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Nathan Nguyen
2024-12-06 12:01:10 -06:00
parent 74e129000b
commit 98c8b94c06
8 changed files with 29 additions and 29 deletions

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@@ -120,9 +120,9 @@ This direct product is in itself an $R$-module.
**Theorem**. For any set $A$, there is a free $R$-module $F(A)$ on $A$ such that $F(A)$ satisfies the universal property: if $M$ is any $R$-module, and $\varphi: A \rightarrow M$ is a map of sets, there eixts a unique $R$-module homomorphism: $\Phi: F(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$ for all $a \in A$.
**Collary**. If $F_1$ and $F_2$ are free modules on $A$, then there is a unique isomorphism between $F_1$ and $F_2$, which is the identity map on A.
**Corollary**. If $F_1$ and $F_2$ are free modules on $A$, then there is a unique isomorphism between $F_1$ and $F_2$, which is the identity map on A.
**Collary**. If $F$ is a free $R$-module with basis $A$, then $F \cong F(A)$.
**Corollary**. If $F$ is a free $R$-module with basis $A$, then $F \cong F(A)$.
**Definition** For a free module $F$ with basis $A$, if $R$ is commutative, tthen the *rank* of $F$ is the cardinality of $A$.