# 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:

1. 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?

2. How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!

• math.stackexchange.com/q/216820 – Jean Marie Jan 8 at 7:42
• In a field, a polynomial of degree $n$ cannot have more than $n$ roots. – Akiva Weinberger Jan 8 at 7:43
• Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $\Bbb C$. – Akiva Weinberger Jan 8 at 7:49
• I added the "field-theory" and "extension-field" tags to your post. Cheers! – Robert Lewis Jan 9 at 0:29
• While $\mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y \varepsilon+z \varepsilon^2$ where $\varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135 – James S. Cook Jan 9 at 0:48

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

$$\mathbf v = -\dfrac{b_1}{2} \pm \sqrt d \in \Bbb R, \tag{18}$$

$$d < 0, \tag{19}$$

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.

• Thanks for such a good explanation. I can't help but accept this answer – Silent Jan 9 at 7:31
• @Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again! – Robert Lewis Jan 9 at 7:34
• Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $\Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this. – Jyrki Lahtonen Jan 12 at 10:17
• @JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating! – Robert Lewis Jan 12 at 12:22
1. 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.
2. 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.