Showing that $(\mathbb{R},+,\cdot)$ and $(\mathcal{M}_2(\mathbb{R}),+,\cdot)$ are not isomorphic rings Prove that $(\mathbb{R},+,\cdot)$ and $(\mathcal{M}_2(\mathbb{R}),+,\cdot)$ are not isomorphic rings.
I came up with the following argument, but I am not sure it works : the equation $x^2=1$ has only two solutions in $\mathbb{R}$, whereas in $\mathcal{M}_2(\mathbb{R})$ it has infinitely many (for instance, any matrix $A=\left(\begin{matrix} 1 & 0 \\ a & -1 \\ \end{matrix} \right)$ where $a$ is an arbitrary real number works). Is this enough to conclude that the two rings are not isomorphic?
EDIT: Thanks everyone and sorry for forgetting to type that minus sign !

The method works like this: any isomorphism of rings preserves addition and multiplication. Therefore, suppose $f$ is an isomorphism between these two rings. Then $f$ preserves addition and multiplication $\implies$ a root of $X^2=1$ in $\mathbb R$ is sent to a root of the same equation in $M_2(\mathbb R)$. The converse is true by the same argument. So if there were an isomorphism, the number of roots of $X^2=1$ must be the same in both rings.
You just need to find a root of $X^2=1$ other than $\pm I$ in the matrix ring, e.g.
$$\left(\matrix{1&0\\0&-1}\right)$$
More generally, if there's any other ring property that one ring has but the other does not, then the rings are not isomorphic.
It remains to make precise what is meant by a "ring property". You can understand a ring property as a property preserved by ring isomorphisms. Here are some examples:


*

*cardinality: because an isomorphism is a bijection between sets. (So that $\mathbb Q$ and $\mathbb R$ are never isomorphic);

*commutativity: just like the other answer suggests;

*invertibility of elements: all nonzero elements in $\mathbb R$ are invertible but this does not hold for the matrix ring;

*being Noetherian, Artinian, unital, simple, semisimple, local, and so on.
You can also generalize this idea to other categories.
