Prove any ring homomorphism between $M_2(\mathbb{R})$ and $\mathbb{R}$ is trivial

Question

Prove any ring homomorphism between $M_2(\mathbb{R})$ ($2\times 2$ matrices) and $\mathbb{R}$ is trivial.

I am not looking for answers, I just want to know how to approach these types of problems (how to prove that any homomorphism must be trivial? What is the typical way of showing these results?) I am guessing we have to show that there is a property in one that is not in the other, but is there a rigorous way to do this?

Answer 1

Like you say, there are properties that ring homomorphisms have to preserve. A common one to check is idempotence. If you know more things about your domain, there are more properties. Say, for example, that your domain is a field. Then you know that any ring homomorphism from our domain also maps to another integral domain.

Answer 2

A lot of ring properties aren't preserved under homomorphism - unless that homomorphism happens to be injective, or maybe surjective. So, to control that, we look at the map as a whole.

The kernel of a ring homomorphism is a two-sided ideal. What are the two-sided ideals of the matrix ring? (Definition: a left ideal is closed under left multiplication by ring elements ($rx\in I$ if $x\in I$ and $r\in R$), a right ideal is closed under right multiplication by ring elements ($xr\in I$ if $x\in I$ and $r\in R$), and a two-sided ideal is closed under both)

Answer 3

You should look at kernel of the homomorphism. Which in turn means, you should look at possible ideals of the (domain) ring. For the homomorphism to be trivial, the kernel should be the entire ring.

In your particular question, it can be shown that $M_{2}(\Bbb{R})$ has only two ideals, namely itself and the zero ideal. Thus either the homomorphism will be one-one or it will be trivial.