Showing that $(\mathbb{R},+,\cdot)$ and $(\mathcal{M}_2(\mathbb{R}),+,\cdot)$ are not isomorphic rings

Question

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 !

Answer 1

One is commutative; the other isn't.

Answer 2

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.

Answer 3

In general, if you define a set in terms of only the properties of the ring - such that "the set of $x\in R$ so that $x^2=1_R$" or "the set of $x\in R$ so that $xy=yx$ for all $y\in R$", it will be preserved under homomorphism - in particular, if you let $$S_R=\{x\in R : x^2 = 1_R\}$$ you can prove that if $f:R\rightarrow T$ is a (unital) ring homomorphism, then $$f(S_R)\subseteq S_T$$ just by noting that if $x^2=1_R$ then $f(x^2)=f(1_R)$ which reduces to $f(x)^2=1_T$.

If $f$ is an isomorphism, we would then have $f(S_R)\subseteq S_T$ and $f^{-1}(S_T)\subseteq S_R$, which would, since $f$ is a bijection between $R$ and $T$, imply that $f$ is a bijection between $S_R$ and $S_T$. Thus, the fact that the set $S_R$ has size $2$ if $R=\mathbb R$ and is infinite if $R=M_2(\mathbb R)$ implies that the rings are not isomorphic. This reasoning extends to a great many varieties of proof where you identify some property that is not shared and can be encoded in a way that is preserved by homomorphisms - and, indeed, this method encompasses looking at the set of pairs of elements that don't commute, and can be applied to structures other than rings.

A more general way to think about this is that the assignment of a ring $R$ to the set $S_R$ is an example of a functor, since maps between rings can be transformed into maps between these sets - so this sort of argument is really passing to a category we understand better (sets) to solve a problem about rings.

(Note that your choice of solutions to $x^2=I$ does not work, but if you just look at matrices conjugate to $\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$ you get an infinite class of roots to $x^2=I$)

Answer 4

The way you approached a problem is correct. If you can find a ring property which is valid in one ring but not in the other, those rings cannot be isomorphic. The property you chose "has exactly two solutions of $x^2=1$" is a good choice, but you have not (as mentioned in comments) really shown that it doesn't hold in $\mathcal{M}_2(\mathbb{R})$.

Let's therefore fix your proof by noticing that there are at least four different matrices that have this property:

$$M_1=\left[\begin{matrix}1&0\\0&1\end{matrix}\right], M_2=\left[\begin{matrix}-1&0\\0&-1\end{matrix}\right], M_3=\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]\text{ and } M_4=\left[\begin{matrix}-1&0\\0&1\end{matrix}\right]$$

How do you now proceed with the actual proof that $\mathbb{R}$ and $\mathcal{M}_2(\mathbb{R})$ are not isomorphic? Suppose the contrary, i.e. that there is an isomorphism $\varphi:\mathbb{R}\mapsto\mathcal{M}_2(\mathbb{R})$. Then, $\psi=\varphi^{-1}:\mathcal{M}_2(\mathbb{R})\mapsto\mathbb{R}$ is also an isomorphism. Isomorphisms "carry over" any properties of elements, so you can easily show that $\psi(M_1)^2=\psi(M_1^2)=\psi(I)=1$ in $\mathbb{R}$, so $x_1=\psi(M_1)\in\mathbb{R}$ satisfies $x_1^2=1$. Similarly, $M_2$, $M_3$ and $M_4$ will give you three more elements of $x_2,x_3,x_4\in\mathbb{R}$ with the same property. Note also, as an isomorphism, $\psi$ is one-to-one, so all those four elements of $\mathbb{R}$ would be different from each other. This would imply that $x^2=1$ has at least four solutions in $\mathbb{R}$, which is a contradiction.