Basic Theorems Regarding Ideals in a Linear Space 1

# Basic Theorems Regarding Ideals in a Linear Space 1

 Proposition 1: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace of $X$. If $u \in X$ is a left (right) modular unit of $J$ then $u$ is a left (right) modular unit of any linear subspace $K$ of $X$ with $J \subseteq K$.
• Proof: Suppose that $u \in X$ is a left modular unit of $J$. Then $(1 - u)X \subseteq J$. But since $J \subseteq K$ we have that $(1 - u)X \subseteq K$ and so $u \in X$ is a right modular unit of $K$. $\blacksquare$
 Proposition 2: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace of $X$. a) If $J$ is a proper left ideal of $X$ and $u$ is a right modular unit of $J$ then $u \not \in J$. b) If $J$ is a proper right ideal of $X$ 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 $X$ we have that $XJ \subseteq J$ and since $u$ is a right modular unit of $J$ we have that $X(1 - u) \subseteq J$. So if $x \in X$ then:
(1)
\begin{align} \quad x = \underbrace{x - xu}_{\in X(1 - u) \subseteq J} + \underbrace{xu}_{\in XJ \subseteq J} \in J \end{align}
• This shows that $X \subseteq J$ and so $X = J$, 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 $X$ we have that $JX \subseteq j$ and since $u$ is a left modular unit of $J$ we have that $(1 - u)X \subseteq J$, so if $x \in X$ then:
(2)
\begin{align} \quad x = \underbrace{x - ux}_{\in (1 - u)X \subseteq J} + \underbrace{ux}_{\in JX \subseteq J} \in J \end{align}
• This shows that $X \subseteq J$ and so $X = J$, again contradicting $J$ being proper. $\blacksquare$
 Proposition 3: Let $X$ be an algebra and let $u \in X$. Then: a) $u$ is left quasi-invertible if and only if $X(1 - u) = X$. b) $u$ is right quasi-invertible if and only if $(1 - u)X = X$.
• Proof of a) $\Rightarrow$ Suppose that $u$ is left quasi-invertible. Then there exists an element $x \in X$ such that $x \cdot u = 0$, i.e., $x + u - xu = 0$, that is:
(3)
\begin{align} \quad u = -x + xu = \underbrace{(-x)}_{\in X} - \underbrace{(-x)u}_{\in Xu} \in X(1 - u) \quad (*) \end{align}
• Let $J = X(1 - u)$. Then $J$ is a left ideal of $X$ since clearly $XJ = XX(1-u) = \subseteq J$. Also we have that $X(1 - u) \subseteq J$, 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 $X$ and so $X(1 - u) = X$.
• $\Leftarrow$ Suppose that $X(1 - u) = X$. Then for $u\ in X$ there exists an $x \in X$ such that $u = x - xu$. So $(-x) + u - (-x)u = 0$, that is, $(-x) \cdot u = 0$. So $-x$ 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 $x \in X$ such that $u \cdot x = 0$, i.e., $u + x - ux = 0$, that is:
(4)
\begin{align} \quad u = (-x) + ux = \underbrace{(-x)}_{\in X} - \underbrace{u(-x)}_{\in uX} \in (1 - u)X \quad (**) \end{align}
• Let $J = (1 - u)X$. Then $J$ is a right ideal of $X$ since $JX = (1-u)XX \subseteq J$. Also we have that $(1 - u)X \subseteq J$ 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 $X$ and so $(1 - u)X = X$.
• $\Leftarrow$ Suppose that $(1 - u)X = X$. Then for $u \in X$ there exists an $x \in X$ such that $u = x - ux$. So $u + (-x) - u(-x) = 0$, that is, $u \cdot (-x) = 0$. So $-x$ is a right quasi-inverse for $u$, i.e., $u$ is right quasi-invertible. $\blacksquare$
 Proposition 4: Let $X$ be a linear space and let $J \subseteq X$ be a linear subspace. Then: a) If $J$ is a proper modular left ideal of $X$ then $J$ is contained in some maximal left ideal $K$. b) If $J$ is a proper modular right ideal of $X$ then $J$ is contained in some maximal right ideal $K$.
• Proof of a) Let $J$ be a proper modular left ideal of $X$. Then $J$ is a left ideal of $X$ with $J \neq X$ and there exists a right modular unit $u$ of $J$, i.e., $X(1 - u) \subseteq J$.
• Let:
(5)
\begin{align} \quad \mathcal F = \left \{ K : K \: \mathrm{is \: a \: proper \: left \: ideal \: of \:} X \: \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 $X$ 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. $\blacksquare$