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!
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.