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?

  • $\begingroup$ The last part of math.stackexchange.com/a/105457/4280 suggests the answer is that there is only one way. I don't see the easy argument now, but it's late (for me)... $\endgroup$ – Henno Brandsma Jan 25 '16 at 21:43
  • $\begingroup$ Technically, you could also define the multiplication by $(u,v) \times (x,y) = (vy - ux,uv + vx)$ (i.e. associating the x-component with the imaginary part and associating the y-component with the real part). $\endgroup$ – Roland Jan 25 '16 at 21:44
  • $\begingroup$ Well, consider (x,y)(z,w) = (f(x,y,z,w),g(x,y,z,w)) where f and g are linear functions and you f(1,0,z,w) = z, g(1,0,z,w) = w and f(1,0,1,0) = 1 ... maybe you can find a system of equations that force f and g. (my comment assume identity = (1,0) which perhaps I shouldn't have.) $\endgroup$ – fleablood Jan 25 '16 at 21:51
  • $\begingroup$ I'm not sure if you're looking for other explicit formulas for multiplication operations on $\mathbb{R}^2$, but here is an abstract argument showing that up to isomorphism yours is the only one. The dimension of $\mathbb{R}^2$ over $\mathbb{R}$ is 2, so a multiplicative structure on $\mathbb{R}^2$ making it into a field will give a field extension of $\mathbb{R}$ of degree 2, and thus isomorphic to the complex numbers. Thus the multiplication structure will be isomorphic to the one defined above. $\endgroup$ – Alex Saad Jan 25 '16 at 21:57

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.


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)

  • 4
    $\begingroup$ The way I understood the question is that, given the definition of addition on $\mathbb{R}^2$, we look for a multiplication making it into a field. I don't see how, just from that, we can infer that the result will be a real vector space of dimension $2$. Wouldn't we need some compatibility with the multiplication of $\mathbb{R}$? $\endgroup$ – Pierre-Guy Plamondon Jan 25 '16 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.