Basic Theorems Regarding Ideals in an Algebra 1
 Table of Contents

# Basic Theorems Regarding Ideals in an Algebra 1

 Proposition 1: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$ be a linear subspace of $\mathfrak{A}$. If $u \in \mathfrak{A}$ is a left (right) modular unit of $J$ then $u$ is a left (right) modular unit of any linear subspace $K$ of $\mathfrak{A}$ with $J \subseteq K$.
• Proof: Suppose that $u \in \mathfrak{A}$ is a left modular unit of $J$. Then $(1 - u)\mathfrak{A} \subseteq J$. But since $J \subseteq K$ we have that $(1 - u)\mathfrak{A} \subseteq K$ and so $u \in \mathfrak{A}$ is a right modular unit of $K$. $\blacksquare$
 Proposition 2: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$ be a linear subspace of $\mathfrak{A}$. a) If $J$ is a proper left ideal of $\mathfrak{A}$ and $u$ is a right modular unit of $J$ then $u \not \in J$. b) If $J$ is a proper right ideal of $\mathfrak{A}$ and $u$ is a left modular unit of $J$ then $u \not \in J$.
• Proof of a): Suppose instead that $u \in J$. Since $J$ is a left ideal of $\mathfrak{A}$ we have that $\mathfrak{A}J \subseteq J$ and since $u$ is a right modular unit of $J$ we have that $\mathfrak{A}(1 - u) \subseteq J$. So if $a \in \mathfrak{A}$ then:
(1)
\begin{align} \quad a = \underbrace{a - au}_{\in \mathfrak{A}(1 - u) \subseteq J} + \underbrace{au}_{\in \mathfrak{A}J \subseteq J} \in J \end{align}
• This shows that $\mathfrak{A} \subseteq J$ and so $\mathfrak{A} = J$, contradicting $J$ being proper. $\blacksquare$
• Proof of b) This is analogous to the proof above. If we suppose that $u \in J$ then since $J$ is a right ideal of $\mathfrak{A}$ we have that $J\mathfrak{A} \subseteq J$ and since $u$ is a left modular unit of $J$ we have that $(1 - u)\mathfrak{A} \subseteq J$, so if $a \in \mathfrak{A}$ then:
(2)
\begin{align} \quad a = \underbrace{a - ua}_{\in (1 - u)\mathfrak{A} \subseteq J} + \underbrace{ua}_{\in J\mathfrak{A} \subseteq J} \in J \end{align}
• This shows that $\mathfrak{A} \subseteq J$ and so $\mathfrak{A} = J$, again contradicting $J$ being proper. $\blacksquare$
 Proposition 3: Let $\mathfrak{A}$ be an algebra and let $u \in \mathfrak{A}$. Then: a) $u$ is left quasi-invertible if and only if $\mathfrak{A}(1 - u) = \mathfrak{A}$. b) $u$ is right quasi-invertible if and only if $(1 - u)\mathfrak{A} = \mathfrak{A}$.
• Proof of a) $\Rightarrow$ Suppose that $u$ is left quasi-invertible. Then there exists an element $a \in \mathfrak{A}$ such that $a \cdot u = 0$, i.e., $a + u - au = 0$, that is:
(3)
\begin{align} \quad u = -a + au = \underbrace{(-a)}_{\in \mathfrak{A}} - \underbrace{(-a)u}_{\in \mathfrak{A}u} \in \mathfrak{A}(1 - u) \quad (*) \end{align}
• Let $J = \mathfrak{A}(1 - u)$. Then $J$ is a left ideal of $\mathfrak{A}$ since clearly $\mathfrak{A}J = \mathfrak{A}\mathfrak{A}(1-u) = \subseteq J$. Also we have that $\mathfrak{A}(1 - u) \subseteq J$, and so $u$ is a right modular unit of $J$. But from $(*)$ we have that $u \in J$. So by the contrapositive of Proposition 2, $J$ cannot be a proper ideal of $\mathfrak{A}$ and so $\mathfrak{A}(1 - u) = \mathfrak{A}$.
