Algebra over Field Embeds into Unitization as Ideal

Theorem

Let $K$ be a field.

Let $A$ be an algebra over $K$ that is not unital.

Let $A_+$ be the unitization of $A$.

Let:

$A_0 = \set {\tuple {x, 0_K} : x \in A} \subseteq A_+$.

Then $A_0$ is an ideal in $A_+$.

It remains to show that for each $u \in A_0$ and $v \in A_+$, we have $u v \in A_0$ and $v u \in A_0$.

Let $x \in A$.

Let $\tuple {y, \lambda} \in A_+$.

Then, we have:

$\ds \tuple {x, 0_K} \tuple {y, \lambda}$

$=$

$\ds \tuple {x y + \lambda x + 0_K y, 0_K \lambda}$

$\ds $

$=$

$\ds \tuple {x y + \lambda x, 0_K}$

$\ds $

$\in$

$\ds A_0$

and:

$\ds \tuple {y, \lambda} \tuple {x, 0_K}$

$=$

$\ds \tuple {y x + 0_K y + \lambda x, 0_K \lambda}$

$\ds $

$=$

$\ds \tuple {y x + \lambda x, 0_K}$

$\ds $

$\in$

$\ds A_0$

$\blacksquare$