I'm trying to prove that $\mathbb R^3$ is not a field with component-wise multiplication and sum defined. I think it's weird, because every properties of a field are inherit from $\mathbb R$. Anyone can help?


  • 9
    $\begingroup$ How do you define multiplication in $\mathbb{R}^3$? $\endgroup$ – Mikko Korhonen Oct 19 '12 at 10:16
  • $\begingroup$ @m.k. the same of $\mathbb R$ component-wise $\endgroup$ – user42912 Oct 19 '12 at 10:34
  • 5
    $\begingroup$ Then it's not even an integral ring. $\endgroup$ – Alexei Averchenko Oct 19 '12 at 12:45
  • 2
    $\begingroup$ Or phrased more simply, how do you define inversion on all non-zero elements? (paying careful attention to the fact (0,0,1) is nonzero) $\endgroup$ – Hurkyl Oct 19 '12 at 15:36
  • 3
    $\begingroup$ @QiaochuYuan The OP went on to clarify that in the second comment. $\endgroup$ – rschwieb Oct 19 '12 at 19:06

I'm assuming you define muliplication component-wise on $\mathbb{R}^3$, i.e. define $(x_1,x_2,x_3)\cdot(y_1,y_2,y_3) = (x_1y_1,x_2y_2,x_3y_3)$.

The field axiom which causes trouble is $\forall x \:((x=0) \lor \exists y\: (xy = 1))$, i.e. the existence of multiplicative inverses. To show that $\mathbb{R}^3$ is not a field, try to find the multiplicative inverse of $x = (1,1,0)$ (note that $x \neq 0$!)...

The axiom about multiplicative inverses is, btw, the only axiom which can cause trouble. You can prove that whenever you have a structure defined purely by equations, i.e. by axioms of the form $\forall x_1 \ldots \forall x_2 t_1(x_1,\ldots,x_n) = t_2(x_1,\ldots,x_n)$ for some terms $t_1$, $t_2$ in $x_1,\ldots,x_n$, then the set-theoretic product of arbitrary many such structure together with component-wise definition of all functions will obey the same axioms. Thus, the product of groups is a group, the product of rings is a ring, ...

For fields this breaks down because the condition $x \neq 0$ in the axiom about multiplicative inverses prevents that axiom from being written as an equation. But this axiom is the only axiom which cannot be written as an equation, and since this axiom is what turns a ring into a field, you get that the product of arbitrarily many fields is a ring.


It's easy to see that there are zero divisors: $(1,0,0)\cdot (0,1,0)=(0,0,0)$, for example.

So, it of course cannot be a field... it is not even an integral domain!


Suppose that we want to put a field structure on the $\mathbb{R}$-vector space $\mathbb{R}^3$. Suppose $K$ is isomorphic to $\mathbb{R}^3$ as an $\mathbb{R}$-vector space and is a field. Then $K$ contains a copy of $\mathbb{R}$ as a subfield. Thus, $\mathbb{R}^3$ is an algebraic extension of $\mathbb{R}$ of degree $3$. But all algebraic extensions of $\mathbb{R}$ are either or degree $1$ or $2$ because all algebraic field extensions of $\mathbb{R}$ can be embedded into $\mathbb{C}$ and $\mathbb{C}$ has dimension $2$ as an $\mathbb{R}$ vector space. Thus, $\mathbb{R}^3$ can not be equipt with a field structure.

In this answer I have assumed that you are viewing $\mathbb{R}^3$ as an $\mathbb{R}$-vector space. If you only want to view it as an abelian group then this argument won't work (I don't think).

  • $\begingroup$ Nice answer! Would you care to explain a bit more though? $\endgroup$ – user1729 Oct 19 '12 at 14:21
  • 1
    $\begingroup$ Yes - but I have to pop out for a few hours. I promise to edit it to make it more clear when I get back. Sorry! $\endgroup$ – Conrad Oct 19 '12 at 14:27

There are two "canonical" products on $\mathbb{R}^3$, the cross product and component-wise multiplication. For the latter it's very easy to prove, that it's not a field, since $$\left(\begin{array}1 1 \\ 0 \\ 0\end{array}\right) \neq \left(\begin{array}1 0 \\ 0 \\ 0\end{array}\right),$$ but it has obviously no inverse element.

For the former you can't find a 1, since the cross product is always orthogonal to both its right and left hand side.

  • 5
    $\begingroup$ The cross-product is also not commutative. $a \times b = -b\times a$ $\endgroup$ – fgp Oct 19 '12 at 10:26
  • 6
    $\begingroup$ Indeed. It's also not associative. But it's something called product and it maps $\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}^3$, which made it a candidate :) $\endgroup$ – filmor Oct 19 '12 at 10:30

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.