Ideals in a Linear Space

We will now state many definitions regarding the concept of an ideal in a linear space. It is important to note that the definitions below are in regards to linear spaces - but we will only really use these concepts for algebras.

Ideals

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace of $X$. 1) $J$ is said to be a Left Ideal of $X$ if $XJ \subseteq J$. 2) $J$ is said to be a Right Ideal of $X$ if $JX \subseteq J$. 3) $J$ is said to be a Two-Sided Ideal (or Bi-Ideal) of $X$ if it is both a left ideal of $X$ and a right ideal of $X$.

Here, $XJ$ denotes the set $XJ = \{ xj : x \in X, j \in J \}$ and $JX$ denotes the set $JX = \{ jx : j \in J, x \in X \}$.

Proposition 1: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$. If $J$ is a left/right/two-sided ideal of $\mathfrak{A}$ then $J$ is a subalgebra of $\mathfrak{A}$.

• Proof: We only prove the case when $J$ is a left ideal. The cases when $J$ is a right ideal and a two-sided ideal are analogous.

• Since $J$ is a left ideal of $\mathfrak{A}$ we have that $J$ is a linear subspace of $\mathfrak{A}$ and so it closed under the operations of addition and scalar multiplication and contains the zero vector. Also, since $J$ is a left ideal of $\mathfrak{A}$ we have that $\mathfrak{A}J = \{ aj : a \in \mathfrak{A}, j \in J \} \subseteq J$, and so:

(1)

\begin{align} \quad JJ = \{ j_1j_1 : j_1, j_1 \in J \} \subseteq J \end{align}

• Which shows that $J$ is closed under the operation of multiplication. Thus $J$ is a subalgebra of $\mathfrak{A}$. $\blacksquare$

Proper Ideals and Maximal Ideals

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a (left/right/two-sided) ideal of $X$. Then $J$ is said to be a Proper (Left/Right/Two-Sided) Ideal of $X$ if $J \neq X$.

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a (left/right/two-sided) ideal of $X$. Then $J$ is said to be a Maximal (Left/Right/Two-Sided) Ideal of $X$ if $J$ is a proper (left/right/two-sided) ideal of $X$ and if $K$ is any other proper (left/right/two-sided) ideal of $X$ then $J \not \subseteq K$.

Modular Ideals and Maximal Modular Ideals

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace of $X$. 1) A point $u \in X$ is said to be a Left Modular Unit of $J$ if $(1 - u)X \subseteq J$. 2) A point $u \in X$ is said to be a Right Modular Unit of $J$ if $X(1 - u) \subseteq J$. 3) A point $u \in X$ is said to be a Two-Sided Modular Unit of $J$ if it is both a left modular unit of $J$ and a right modular unit of $J$.

Here, $(1 - u)J$ denotes the set $(1 - u)X = \{ x - ux : x \in X \}$ and $X(1 - u)$ denotes the set $X(1 - u) = \{ x - xu : x \in X \}$.

Proposition 2: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$ be a linear subspace of $\mathfrak{A}$. If $\mathfrak{A}$ is an algebra with unit then $1$ is a left/right/two-sided modular unit of $J$.

• Proof: If $\mathfrak{A}$ is an algebra with unit $1$ then since $J$ is a subspace (and hence contains the zero vector) we have that:

(2)

\begin{align} \quad (1 - 1)\mathfrak{A} = \{ a - a : a \in \mathfrak{A} \} = \{ 0 \} \subseteq J \end{align}

(3)

\begin{align} \quad \mathfrak{A}(1 - 1) = \{ a - a : a \in \mathfrak{A} \} = \{ 0 \} \subseteq J \end{align}

• So $1$ is a left/right/two-sided modular unit of $J$. $\blacksquare$

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace of $X$. 1) $J$ is said to be a Left Modular Ideal if $J$ is a left ideal and has a right modular unit. 2) $J$ is said to be a Right Modular Ideal if $J$ is a right ideal and has a left modular unit.

Definition: Let $X$ be a linear space and let $J \subseteq X$ be a (left/right/two-sided) ideal of $X$. Then $J$ is said to be a Maximal Modular (Left/Right/Two-Sided) Ideal of $X$ if $J$ is a proper (left/right/two-sided) ideal of $X$; if $K$ is any other proper (left/right/two-sided) ideal of $X$ then $J \not \subseteq K$; and $J$ is a (left/right/two-sided) modular ideal.