Isomorphism from R^n via n-Term Sequence
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Theorem
Let $G$ be a unitary $R$-module.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $R^n$ be the $R$-module $R^n$.
Let $\psi: R^n \to G$ be defined as:
- $\ds \map \psi {\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n} } = \sum_{k \mathop = 1}^n \lambda_k a_k$
Then $\psi$ is an isomorphism.
Proof
By Unique Representation by Ordered Basis, $\psi$ is a bijection.
We have:
\(\ds \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} a_k\) |
and we have:
\(\ds \beta \sum_{k \mathop = 1}^n \lambda_k a_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \beta \paren {\lambda_k a_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \paren {\beta \lambda_k} a_k\) |
thus proving that $\psi$ is also a homomorphism.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.5$