# Does there exist a nonzero ring homomorphism from the ring of square rational matrices to the ring of rational numbers?

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!

• What will you map the nilpotent elements of $M_n(\mathbb Q)$ to? $\mathbb Q$ is a field, the matrix ring isn't. – Don Thousand Nov 8 at 20:39
• @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. – Matt Samuel Nov 8 at 20:46
• 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). – user26857 Nov 8 at 21:17

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 $$A=0$$.

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 $$1$$.

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.

• 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 $f(b)=0$. – Gae. S. Nov 8 at 21:47
• @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. – rschwieb Nov 9 at 0:06
• I agree with what you say. – Gae. S. Nov 9 at 6:49

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 $$n=1$$.