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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.


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.

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Please use MathJax to write all mathematical expressions. I’m pretty sure that loving math is just as important as loving typography. –  Haskell Curry Mar 7 '13 at 4:00
    
Hint: Triangle inequality for q2. q3 is the same as q2, since it didn't matter whether you picked $i$ or $j$. For q4, since $i$ and $j$ are perpendicular, so is $i^2 $ and $ij$. For q5, you need to show that $(ij)^2$ = -1$, but this follows from q4. –  Calvin Lin Mar 7 '13 at 4:03
    
Still need some more great ideas and insights!! –  sarah Mar 7 '13 at 4:22
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@mathlover: Is this a currently assigned homework question for a course you're taking? If so, which course? –  Pete L. Clark Mar 7 '13 at 6:26
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The latter parts of the question are instructive in proving the result. Some of the answers refer to the deleted parts. The question is not diminished by their presence, yet some of the answers are by their absence. Unless you have a good reason to delete them, it would be best to leave them in. –  robjohn Mar 8 '13 at 1:08

3 Answers 3

up vote 2 down vote accepted

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.

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To be a bit more explicit about this: the above proof is at the undergraduate level, but not at the level of a typical first year (American) undergraduate. Like Gerry Myerson, I am skeptical that a classification of the finite dimensional extension fields of $\mathbb{R}$ can be proven using only high school algebra and vector geometry. –  Pete L. Clark Mar 7 '13 at 6:23

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.

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The theorem you are trying to prove may also be too hard for a first year math course. Is there anything that makes you think otherwise? –  Gerry Myerson Mar 7 '13 at 5:21
    
I have done this when I was an undergrad, and it was elementary. I don't recall a single bit of the argument, unfortunately. Way too long ago. –  1015 Mar 7 '13 at 5:33
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Your "answers" include statements like, "i dont know how to show $i^2=-1$," "i dont know how to do this," "Still not sure why $1$ and $ij$ are perpendicular," so it's hard to take this as evidence of the level of the question. In any event, think of the Frobenius theorem as something to look forward to. –  Gerry Myerson Mar 7 '13 at 5:41
    
I made a copy of May's note available here: math.uga.edu/~pete/May66.pdf. Note that in his setup he embeds $\mathbb{C}$ into $\mathbb{R}^3$ as the set of all points $(x,y,0)$ and assumes that the algebra structure on $\mathbb{R}^3$ reduces to the standard multiplication on $\mathbb{C}$. From the perspective of field theory or even linear algebra this is a trivial case, but this may be similar to the setup that the OP wants. –  Pete L. Clark Mar 7 '13 at 6:38
    
The copy jstor.org/stable/2315349 appears freely accessible too, but please check, it might be that the site recognizes I am browsing from my Uni. –  Andreas Caranti Mar 7 '13 at 10:25

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 $n=2$.

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