If $R$ is a domain but not a field and $Q=Frac(R)$, prove that $Hom_{R}(Q,R)=\{0\}$. I thought this problem is really easy and give the following 5 line proof:
We assume $R$ is commutative as the book assumed. Suppose we have a non-zero homomorphism $\phi$, then $\ker(\phi)$ is a proper ideal of $Q$. But $Q$ is a field and has only two ideals. Since we assume $\ker(\phi)\not=Q$, $\phi$ must be injective. But this is impossible, since $\phi\in Hom(Q,R)$ and $1$ must be mapped to 1, every non-zero element in $r$ now has an inverse from $\phi(1/r)$. Since $R$ is not a field this is absurd.
However the professor give me zero mark on this problem. I think I might have ignored something really, really basic. So I venture to ask at here. I apologize for my shallowness. I read this again, and other than $R$ could be non-commutative I do not see where it went wrong.  
 A: This $\phi$ is a homomorphism of $R$-modules, not of rings. As such, $\ker\phi$ is an $R$-submodule of $Q$, and is not necessarily an ideal of $Q$ -- it needs to be closed only under multiplication by elements of $R$, not by all elements of $Q$. As Andrew points out, it is not immediate that it must send $1$ to $1$.
A: The important fact here is that $Q$ is a divisible $R$-module, while $R$ is divisible (as $R$-module) iff $R$ is a field. 
Take $f\in\operatorname{Hom}_R(Q,R)$ and set $f(1)=r\in R$. Assume that $r\neq 0$. You get $$r=f(a\cdot\frac{1}{a})=af(\frac{1}{a})\;\;\;\; (*)$$ 
for all $a\in R$, $a\neq 0$. If you take now $a=rx$, with $x\in R$, $x\neq 0$, you get $r=(rx)f(\frac{1}{rx})$, so $1=xf(\frac{1}{rx})$, that is, $x$ is invertible. But $R$ is not a field, contradiction. It remains that $f(1)=0$. Then, from $(*)$, it follows $f(\frac{1}{a})=0$ for all $a\in R$, so $f(\frac{b}{a})=bf(\frac{1}{a})=0$, that is, $f=0$.
A: I do not know if the following proof works:
Let $\phi$ be a non-trivial homomorphism between $Q$ and $R$. Then $\ker(\phi)=K$ is not equal to $Q$, thus we may select element $q=\frac{r}{s}$ in $Q$ that is not in $K$. Now assume $\frac{a}{b}$ is in $K$, then we have $$\frac{a}{b}*\frac{br}{as}=q$$ This implies that $$\phi(\frac{a}{b})*\phi(\frac{br}{as})=\phi(q)$$ while we know $\phi(\frac{a}{b})=0$, $\phi(q)\not=0$. This contradiction showed $\ker(\phi)$ must be $Q$.
