Turning $\mathbb R^n$ into field I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:

If it were possible to define multiplication in $\mathbb R^3$ so as to make $\mathbb R^3$ a field including $\mathbb C,$ we could argue as follows: for every $\bf x$ in $\mathbb R^3$, the vectors $1,\bf x,\bf x^2,\bf x^3$ would be linearly dependent. Hence for each $\bf x$ in $\mathbb R^3,$ a relation of the form $a_0+a_1{\bf x}+a_2{\bf x^2}+a_3{\bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.

I have these couple of questions:


*

*Does above argument show that $\mathbb R^3$ can't be made a field?  Or just that $\mathbb R^3$ can't be field such that $\mathbb C$ is its subfield?

*How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!
 A: We can show that $\Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $\Bbb R$ without assuming such a field contains a subfield isomorphic to $\Bbb C$ as follows:
If $\Bbb R^3$ were such a field, we would have
$[\Bbb R^3:\Bbb R] = 3; \tag 1$
being an extension field of $\Bbb R$, $\Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1\Bbb R = \Bbb R1$ isomorphic to $\Bbb R$ in the usual manner, that is
$\Bbb R \ni r \leftrightarrow r1 \in 1\Bbb R \subsetneq \Bbb R^3; \tag 2$
by virtue of (1), there exists 
$\mathbf v \in \Bbb R^3 \setminus \Bbb R1 \tag 3$
such that $1, \mathbf v, \mathbf v^2, \mathbf v^3$ are linearly dependent over $\Bbb R1 \cong \Bbb R$; that is
$\exists c_i \in \Bbb R, \; 0 \le i \le 3, \tag 4$
not all $c_i$ zero, with
$c_3 \mathbf v^3 +c_2 \mathbf v^2 +  c_1 \mathbf v + c_0 = 0; \tag 4$
let us first consider the case
$c_3 = 0; \tag 5$
then
$c_2 \mathbf v^2 +  c_1 \mathbf v + c_0 = 0; \tag 6$
now if
$c_2 = 0, \tag 7$
then if 
$c_1 = 0 \tag 8$
as well, we find
$c_0 = 0, \tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 \ne 0 \tag{10}$
we may write
$\mathbf v = -\dfrac{c_0}{c_1} \in \Bbb R 1 \cong \Bbb R, \tag{11}$
which contradicts (3); thus we have that
$c_2 \ne 0, \tag{12}$
and we may write (6) as
$\mathbf v^2 + b_1 \mathbf v + b_0 = 0, \tag{13}$
where
$b_i = \dfrac{c_i}{c_2} \in \Bbb R; \tag{14}$
we write (13) as
$\mathbf v^2 + b_1 \mathbf v = -b_0, \tag{15}$
and complete the square:
$\left (\mathbf v + \dfrac{b_1}{2} \right )^2 = \mathbf v^2 + b_1 \mathbf v + \dfrac{b_1^2}{4}  = \dfrac{b_1^2}{4} - b_0 = d; \tag{16}$
if
$d \ge 0, \tag{17}$
(16) yields
$\mathbf v = -\dfrac{b_1}{2} \pm \sqrt d \in \Bbb R, \tag{18}$
in contradiction to (3); thus,
$d < 0, \tag{19}$
and (16) becomes
$\dfrac{1}{{\sqrt{-d}}^2} \left (\mathbf v + \dfrac{b_1}{2} \right )^2 = -1, \tag{20}$
which shows the existence of an element 
$\mathbf i \in \Bbb R^3 \tag{21}$
with
$\mathbf i^2 = -1, \tag {22}$
and in the usual manner we see that the subalgebra
$\Bbb R + \Bbb R \mathbf i = \{ s + t \mathbf i \mid s, t \in \Bbb R \} \cong \Bbb C \tag{23}$
is a subfield of $\Bbb R^3$ with
$[\Bbb C: \Bbb R] = 2; \tag{24}$
but this is impossible since it implies
$3 = [\Bbb R^3:\Bbb R] =[\Bbb R^3:\Bbb C] [\Bbb C: \Bbb R] = 2[\Bbb R^3:\Bbb C]; \tag{25}$
but $2 \not \mid 3$; we conclude then that no such $\mathbf v$ satisfying (6), (13) can exist in $\Bbb R^3$.
Now if
$c_3 \ne 0, \tag{26}$
then $\mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = \dfrac{c_i}{c_3}, \; 0 \le i \le 2, \tag{27}$
we obtain the real monic cubic
$p(\mathbf v) = \mathbf v^3 +b_2 \mathbf v^2 +  b_1 \mathbf v + b_0 = 0, \tag{28}$
which as is well-known always has a root
$r \in \Bbb R, \tag{29}$
whence
$p(\mathbf v) = (\mathbf v - r)q(\mathbf v) \tag{30}$
for some monic real quadratic polynomial $q(\mathbf v)$; thus,
$(\mathbf v - r)q(\mathbf v) =p(\mathbf v) = 0; \tag{31}$
but 
$\mathbf v - r \ne 0 \tag{32}$ 
since
$\mathbf v \notin \Bbb R; \tag{33}$
it follows that
$q(\mathbf v) = 0, \tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $\Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $\Bbb R^3$ contains a subfield isomorphic to $\Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $\Bbb R^3$ is an extension field of $\Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $\Bbb R^3$ has a subfield isomorphic to $\Bbb C$ to show that $\Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $\mathbf x$ than the usual complex numbers falls once we have $\Bbb C \subset \Bbb R^3$, for then the familiar factorizations in $\Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $\mathbf x$ must lie in $\Bbb C$; we need look no further.
A: *

*Just a field containing $\Bbb C$. Which has problems all on its own, as $\Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $\Bbb C$. So there are many reasons this wouldn't work.

*It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $\Bbb R^3$ field doesn't give us any numbers $\Bbb C$ doesn't already have (which is impossible: $\Bbb C\cong\Bbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.

