Whats wrong with this 3d extension of the reals? If I take a 3d number, which I'll write as $A=<a,b,c>$ (with $A_a = a, A_b = b, A_c = c$) , then I can define addition of $$A+B = \,<A_a+B_a,A_b+B_b,A_c+B_c>$$ and I'll get a group which nicely satisfies the required abelian properties. Its also an extension of the reals, since its identity is $<0,0,0>$, and $<a_1,0,0> + <a_2,0,0>\,\, =\,\, <a_1+a_2,0,0>$. 
If I then created multiplication according to the following rule $$A*B=\,<A_aB_a,A_aB_b+B_aA_b,A_aB_c+A_cB_a>$$ then I get something which is closed, associative, has an identity element, is commutative, and invertible (by multiplying by $<\frac{1}{a},-\frac{b}{a^2},-\frac{c}{a^2}>$. It also seems like its an extension of the reals, since its identity element is $<1,0,0>$, and multiplication of $<a_1,0,0> * <a_2,0,0> = <a_1a_2,0,0>$. Furthermore, division also has the same properties (except that of course you can't divide by 0 (so $a_2 \neq 0$), so its closed except for $<0,b_1,c_1>$). 
But I know that I have done something wrong, since I've read that there doesn't exist a 3d extension of the reals. What did I do wrong?
 A: Without even checking to see if your proposed operation is a sensible multiplication, you can see right away that $(0, 1, 0)^2=(0,0,0)$, so it transparently cannot have inverses for all of its nonzero elements.  (You could multiply this example by the inverse of $(0,1,0)$ on the left and arrive at $(0,1,0)=(0,0,0)$, an absurdity.) 
Your first clue would have been that you needed $a\neq 0$ for all your proposed inverses. 
Anyhow, it still could be a ring containing a copy of $\mathbb R$, even if it isn't a division ring. The result you are thinking of talks about finite dimensional $\mathbb R$ division algebras.
There are plenty of $3$-dimensional $\mathbb R$-algebras. For example, $\begin{bmatrix}\mathbb R & \mathbb R \\ 0 & \mathbb R\end{bmatrix}$.
In fact your ring is isomorphic to the subring of $M_3(\mathbb R)$ of elements of the form $\begin{bmatrix}a&0&0 \\ b&a&0\\ c&0&a\end{bmatrix}$ via the mapping $(a,b,c)\mapsto \begin{bmatrix}a&0&0 \\ b&a&0\\ c&0&a\end{bmatrix}$, a commutative local $3$-dimensional $\mathbb R$ algebra with a $2$-dimensional nilradical.
A: There doesn't exist a 3D field extension of the reals. And in a field, every nonzero element has a multiplicative inverse.
So if what you have is really a counterexample of that, then one must ask: What's the multiplicative inverse of elements of the form $\langle 0,b,c\rangle$?
