I am going through Hassani's 'Mathematical Physics':
He defines an algebra $\mathcal{A}$ as a vector space together with a multiplication map $\mathcal{A}\times\mathcal{A}\rightarrow\mathcal{A}$.
He defines a left ideal $\mathcal{L}$ as a vector subspace of $\mathcal{A}$ such that $\mathcal{A}\mathcal{L}\subset\mathcal{L}$.
An ideal $\mathcal{L}$ is minimal if every left ideal of $\mathcal{A}$ contained in $\mathcal{L}$ coincides with $\mathcal{A}$.
He then goes on to state that a left ideal $\mathcal{L}$ is minimal if and only if $\mathcal{A}l=\mathcal{L}$ for all $l\in \mathcal{L}$. It seems to me like the forward direction of this isn't correct. For example, let $\mathcal{A}$ be the algebra spanned by $x,y,w$ with multiplication relations:
- $yx=y$, $xy=x$.
- $wx=wy=w$
- $xw = x$, $yw = x$, $ww=w$.
Then the only ideal is $\mathcal{A}$ iteself, which is therefore minimal. However, $\mathcal{A}w=\{x,w\}\ne \mathcal{A}$. I can't see what I am missing. Any help would be appreciated.
EDIT: On further thinking, I believe the flaw in my 'counterexample' is that I haven't shown there isn't an invariant subspace whose basis elements are linear combinations of the elements $x, y, w$. Any hints as to how to prove the theorem though would still be appreciated.
The example you gave is inconsistent.
$$y(xw) = yx = y$$ $$(yx)w = yw = x$$
$\mathcal{L}$ is minimal if and only if $\mathcal{A}l=\mathcal{L}$ for all $l\in \mathcal{L}$.
"$\Rightarrow$": If $\mathcal{A}l \subsetneq \mathcal{L}$, then $\mathcal{L}$ is not minimal.
"$\Leftarrow$": If $l \in \mathcal{M} \subsetneq \mathcal{L}$, then $\mathcal{A}l \subseteq \mathcal{M} \neq \mathcal{L}$.