Unitization of Algebra over Field preserves Subalgebra Relation

Theorem

Let $K$ be a field.

Let $B$ be a subalgebra of $A$.

Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.

Then $B_+$ is a unital subalgebra of $A_+$.

Let $\struct {x, s}, \tuple {y, t} \in B_+$.

Let $\lambda \in K$.

Then, we have:

$\tuple {x, s} + \lambda \tuple {y, t} = \tuple {x + \lambda y, s + \lambda t}$

Since $\tuple {x, s}, \tuple {y, t} \in B_+$, we have $x, y \in B$.

Since $B$ is a subalgebra of $A$, we have that:

$B$ is a vector subspace of $A$

and hence:

$x + \lambda y \in B$

Hence, we have:

$\tuple {x + \lambda y, s + \lambda t} \in B_+$

So, we have:

$\tuple {x, s} + \lambda \tuple {y, t} \in B_+$

From One-Step Vector Subspace Test, we therefore have that $B_+$ is a vector subspace of $A_+$.

To show that $B_+$ is a subalgebra of $A_+$, it remains to verify that:

$\tuple {x, s} \tuple {y, t} \in B_+$ for all $\tuple {x, s}, \tuple {y, t} \in B_+$.

Let $\tuple {x, s}, \tuple {y, t} \in B_+$.

Then $x, y \in B$.

From the definition of the unitization, we have:

$\tuple {x, s} \tuple {y, t} = \tuple {x y + s y + t x, s t}$

Since $B$ is a subalgebra of $A$, we have:

$x y \in B$

Since $B$ is a vector subspace of $A$, we have:

$x y + s y + t x \in B$

Hence, we have:

$\tuple {x y + s y + t x, s t} \in B_+$

and so:

$\tuple {x, s} \tuple {y, t} \in B_+$

We therefore have that $B_+$ is a subalgebra of $A_+$.

It remains to show that $B_+$ is unital.

Since $B$ is a vector subspace of $A$, we have that:

${\mathbf 0}_A \in B$

Hence, we have:

$\tuple { {\mathbf 0}_A, 1_K} \in B_+$

$\blacksquare$