Is there only one way to make $\mathbb R^2$ a field?

Question

I think I read an answer to this question before but I can't find it by searching.

We can make $\mathbb R^2$ a field by defining addition as normal and defining multiplication by complex multiplication so $(u,v) \times (x,y) = (ux-vy,uy+vx)$ and it satisfies the field axioms. I know defining multiplication by $(u,v) \times (x,y) = (ux,vy)$ does not work as $(0,1)$ for instance has no inverse. But is there another way in which we could define multiplication in $\mathbb R^2$ that would satisfy the field axioms, while keeping normal vector addition?

Answer 1

The answer of Hagen von Eitzen gives the correct answer if we demand that the multiplication that we are constructing is compatible with the usual multiplication on $\mathbb{R}$. However, if we do not make that assumption, then there are many non-isomorphic field structures on $\mathbb{R}^2$.

To see this, let $F$ be any field of characteristic zero and cardinality $|\mathbb{R}|$. We claim that $F$ is isomorphic as a field to $\mathbb{R}^2$ for a suitable choice of multiplication for $\mathbb{R}^2$.

To see this, we note that both $F$ and $\mathbb{R}^2$ have the structure of a $\mathbb{Q}$-vector space. Moreover, looking at the cardinalities, each has dimension $|\mathbb{R}|$ over $\mathbb{Q}$. Since vector spaces with the same dimension are isomorphic, this means that there is a linear isomorphism $\phi: F \to \mathbb{R}^2$.

Since $\phi$ is an isomorphism, we can use it to define a multiplication $\cdot : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2$ by transport of structure: we simply define $a \cdot b = \phi(\phi^{-1}(a) \cdot \phi^{-1}(b))$ for all $a, b \in \mathbb{R}^2$. Now $\phi$ preserves not only addition (which it already did because it is linear), but also multiplication (by construction). Since $F$ is a field, and $\phi$ is a bijection, this proves directly that $\mathbb{R}^2$ with this multiplication and the usual addition is a field, and in fact isomorphic to $F$.

A surprising consequence is that $\mathbb{R}^2$ has a multiplication such that it is isomorphic to $\mathbb{R}$! Unfortunately, we have shown the existence of this multiplication using the axiom of choice, so it might not be possible to give a 'direct' description of this multiplication.

Answer 2

Up to isomorphism, there is only one field that is vector space of dimension two over the reals: If $F$ is such a field, then $1_F\cdot \Bbb R$ is a subfield isomorphic to $\Bbb R$ (henceforth identified with $\Bbb R$) and for any $\alpha\in F\setminus \Bbb R$, we know that $1,\alpha,\alpha^2$ are $\Bbb R$-linearly dependent, i.e., $\alpha$ is the root of a quadratic polynomial with real coefficients. This allows us to identify $\alpha$ with either of the two roots that polynomial has in $\Bbb C$, which leads to an isomorphism $F\to \Bbb C$.

So to define a multiplication in $\Bbb R^2$ that turns it into a field, we have to

• pick a basis $e_1,e_2$ of $\Bbb R^2$
• consider the vector space isomorphism $\Bbb R^2\to \Bbb C$ given by $e_1\mapsto 1$, $e_2\mapsto i$.
• define a multiplication $\odot $ in $\Bbb R^2$ by $v\odot w = f^{-1}(f(v)\cdot f(w))$

Any choice of basis produces a valid multiplication, and distinct bases mostly lead to distinct multiplications (except that $(e_1,e_2)$ and $(e_1,-e_2)$ lead to the same multiplication)