Ring homomorphism from $M_3(\mathbb{R})$ into $\mathbb{R}$ I was working on this problem 

Find all ring homomorphisms from $M_3(\mathbb{R})$ into $\mathbb{R}$.

My attempt:-
I found that if we have any ring homomorphism $\phi$, then $\ker(\phi)$ should be either zero or the entire ring (since $M_3(\mathbb{R})$ is simple) and in case the ideal is the entire ring the ring homomorphism should be zero mapping. But I am  not sure if we have case where $\ker(\phi)={0}$. 
 A: There is no injective ring homomorphism $\phi:M_3(\mathbb R)\to\mathbb R$ for simple reasons: consider the matrix $A=\pmatrix{0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0}$. Notice that $A^3=0$. Then $\phi(A)^3=0$, so $\phi(A)=0$. Since $\phi$ is injective we get $A=0$, false.
A: A ring homomorphism from any matrix ring containing $\mathbb R I$ to $\mathbb R$ is also an algebra homomorphism: 


*

*If $\phi$ is not identically $0$, $\phi(I) = 1$ because $\phi(x) = \phi(I x) = \phi(I) \phi(x)$.  

*If $n \in \mathbb Z$, we then get $\phi(nI) = n \phi(I) = n$.

*If $r$ is rational, $\phi(rI) = rI$.

*If $0 \le c \in \mathbb R$, then $\phi(cI) = \phi((\sqrt{c}I)^2) = \phi(\sqrt{c}I)^2 \ge 0$.

*If $c \in \mathbb R$, take sequences of rationals $a_n, b_n \to c$ with $a_n \le c \le b_n$.  Then $a_n = \phi(a_n I) \le \phi(c I) \le \phi(b_n I) = b)n$, from which we conclude $\phi(cI) = c$.

*Thus $\phi(c x) = \phi(cI) \phi(x) = c \phi(x)$, i.e. $\phi$ is an algebra homomorphism.


In particular it is a linear mapping over $\mathbb R$.
Since $\dim M_3(\mathbb R) > \dim \mathbb R$ (as vector spaces over $\mathbb R$), it's impossible for the kernel to be $\{0\}$.
A: There is no injective ring homomorphism since every matrix of the form $AB-BA$ must be mapped into $0$. To conclude, it is well known that there exist some $AB-BA \neq 0$.
A: It is known that every injective ring homomorphism $\phi:M_n(K)\rightarrow M_m(K)$ must satisfy $n\le m$, see for example Lemma $3.2$ here. For $m=1$ and $n=3$ this is a contradiction. One could also use the Amitsur-Levitzki theorem.
