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I am wondering if it is possible to construct a nonzero ring homomorphism from $M_n(\mathbb{Q})$ to $\mathbb{Q}$.

So far, I've been unsuccessful in constructing such a nonzero ring homomorphism. Is there a possible construction? If not, how can we prove this?

Thanks!

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    $\begingroup$ What will you map the nilpotent elements of $M_n(\mathbb Q)$ to? $\mathbb Q$ is a field, the matrix ring isn't. $\endgroup$ – Don Thousand Nov 8 at 20:39
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    $\begingroup$ @Don The nilpotent elements must go to zero. There are certainly ring homomorphisms from rings with zero divisors to fields, for example projection $F\times F\to F$. I don't think it can happen with this ring though because of lack of ideals. $\endgroup$ – Matt Samuel Nov 8 at 20:46
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    $\begingroup$ If $n\ge 2$ and $f:M_n(\mathbb Q)\to\mathbb Q$ is a non-zero ring homomorphism, then $\ker f=(0)$, so $M_n(\mathbb Q)$ is isomorphic to a subring of $\mathbb Q$ (which doesn't contain non-commutative subrings). $\endgroup$ – user26857 Nov 8 at 21:17
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Here is yet another argument. Let $\theta:M _n (\mathbb Q)\to \mathbb Q $ be linear. It is straightforward to check then that $\theta=\operatorname {Tr}(A\cdot) $ for some $A\in M _n (\mathbb Q)$.

If $\theta $ is multiplicative, then in particular $\theta (BC)=\theta (B)\theta (C)=\theta (CB) $ for all $B,C $. Then $$ \operatorname {Tr}(ABC)=\operatorname {Tr}(ACB)=\operatorname {Tr}(BAC). $$ So $\operatorname {Tr}((AB-BA)C)=0$ for all $B,C $. Taking $C=(AB-BA)^T $ we obtain $AB-BA =0$. So $A $ commutes with all matrices, making it a scalar multiple of the identity. Thus $\theta $ is a scalar multiple of the trace; for $n\geq2$ it is easy to check that it can only be multiplicative if $A=0$.

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Notice that, as $\Bbb Z$-module, $M_n(\Bbb Q)$ is generated by the matrices of rank $1$, all of which must necessarily be mapped to $0$ (provided $n\ge2$). This is the case, for instance, because if $\operatorname{rk}A=1$, then there are invertible matrices $L$ and $R$ such that $LAR^{-1}$ is nilpotent. Therefore $0$ is the only multiplicative and additive map $M_n(\Bbb Q)\to \Bbb Q$. It might be worth mentioning that it is not a homomorphism of unital rings, because it does not map $1$ to $1$.

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Yes, if $n=1$: the identity homomorphism.

Otherwise, no.

$M_n(\mathbb Q)$ is simple, so any nonzero ring homomorphism leaving it is injective.

But then $M_n(\mathbb Q)$ has many zero divisors if $n>1$, and those would have to map to zero divisors in $\mathbb Q$, of which there are $0$ or $1$, depending on how you like to count.

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    $\begingroup$ In general, if $a$ is a zero divisor, it is still possible for $f(a)$ not to be a zero divisor, as long as all $b$ such that $ba=0$ satisfy $f(b)=0$. $\endgroup$ – Gae. S. Nov 8 at 21:47
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    $\begingroup$ @Gae.S. I think you missed that phrase where I said the map is injective, because what you said is not an issue then. All such $b$ are equal to $0$. Of course what you said is true sometimes when the map isn’t injective. $\endgroup$ – rschwieb Nov 9 at 0:06
  • $\begingroup$ I agree with what you say. $\endgroup$ – Gae. S. Nov 9 at 6:49
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Another easy argument comes from looking at idempotents. If $\theta $ is the homomorphism and we consider the usual matrix units $\{E_{kj}\} $, from $E_{11}=E_{1k}E_{k1} $ and $E_{kk}=E_{k1}E_{1k} $ we get $$\theta (E_{kk})=\theta (E_{k1})\theta (E_{1k})=\theta (E_{1k})\theta (E_{k1})=\theta (E_{11}),\ \ \ k=1,\ldots,n$$ If $\theta\ne0$ then $$1=\theta (I_n)=\theta (\sum_kE_{kk})=n\theta (E_{11}). $$ As $\theta (E_{11})=1$ (because $E_{11}^2=E_{11}$ and $\theta \ne0$), we obtain $n=1$.

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