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

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?

• 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. – TrostAft Dec 2 '18 at 7:12
• Wait, is this true? What about the determinant? Oh no nevermind that's only a group homomorphism. – TrostAft Dec 2 '18 at 7:22

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)
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.