• $\Leftarrow$ Suppose that $\mathfrak{A}(1 - u) = \mathfrak{A}$. Then for $u \in \mathfrak{A}$ there exists an $a \in \mathfrak{A}$ such that $u = a - au$. So $(-a) + u - (-a)u = 0$, that is, $(-a) \cdot u = 0$. So $-a$ is a left quasi-inverse for $u$, i.e., $u$ is left quasi-invertible. $\blacksquare$
• Proof of b) $\Rightarrow$ Suppose that $u$ is right quasi-invertible. Then there exists an element $a \in \mathfrak{A}$ such that $u \cdot a = 0$, i.e., $u + a - ua = 0$, that is:
(4)
\begin{align} \quad u = (-a) + ua = \underbrace{(-a)}_{\in \mathfrak{A}} - \underbrace{u(-a)}_{\in u\mathfrak{A}} \in (1 - u)\mathfrak{A} \quad (**) \end{align}
• Let $J = (1 - u)\mathfrak{A}$. Then $J$ is a right ideal of $\mathfrak{A}$ since $J\mathfrak{A} = (1-u)\mathfrak{A}\mathfrak{A} \subseteq J$. Also we have that $(1 - u)\mathfrak{A} \subseteq J$ and so $u$ is a left modular unit of $J$. But from $(**)$ we have that $u \in J$. So by the contrapositive of Proposition 2, $J$ cannot be a proper ideal of $\mathfrak{A}$ and so $(1 - u)\mathfrak{A} = \mathfrak{A}$.
• $\Leftarrow$ Suppose that $(1 - u)\mathfrak{A} = \mathfrak{A}$. Then for $u \in \mathfrak{A}$ there exists an $a \in \mathfrak{A}$ such that $u = a - ua$. So $u + (-a) - u(-a) = 0$, that is, $u \cdot (-a) = 0$. So $-a$ is a right quasi-inverse for $u$, i.e., $u$ is right quasi-invertible. $\blacksquare$
 Proposition 4: Let $\mathfrak{A}$ be an algebra and let $J \subseteq \mathfrak{A}$ be a linear subspace. Then: a) If $J$ is a proper modular left ideal of $\mathfrak{A}$ then $J$ is contained in some maximal left ideal $K$. b) If $J$ is a proper modular right ideal of $\mathfrak{A}$ then $J$ is contained in some maximal right ideal $K$.
• Proof of a) Let $J$ be a proper modular left ideal of $\mathfrak{A}$. Then $J$ is a left ideal of $\mathfrak{A}$ with $J \neq \mathfrak{A}$ and there exists a right modular unit $u$ of $J$, i.e., $\mathfrak{A}(1 - u) \subseteq J$.
• Let:
(5)
\begin{align} \quad \mathcal F = \left \{ K : K \: \mathrm{is \: a \: proper \: left \: ideal \: of \:} \mathfrak{A} \: \mathrm{and} \: J \subseteq K \right \} \end{align}
• By Proposition 1, since $u$ is a right modular unit of $J$ we have that $u$ is a right modular unit of $K$ for every $K \in \mathcal F$ and so by Proposition 2, $u \not \in K$ for every $K \in \mathcal F$.
• Let $\mathcal F$ be partially ordered by inclusion and let $\mathcal C \subseteq \mathcal F$ be a chain in $\mathcal F$. Let $U = \bigcup_{K \in \mathcal C} K$. Since $\mathcal C$ is a chain ordered by inclusion we see that $U$ is a subspace and is itself a proper left ideal of $\mathfrak{A}$ such that $u \not \in U$. It is clearly an upper bound for $\mathcal C$.
• So by Zorn's Lemma, $\mathcal F$ must have a maximal element, i.e., $J$ is contained in some maximal left ideal. $\blacksquare$
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