Given a field $K$, we know that $K$ can be viewed as a 1-dimensional vector space over itself. Can this be extended to modules? In other words if I were to take a ring $R$ with unity and considered the module of $R$ over itself, then is this module necessarily free? In particular I was wondering if the ring of Quaternions over themselves would form a free module.
2026-04-24 22:14:53.1777068893
Can Rings (with unity) over themselves always be viewed as free modules?
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A free $R$-module $M$ is one that admits a basis. You can easily show that this is equivalent to saying that $M \cong \oplus_I R$, where $I$ is the cardinality of some basis. (For noncommutative rings, different bases can have different cardinalities and it could happen in particular that $\oplus_I R \cong \oplus_J R$ for different indexing sets $I$ and $J$)
In particular, $R$ is free, viewed as a (left/right) $R$-module. $\{1\}$ is a basis.