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)... $\endgroup$ Jan 25, 2016 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, 2016 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, 2016 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, 2016 at 21:57

2 Answers 2


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.

  • $\begingroup$ All of this can be avoided if you assume there is no Q-basis for R (which is consistent with ZF) $\endgroup$ Sep 25 at 17:39

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)

  • 7
    $\begingroup$ 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 $\mathbb{R}$? $\endgroup$ Jan 25, 2016 at 22:16

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