Left Ideals are Left A-Modules, Right Ideals are Right A-Modules

Examples of Modules - Left Ideals J are Left X-Modules, Right Ideals J are Right X-Modules

Let $\mathfrak{A}$ be an algebra and let $J$ be a left-ideal of $\mathfrak{A}$. Then $\mathfrak{A}J \subseteq J$. Let $f : \mathfrak{A} \times J \to J$ be the module multiplication defined by $f(a, j) = aj$, that is, the product is simply the multiplication given by the multiplication in $\mathfrak{A}$.

For each fixed $a \in \mathfrak{A}$ the mapping $f_a : J \to J$ defined by $f_a(j) = aj$ is clearly linear, since if $j_1, j_2 \in J$ then $a(j_1 + j_2) = aj_1 + aj_2$ by distributivity in $\mathfrak{A}$. So axiom $LM1$ is satisfied.

For each fixed $j \in J$ the mapping $f_j : \mathfrak{A} \to J$ defined by $f_j(a) = aj$ is also linear since if $a_1, a_2 \in \mathfrak{A}$ then $(a_1 + a_2)j = a_1j + a_2j$, again by distributivity in $\mathfrak{A}$. So axiom $LM2$ is satisfied.

Lastly, for all $a_1, a_2 \in \mathfrak{A}$ and all $j \in J$ we certainly have by the associativity of multiplication in $\mathfrak{A}$ that:

(1)

\begin{align} \quad a_1(a_2j) = (a_1a_2)j \end{align}

So axiom $LM3$ is satisfied. Thus any left-ideal $J$ of $\mathfrak{A}$ is a left $\mathfrak{A}$-module.

In a similar fashion it can be shown that any right ideal of $\mathfrak{A}$ is a right $\mathfrak{A}$-module, and that any two-sided ideal of $\mathfrak{A}$ is an $\mathfrak{A}$-bimodule.