0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Typically when approaching a question that regards any map, I try and consider an arbitrary one, and then try to manipulate a contradiction / or force a conclusion. Like here I might try and assume I have a nontrivial map and try to find a contradiction. Looking at the rings, notice that somehow multiplication on $M_2(\mathbb{R})$ must be respected under the hom. on $\mathbb{R}$ which seems very restrictive, since one is a ring, and another a field. $\endgroup$ – TrostAft Dec 2 '18 at 7:12
  • $\begingroup$ Wait, is this true? What about the determinant? Oh no nevermind that's only a group homomorphism. $\endgroup$ – TrostAft Dec 2 '18 at 7:22
1
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.