I am reading Sadri Hassani, Mathematical Physics (2nd edition).
He defines
an algebra $\mathcal{A}$ as a vector space with a composition $\mathcal{A}\times\mathcal{A}\to\mathcal{A}$, here denoted by juxtaposition. Only distributivity wrt. linear combinations is assumed, but e.g. commutativity and associativity are not.
a left ideal $\mathcal{L}$ of $\mathcal{A}$ as a subspace satisfying that for any $\mathbf{a}\in\mathcal{A}$ and $\mathbf{x}\in\mathcal{L}$, we have $\mathbf{a}\mathbf{x}\in\mathcal{L}$, which is written $\mathcal{A}\mathcal{L}\subseteq\mathcal{L}$.
that a left ideal $\mathcal{M}$ of $\mathcal{A}$ is minimal if every left ideal of $\mathcal{A}$ contained in $\mathcal{M}$ coincides with $\mathcal{M}$.
Then he states the following theorem (Theorem 3.2.6 in the book): Let $\mathcal{L}$ be a left ideal of $\mathcal{A}$. Then the following statements are equivalent:
- (a) $\mathcal{L}$ is a minimal left ideal.
- (b) $\mathcal{A}\mathbf{x} = \mathcal{L}$ for all $\mathbf{x}\in\mathcal{L}$.
- (c) $\mathcal{L}\mathbf{x} = \mathcal{L}$ for all $\mathbf{x}\in\mathcal{L}$.
(where $\mathcal{B}\mathbf{x}$ denotes the set of all elements of the form $\mathbf{b}\mathbf{x}$; above $\mathcal{B}$ is either $\mathcal{A}$ or $\mathcal{L}$)
I've concluded so far that this theorem is not true the way it's written in the book, i.e. it either needs modifications or assumptions (I'll elaborate below). My question then is, what is true about minimal left ideals in the context above - how to modify the statements and/or assumptions in order for the theorem to be true? And how to prove it?
Elaboration:
This question seems to be about the same theorem, but I think I've made further progress and am able to pinpoint the problem more specifically: If $\mathcal{L}$ is a minimal left ideal of an algebra $\mathcal{A}$, then $\mathcal{A}l=\mathcal{L}$ for all $l\in \mathcal{L}$?
(1) First of all (and I think this is what egreg is hinting at in his/her comment on the question above), if following the definition given in the book the zero subspace $\{\mathbf{0}\}$ is the only minimal left ideal of any algebra. Which would make for some pretty dull statements about minimal ideals, I guess.
Therefore I'll assume it's a mistake on the author's side, and that the proper definition should be
that a left non-zero ideal $\mathcal{M}$ of $\mathcal{A}$ is minimal if every left non-zero ideal of $\mathcal{A}$ contained in $\mathcal{M}$ coincides with $\mathcal{M}$.
Also, item (b) and (c) should probably be modified to
... for all $\mathcal{x} \in \mathcal{L}\setminus\{\mathbf{0}\}$.
(2) Secondly, I think the theorem should have assumed the algebra $\mathcal{A}$ to be associative. That's my conclusion from Ruvi Lecamwasam's example in his question. Another simple counterexample could be the algebra $\mathcal{A} = (\mathbb{R}^3,\times)$ (3D space with vector cross product) with the Cartesian unit vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ as basis; its only (non-zero) left ideal (AFAICT) is $\mathcal{A}$ itself so $\mathcal{A}$ is a minimal by defintion, but e.g. $\mathcal{A}\mathbf{x} = \text{span}(\mathbf{y}, \mathbf{z}) \neq \mathcal{A}$.
So I also assume that the theorem should only apply to associative algebras;
Let $\mathcal{L}$ be a left ideal of an associative algebra $\mathcal{A}$...
(3) Even with these corrections, the theorem is not true as it stands, because Hassani has an example of an algebra (4-dimensional) with a minimal left ideal $\mathcal{L}_1$ for which (c) of the theorem doesn't hold; for one basis vector $\mathbf{f}_1 \in \mathcal{L}_1$ we have $\mathcal{L}_1\mathbf{f}_1 = \mathcal{L}_1$ but for the other basis vector $\mathbf{f}_4 \in \mathcal{L}_1$ we get $\mathcal{L}_1\mathbf{f}_4 = \{\mathbf{0}\}$.
Therefore I speculate that maybe item (c) should be changed to something like
(c) For all $\mathbf{x}\in\mathcal{L}$; $\mathcal{L}\mathbf{x} = \mathcal{L}$ or $\mathcal{L}\mathbf{x} = \{\mathbf{0}\}$.
I think I've successfully proven the implications (b) $\Rightarrow$ (a) and (a) $\Rightarrow$ (c), assuming associativity, the "non-zero" modification to the definition of "minimal" and the modification to item (c).
I was able to prove the implication (c) $\Rightarrow$ (b) in the form (c) is written in the book, but I'm not sure how to do this (if that's the way to go at all) with the "$\mathcal{L}\mathbf{x} = \{\mathbf{0}\}$" modification to item (c).
This is what I've been able to conclude so far. Please point out any mistakes I may have made. I'll also be happy to elaborate further on my preliminary proofs, examples, etc., but my post is already rather long! Anyway, just let me know in the comments.