Unitization of Algebra over Field preserves Subalgebra Relation

Theorem
Let $K$ be a field.

Let $A$ be a non-unital commutative algebra over $K$.

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 $A_+$.

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

From Unitization of Algebra over Field is Unital Algebra over Field, we conclude: $B_+$ is a unital subalgebra of $A_+$.