Ideal of 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 $I$ be an ideal of $A$.

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

Let:

$I_0 = \set {\tuple {x, 0_K} : x \in I}$

Then $I_0$ is an ideal of $A_+$.

Proof

From the definition of an ideal, $I$ is a vector subspace of $A$.

Hence from Vector Subspace of Algebra over Field Embeds into Unitization as Vector Subspace, $I_0$ is a vector subspace of $A_+$.

Let $\tuple {x, 0_K} \in I$ and $\tuple {y, \lambda} \in A_+$.

Then we have:

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

Since $I$ is an ideal, we have $x y \in I$.

Since $I$ is a vector subspace of $A$, we have $x y + \lambda x \in I$.

Hence we have $\tuple {x, 0_K} \tuple {y, \lambda} \in I_0$.

Similarly we have:

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

Since $I$ is an ideal, we have $y x \in I$.

Since $I$ is a vector subspace of $A$, we have $y x + \lambda x \in I$.

Hence we have $\tuple {y, \lambda} \tuple {x, 0_K} \in I_0$.

$\blacksquare$