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Let $A$ be a complex unital algebra and let $M$ be a proper left ideal in $A$. Furthermore, suppose that $\{L_i\colon i \in I\}$ is an uncountable family of left-ideals such that

1) for each $i$ the $L_i\subseteq M$ and $\mbox{dim }M/L_i = 2$

2) for $s\neq t$ neither $L_s\subseteq L_t$ nor $L_t\subseteq L_s$.

Can $M$ be finitely generated?

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1 Answer 1

Yes. $M$ can be finitely generated.

Let $M=\mathbb{C}^{2 \times 2}$ denote $2 \times 2$ complex matrices. Let $A = \mathbb{C}^{2 \times 2} \oplus \mathbb{C}$ (a 5 dimensional complex vector space). Give $A$ the multiplication: $(P,s)(Q,t)=(PQ,st)$. Then your identity is $(I_2,1)$.

We see that $M$ is an ideal of $A$ (not just left but actually 2-sided).

Let $m\in\mathbb{C}$ and

$$L_m = \left\{ \begin{pmatrix} a & ma \\ b & mb \end{pmatrix} \,{\Huge|}\, a,b \in \mathbb{C} \right\}$$

This is certainly an uncountable set of left ideals (these matrices are preserved under row operations). Also, $\mathrm{dim}(L_m)=2$ so $\mathrm{dim}(M/L_m)=4-2=2$.

Notice that $L_m \cap L_n = \{ 0_{2 \times 2} \}$ [Note: You can think of $L_m$ sort of like lines through the origin.]

Everything is finitely generated.

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