $\mathbb{R}^n$ into a field by changing the multiplication?

Jack
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    Does this answer your question? The crucial hypothesis is that there's an isomorphic copy of $\mathbb{R}$ inside your larger field and that the operations are compatible (like they are for the complex numbers we know and love: e.g., $2 \cdot 3 = 6$, whether we think of $2 \in \mathbb{R}$ or $2 = 2 + 0i \in \mathbb{C}$, and analogously for $3$). – Sammy Black Jul 18 '24 at 14:35
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    One can prove (with heavy use of the axiom of choice) that all the additive groups $\mathbb R^n$ are isomorphic, and in fact there are many isomorphisms between each pair of them. Using such isomorphisms, you can transport the multiplication of reals (or of complex numbers) from $\mathbb R$ (or $\mathbb R^2$) to get many field structures on $\mathbb R^n$. Unfortunately, these multiplication operations will be really bad: discontinuous, not even measurable, possibly not definable (depending on your set-theoretic universe). – Andreas Blass Jul 18 '24 at 16:27
  • If you want decently behaved field operations, say continuous with respect to the usual topologies, then $\mathbb R$ and $\mathbb R^2$, with their ordinary field structures as $\mathbb R$ and $\mathbb C$ are the only ones. (If you allow non-commutative fields, i.e., division rings, then you also get the structure of quaternions on $\mathbb R^4$, but nothing more.) – Andreas Blass Jul 18 '24 at 16:31

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It looks like Sammy Black's link addresses this, but I will still elaborate a bit:

I suspect you may be interested in imposing operations which preserve the original structure of $\mathbb{R}^2$ somehow, or which themselves are related to $\mathbb{R^2}$. For example, maybe letting the addition operation be defined by $(x,y) \mapsto (Q(x,y), P(x,y))$ with $Q$ and $P$ regular functions.

If you have nothing in mind for stipulations on the operations and don't care if they respect the structure of $\mathbb{R}^2$, then you can impose any field structure provided the field is uncountable. The reason is that if a field $F$ is uncountable, then we can think of elements of $\mathbb{R}^2$ as really being elements of $F$ at the level of sets as follows:

Since $F$ is uncountable there is a bijection of underlying sets $\phi: \mathbb{R}^2 \rightarrow F$. Now we can define that for all $a, b \in \mathbb{R}^2$, let $a + b = \phi^{-1}(\phi(a) + \phi(b))$ and $a \cdot b = \phi^{-1}(\phi(a) \cdot \phi(b))$. Here $\phi(a) + \phi(b)$ takes $a,b \in \mathbb{R}^2$, and "turns them into elements of $F$," and then applies the addition operation in $F$. Finally the $\phi^{-1}$ part pulls back the sum in $F$ to $\mathbb{R}^2$.

Notice this requires we forget the structure of $\mathbb{R}^2$ entirely, and just think of the underlying set of $\mathbb{R}^2$ as the underlying set of $F$ by relabeling with $\phi$. If this way of getting a field structure feels like cheating then you need to pinpoint precisely why you feel that way, e.g. it failed to respect some structure you had in mind for $\mathbb{R}^2$, or maybe the operations weren't what you had in mind.

McNugget
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