Help with proof of Frobenius Theorem Let $D$ be a finite dimensional associative division algebra over $\mathbb{R}$ with unit $1_{D}$. For all $a\in\mathbb{R}$ we identify the elements of the form $a\cdot1_{D}$ with $a$ itself and consider $\mathbb{R}$ to be a sub-ring of $D$. We will also idenitfy $1_{D}$ with $1\in\mathbb{R}.$
Theorem (Frobenius): Every finite dimensional associative division algebra over $\mathbb{R}$
  is isomorphic to $\mathbb{R},
 \mathbb{C}$
 or $\mathbb{H}$
  (the quaternions)
Guidance:


*

*Given $x\in D$
show than the mapping $m_{x}:D\to D$
defined by $m_{x}\left(d\right)=x\cdot d$
is a linear transformation. Conclude using Cayley-Hamilton Theorem that $x$
is a zero of a polynomial with real coefficients. (Done)

*Use the Fundamental Theorem of Algebra to show $x$
is a zero a real polynomial of degree 1 or 2 (Done)

*Show that if each $x\in D$
is a zero of real polynomial of degree 1 then $D\cong\mathbb{R}$
(Done)

*Show than if the following claim is unture then there is an $x\in D$
satisfying the equation $x^{2}=-1$
(identifying $-1$ of the ring with the real $-1$
) - (No clue)

*Show that if in the conditions of the previous claim $D$
is a vector space of dimension $2$ over $\mathbb{R}$ then $D\cong\mathbb{C}$
(Done)

*Show that if $\dim D>2$
then the collection $V=\left\{ x\in D\,|\, x^{2}\leq0\right\}$
is a sub-space of $D$ of codimension $1$. (When writing for $d\in D$
that $d\leq0$
we assume that $d$
is in $\mathbb{R}$
as a subring of $D$
) - (No idea how to show the part about the dimension)

*Show that the mapping $B:V\to V$
given by $B\left(a,b\right)=-ab-ba$ is an inner product on $V$
(Done).

*Given an orthonormal basis of $V$
relative to $B$
denoted $\left\{ e_{1},....,e_{n}\right\} 
  (n\geq2)$
if $n=3$
then $D\cong \mathbb{H}$
and otherwise there is no such algebra $D$
for which the conditions of the theorem hold (No clue)
I'd really appreciate help (or really a proof) for parts 4,6,8.
Thanks in advance!
 A: The HW tag is making me a little uncomfortable giving you a full solution in spite of your request. Here is a fairly detailed indication of the right lines of thought, but I've tried to save you a few key steps / calculations.
Both (4) and (6) hang on the question,

How do you take an arbitrary element of $D$ that's not in $\mathbb{R}$, and change it by a real number to put it in $V$ (i.e. to make it have a negative real square)?

This question can be answered by taking any element $x\in D\setminus\mathbb{R}$ and considering the quadratic polynomial it's a root of. The discriminant must be negative because otherwise the polynomial would factor over $\mathbb{R}$, making $x$ the root of a linear polynomial, and then it would lie in $\mathbb{R}$. Complete the square to find a real number you can add to $x$ to make its square equal to (a positive multiple of) the discriminant.
For (4), you are assuming there is at least one such $x$. The above gives you a program to use it to find an element of $D$ with negative real square. You can then normalize by some real factor to make the square $-1$.
For (6), the same method tells you that every element of $D$ that's not in $\mathbb{R}$ can be changed by an element of $\mathbb{R}$ to lie in $V$. This shows that $D$ is the direct sum of $\mathbb{R}$ and $V$.
For (8), the key is that the fact that $e_1,\dots,e_n$ is an orthonormal basis for $B$ tells you that $e_i^2 = -1/2$ and $e_ie_j=-e_je_i$. Actually I would prefer aesthetically to renormalize $B$: using $B(x,y)=-(xy+yx)/2$ instead, an orthonormal basis has $e_i^2=-1$. Note that this means $(e_ie_j)^2 = -e_ie_j^2e_i = +e_i^2=-1$, and $(e_ie_je_k)^2=-(e_ie_je_k)(e_ke_je_i)=+1$ similarly. An element of a real division algebra whose square is $1$ must be $\pm 1$, because it satisfies $x^2-1$, which factors over the reals. It follows that $e_ie_je_k=\pm 1$ for distinct $i,j,k$.
Wikipedia takes it from here by passing to a minimal subset of the $e_i$'s that generate $D$ as an $\mathbb{R}$-algebra, and noting that if there is one $e_i$ in this set we have $\mathbb{C}$; if two, we have $\mathbb{H}$; and if more than two, we have a contradiction. I prefer a more direct calculation with all the $e_i$'s. If $n=1$, we have $\mathbb{C}$. If $n=2$ then $e_1e_2$ is a linear combination of $e_1$ and $e_2$ because it has a negative square; this can yield a contradiction. If $n=3$ we get the quaternions. If $n>3$, then $e_1e_2e_3$ and $e_1e_2e_4$ are both $\pm 1$, and this contradicts the linear independence of $e_3$ and $e_4$ together with the fact that $D$ is a division algebra.
