Explicit Form for Generated Subalgebra

Theorem

Let $K$ be a field.

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

Let $S \subseteq A$ be a non-empty set.

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

Then:

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

Proof

Let:

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

First, for each $x \in S$ we have:

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

Hence $S \subseteq B$.

We show that $B$ is a subalgebra.

From Linear Span is Linear Subspace, $B$ is a linear subspace of $A$.

Now let $x, y \in B$.

Since $B$ is a linear subspace, it suffices to consider $x, y$ of the form:

$x = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n}$

and:

$y = y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$

We have:

$x y = x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} y_1^{s_1} y_2^{s_2} \ldots y_m^{s_m}$

By relabelling $x_{n + t} = y_t$ and $k_{n + t} = s_t$ for $1 \le t \le m$ we obtain that $x y \in B$.

Hence $B$ is subalgebra of $A$.

Since $S \subseteq B$, we obtain $K \sqbrk S \subseteq B$ from the definition of the subalgebra generated by $A$.

Conversely, note that for each $x_1, \ldots, x_n \in S$ we have $x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} \in S$ from the definition of a subalgebra.

Since $K \sqbrk S$ is a linear subspace of $A$, we have $B = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1} \subseteq K \sqbrk S$.

Hence $B = K \sqbrk S$.

$\blacksquare$