An integral domain and its field of fractions.

Question

I'm trying to solve the following problem:

Let $R$ be a integral domain which is not a field and $K$ its fractions field. Show that a non-zero module $R$-homomorphism from $K$ to $R$ does not exist.

But I get stuck, someone can give some hint?

Thanks a lot!

Answer 1

Suppose $a \in R$ is a non-invertible element. Suppose $f: K \longrightarrow R$ is an $R$-linear map such that $f(1) \neq 0$. Then $$ af(1)f \left( \frac{1}{af(1)} \right) = f \left( \frac{af(1)}{af(1)} \right) = f(1) $$ hence $1= a f \left( \frac{1}{af(1)} \right)$ and this contradicts that $a$ is not invertible.

So it must be $f(1)=0$. And now for all $x, y \in R \setminus 0$ $$ y f \left( \frac{x}{y} \right) = f(x) = x f(1) =0 $$ so $f$ must be $0$.

Answer 2

Hint: let $f\colon K\to R$ be a homomorphism of $R$ modules; suppose $f\ne0$ and take $a/b\in K$ with $f(a/b)=r\ne0$. Then $f(a)=f(b(a/b))=bf(a/b)=br\ne0$. Next $br=f(a)=f(br(a/br))=brf(a/br)$, so $f(a/br)=1$. Can you go on?