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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, ...


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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 ...


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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, ...



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