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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 $x, ... 1 Let$M$be a non-zero divisible$R$module, finitely generated by$m_1,\ldots,m_n$. Say$r\in R$is non-zero. We need to show$r$has an inverse in$R$. Since$M$is divisible, there exist elements$m’_1,\ldots,m’_n\in M$such that$rm’_i=m_i$for$i=1,2,\ldots,n$. We can write each$m’_i$as an$R$-linear combination of the$m_j$, say$$m’_i=\sum_{j=1}^n ... 1 Proof in short: If we can construct a divisible cyclic submodule, it must be isomorphic to$R$, and therefore$R$is a field. We will do this by showing that any element of a minimal set of generators of a divisible$R$-module generates a cyclic divisible submodule by showing that otherwise we obtain a smaller set of generators. Proof: Suppose$x_1,\ldots, ...