Subalgebra Generated by Commuting Elements is Commutative

Theorem

Let $K$ be a field.

Let $A$ be an algebra over $K$.

Let $S \subseteq A$ be a non-empty set such that:

for all $x, y \in S$ we have $x y = y x$.

Let $K \sqbrk S$ be the subalgebra generated by $S$.

Then $K \sqbrk S$ is a commutative algebra.

Proof

From Explicit Form for Generated Subalgebra, we have:

$K \sqbrk S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$

and $K \sqbrk S$ is a subalgebra of $A$.

Let $x, y \in K \sqbrk S$.

Then:

$\ds x = \sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} }$

and:

$\ds y = \sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$

where each $x_{j, k}$ and $x_{i, k}$ is in $S$.

Then we have:

$\ds x y = \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$

Since $K$ is a field and every element in $S$ commutes, we have:

$\ds \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } = \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } }$

Then:

$\ds \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = \paren {\sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = y x$

Hence $K \sqbrk S$ is a commutative algebra.

$\blacksquare$