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

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 disitinct multiplications (except that $(e_1,e_2)$ and $(e_1,-e_2)$ lead to the same multiplication)

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 isomorpic 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' desciption of this multiplication.