There isn't a product operation that is commmutative on $ \mathbb{R}^{n} $ that satisfies all the field axioms for $ n \geq 3 $.

Question

This proof is broken down into simple easy algebra and vector questions. I would like to discuss different answers and approaches.

Please see pg 162-163 on books.google.ca/books?isbn=0387290524

There are 5 questions from 7.6.3 - 7.6.7. You can read the paragraph above 7.6.3. You can also read the first part on quarternions. Exclude the "rotations of ijk space section.

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Here is what I have tried.

Q1: I used the Pythagorean Theorem to get the norm equal to $ \sqrt{2} $. Then I used the property $ \text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v) $ to show that that $ 2 = \text{Norm}(1 - i^{2}) $. Am I right?

Q2: I used the property $ \text{Norm}(uv) = \text{Norm}(u) \text{Norm}(v) $ since $ \text{Norm}(i) = 1 $. But I don’t know how to show $ i^{2} = -1 $. One says to use the Triangle Inequality. I guess the equality implies that $ i^{2} $ and $ 1 $ are collinear.

Q3: And thus I don’t know how to do this.

Q4: The map $ p \longmapsto pi $ multiplies all distances in $ \mathbb{R}^{n} $ by $ |i| = 1 $, since $ |pi| = |p||i| $. For any points $ p_{1} $ and $ p_{2} $ in $ \mathbb{R}^{n} $, $ |p_{1} * i - p_{2} * i| = |(p_{1} - p_{2})i| = |p_{1} - p_{2}||i| $. Therefore, the distance $ |p_{1} - p_{2}| $ between any two points is multiplied by $ |i| = 1 $. Therefore, the map is an isometry of $ \mathbb{R}^{n} $. Therefore, since $ i $ and $ j $ are perpendicular directions, $ i * i $ and $ i * j $ are still perpendicular by the isometry. (An isometry preserves the distance between points.)

Still not sure why $ \mathbf{1} $ and $ ij $ are perpendicular.

Q5: From $ jiij= j i^{2} j = jj i^{2} = j^{2} i^{2} = - \mathbf{1} * - \mathbf{1} = \mathbf{1} $, therefore $ 1 = -1 $, which is a contradiction.

Answer 1

Here is a field theory fact that I regard as being at the (advanced) undergraduate level:

Fact: Let $F$ be a field, let $K/F$ be a finite degree field extension, and let $\overline{F}$ be an algebraically closed extension of $F$. Then there is a homomorphism $\varphi: K \rightarrow \overline{F}$ such that $\varphi(x) = x$ for all $x \in F$.

(Proof: $K/F$ can be decomposed into a tower of simple extensions, so by induction it suffices to assume that $K/F$ is simple, i.e., $K = F[t]/(f(t))$ for some irreducible polynomial $f$. Since $\overline{F}$ is algebraically closed, $f(t)$ has a root, say $\alpha$, in $\overline{F}$, and then sending $p(t) \in F[t] \mapsto p(\alpha)$ gives a homomorphism $F[t] \rightarrow \overline{F}$ with kernel $(f(t))$, so induces a field embedding $\varphi: K \rightarrow \overline{F}$ such that $\varphi(x) = x$ for all $x \in F$.)

Using this fact, the assertion that for all $n \geq 3$ there is no $\mathbb{R}$-algebra structure on $\mathbb{R}^n$ which makes it into a field is seen to be equivalent to the Fundamental Theorem of Algebra: $\mathbb{C}$ is algebraically closed. Indeed, if $\mathbb{C}$ were not algebraically closed it would admit a finite extension $K/\mathbb{C}$ of degree $d > 1$, and then $K/\mathbb{R}$ would be an $\mathbb{R}$-algebra of dimension $2d > 2$ which is a field. The fact above shows the converse: any finite degree field extension $K/\mathbb{R}$ must embed into the algebraically closed field $\mathbb{C}$, giving a tower $\mathbb{C}/K/\mathbb{R}$. Comparing degrees gives $K = \mathbb{R}$ or $K = \mathbb{C}$.

Since there are many undergraduate level proofs of the Fundamental Theorem of Algebra -- my favorite uses a little Galois theory and Sylow theory to establish a more general, purely field-theoretic result: see the implication (ii) $\implies$ (iii) in the Grand Artin-Schreier Theorem in these notes -- this reduction serves to give an undergraduate level proof of the theorem at hand. (Note also that the proof of the -- deeper, I would say -- theorem of Frobenius that Gerry Myerson cites also uses the Fundamental Theorem of Algebra.)

This argument also shows that if you want to classify the finite degree field extensions of $\mathbb{R}$ you have to prove the Fundamental Theorem of Algebra, which is certainly not trivial and is, so far as I know, not amenable to a proof using only the tools of pre-calculus, elementary vector geometry and single variable calculus. The OP's question is a little weaker than this because it also postulates a multiplicative norm on the field. Unfortunately I didn't understand exactly what properties of the norm the OP wants to assume: merely assuming $|xy| = |x| |y|$ and $|1| = 1$ on a finite-dimensional field extension of $\mathbb{R}$ is not enough to imply, for instance, that $|1+i| = \sqrt{2}$: perhaps the norm maps every element to $1$! So to me it is simpler to ignore the norm entirely.

Answer 2

Every finite-dimensional associative division algebra over the real numbers is isomorphic to the reals, the complex numbers, or the quaternions, by a theorem of Frobenius. A proof is here.

EDIT: Maybe a simpler reference would be Kenneth O May, The impossibility of a division algebra of vectors in three dimensional space, The American Mathematical Monthly, Vol. 73, No. 3, Mar., 1966, pages 289 to 291. I don't know whether this is freely accessible on the web, but libraries should have access or be able to get access.

Answer 3

Here is a possible shortcut.

From Q1 and Q2, you have an element $i$ such that $i^2=-1$.

So $\mathbb{R}^n$ contains an isomorphic copy of $\mathbb{C}$.

Hence $\mathbb{R}^n$ is a finite extension of $\mathbb{C}$.

Since $\mathbb{C}$ is algebraically closed, $\mathbb{R}^n$ is isomorphic to $\mathbb{C}$. And $n=2$.