**Proof.**
It is clear that (2) implies (1). Conversely, suppose that $g \otimes 1 = 1 \otimes g$. Choose generators $\{ s_ i\} _{i \in I}$ of $S$ as an $R$-module with $0, 1 \in I$ and $s_0 = 1$ and $s_1 = g$. Apply Lemma 10.107.10 to the relation $g \otimes s_0 + (-1) \otimes s_1 = 0$. We see that there exist $a_{i, j} \in R$ such that $g = \sum _ i a_{i, 0} s_ i$, $-1 = \sum _ i a_{i, 1} s_ i$, and for $j \not= 0, 1$ we have $0 = \sum _ i a_{i, j} s_ i$, and moreover for all $i$ we have $\sum _ j a_{i, j}s_ j = 0$. Then we have

\[ \sum \nolimits _{i, j \not= 0} a_{i, j} s_ i s_ j = -g + a_{0, 0} \]

and for each $j \not= 0$ we have $\sum _{i \not= 0} a_{i, j}s_ i \in R$. This proves that $-g + a_{0, 0}$ can be written as in (2). It follows that $g$ can be written as in (2). Details omitted. Hint: Show that the set of elements of $S$ which have an expression as in (2) form an $R$-subalgebra of $S$.
$\square$

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