Show that two rings are not isomorphic I don't know how to show (or why) 
$M_{2\times2}\mathbb{(R)}$ is not isomorphic to $\mathbb{R}[x]/(x^4-1)$
does it have something to do with the order of coset representatives of the quotient group?
 A: I'm sure there are plenty of ways to demonstrate this, but I think that Cameron's suggestion is probably easiest. Since the quotient of any commutative ring is commutative, $\frac{\mathbb{R}[x]}{x^4-1}$ is commutative.
But $M_{2\times 2}(\mathbb{R})$ is not commutative.
For instance,
$$ \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}$$
but $$ \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}_.$$
A: Here is a completely different approach: (and certainly overkill)
Every $M\in M_{2\times 2}(\mathbb{R})$ satisfies a polynomial of degree $2$ over $\mathbb{R}$.  But then the $\mathbb{R}$-algebra generated by $M$ has dimension at most $2$ over $\mathbb{R}$, while $\mathbb{R}[X]/(X^4-1)$ (which is generated by the single element $X$) has dimension $4$.
A: Can't beat the commutative/noncommutative observation, but there is another way: the quotient has nontrivial ideals (eg $(x^2-1)/(x^4-1))$ whereas the matrix ring has no nontrivial ideals. 
