If $R$ is an integral domain and $\mathrm{Frac}(R)$ is its field of quotients, the canonical map $\psi: R \rightarrow Frac(R)$ is a bijection iff $R$ is a field? Why? How do I show this?
Elements of $\mathrm{Frac}(R)$ are $[a,b] = \{(c,d) | ad = bc, c\in R, d\in R - \{0\}\}$ with $a,b \in R$, and $b$ nonzero. The mapping here is $\psi(r) = [r,1]$.
If $R$ is an integral domain, then $\psi:R \to \mathrm{Frac}(R)$ is an injection automatically.
If it were a surjection, then $\mathrm{Frac}(R) \cong R/\{0\}\cong R$, so $R$ was a field.
On the other hand, if $A$ is a field containing $R$, then it contains a subfield isomorphic to $\mathrm{Frac}(R)$. Hence, if $R$ is a field, it contains a subfield isomorphic to $\mathrm{Frac}(R)$, so there is an injection $\mathrm{Frac(R)} \to R$ that factors through $\psi$, so it is in fact a right inverse, so $\psi$ was a surjection.