----- > [!proposition] Proposition. ([[module is free iff admits basis]]) > Let $R$ be a [[ring]] and $M$ an $R$-[[module]]. Then $M$ is [[free module|free]] if and only if it admits a [[basis]]. In fact, $B \subset M$ is a [[basis]] if and only if the [[free module|natural (i.e., via the universal property of free modules)]] [[linear map|homomorphism]] $R^{\oplus B} \to M$ is an [[module isomorphism|isomorphism]]. > > So the choice of [[basis]] $B$ for a [[free module]] amounts to a choice of [[isomorphism]] $R^{\oplus B} \to M$, called a **coordinate isomorphism** [^2]. Note that [^1] $b \in M$ is identified with the indicator $\mathbb{1}_{\{ b \}}$ under said [[isomorphism]]. ^proposition > [!specialization] > - [[every vector space has a basis]] ^specialization [^1]: Refresh on the [[free module|universal property of free modules]] if confused. Basically: Assuming finitely-generated for simplicity, recall that $j:B=\{ b_{j} \}_{j=1}^{n} \to R^{\oplus B}$ maps $b_{j} \mapsto (0,\dots,0,\overbrace{1}^{j^{th }\text{ entry}},0,\dots,0)$, and diagram commutativity enforces $\varphi \circ j(b_{j})=\iota(b_{j})$, i.e., $\varphi\big( 0,\dots,0,1,0,\dots,0 \big)=b_{j}$. This is what I mean by saying the [[isomorphism]] $\varphi$ 'identifies' $b_{j}$ with its indicator. [^2]: I do have a full note on this from Math 217: [[coordinate isomorphism]]. It is basically the same as what is going on here, but it is also old and not too refined. > [!proof]- Proof. ([[module is free iff admits basis]]) > ~ > Let $\varphi$ be the natural homomorphism in question. > Suppose $M$ admits a [[basis]] $B$. Then the definition of [[linearly independent]] (which says $\varphi$ is an [[injection]]) and the definition of [[submodule generated by a subset|generating subset]] (which says $\varphi$ is a [[surjection]]) immediately imply $\varphi$ is an [[isomorphism]]. Hence $M$ is isomorphic to the free module $R^{\oplus B}$ and is thus itself free. > Conversely, suppose $\varphi$ is an [[isomorphism]]. Then $B$ is identified with a subset of $M$ the generates $M$ ($\varphi$ is a [[surjection]]) and is [[linearly independent]] ($\varphi$ is an [[injection]]). > ~~suppose $M$ is free. [[terminal objects are unique up to a unique isomorphism|Abstract nonsense]] guarantees $M \cong R^{\oplus B}$; $\varphi$ is the ~~ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```