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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 $\ker(\phi)={0}$.

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  • $\begingroup$ Is $M_3(\mathbb{R})$ the algebra of $3\times3$ real matrices? $\endgroup$ – user228113 Apr 14 '15 at 19:55
  • $\begingroup$ @G.Sassatelli its matrix algebra or matrix ring. $\endgroup$ – henry Apr 14 '15 at 19:56
  • $\begingroup$ It makes a big difference whether it's an algebra or a ring. If you work with rings, then $\ker \phi$ can never be $0$ (use the fundamental isomorphism theorem), so your question is quite easy in this setting. $\endgroup$ – Alex M. Apr 14 '15 at 20:51
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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.

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A ring homomorphism from any matrix ring containing $\mathbb R I$ to $\mathbb R$ is also an algebra homomorphism:

  1. If $\phi$ is not identically $0$, $\phi(I) = 1$ because $\phi(x) = \phi(I x) = \phi(I) \phi(x)$.
  2. If $n \in \mathbb Z$, we then get $\phi(nI) = n \phi(I) = n$.
  3. If $r$ is rational, $\phi(rI) = rI$.
  4. If $0 \le c \in \mathbb R$, then $\phi(cI) = \phi((\sqrt{c}I)^2) = \phi(\sqrt{c}I)^2 \ge 0$.
  5. 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 $\phi(cI) = c$.
  6. 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 $\{0\}$.

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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 $AB-BA \neq 0$.

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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.

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  • $\begingroup$ Your answer make me wonder: Did we have $\mathbb{R}$-algebra homormphism $\phi:M_n(K)\rightarrow M_m(K)$ other than zero mapping ? and thanks for that reference its really useful. $\endgroup$ – henry Apr 14 '15 at 20:24
  • $\begingroup$ thanks,can you give me any reference to this or did this have any name. $\endgroup$ – henry Apr 14 '15 at 20:30
  • $\begingroup$ You mean, a name other than Lemma $3.2$ ? I don't know any name except for the Amitsur-Levitzki theorem in this context. $\endgroup$ – Dietrich Burde Apr 15 '15 at 8:25

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