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

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?

• 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)... Jan 25, 2016 at 21:43
• 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). Jan 25, 2016 at 21:44
• 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.) Jan 25, 2016 at 21:51
• 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. Jan 25, 2016 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 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.

• All of this can be avoided if you assume there is no Q-basis for R (which is consistent with ZF) 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)

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