Consider the free monoid $2^{<\omega}$ on two generators, whose elements are finite binary strings, and whose operation is concatenation. A left ideal is a subset $L \subseteq 2^{<\omega}$ such that for all $\lambda \in L$ and $\sigma \in 2^{<\omega}$, we have $\lambda \sigma \in L$. Similarly, a right ideal is a subset $R \subseteq 2^{<\omega}$ such that for all $\sigma \in 2^{<\omega}$ and $\rho \in R$, we have $\sigma \rho \in R$.
An obvious way to generate left ideals is to take all the strings having a certain prefix, or set of prefixes. For example, all strings starting in $1$ form a left ideal, as do all strings starting in either $01$ or $10$. We can also form right ideals by taking all strings ending in a certain set of suffixes.
I have a gut feeling that all left [right] ideals of $2^{<\omega}$ have this form, and in fact, the set of prefixes [suffixes] can be chosen to be finite (because the alphabet $2 = \{ 0, 1 \}$ is finite). So, can anyone prove or dispute:
for every left ideal $L \subseteq 2^{<\omega}$, there is a finite subset $F \subseteq 2^{<\omega}$ such that $L = \{ \varphi \sigma : \varphi \in F,\ \sigma \in 2^{<\omega} \}$?
I think the result for left ideals is equivalent to the result for right ideals, since $2^{<\omega}$ has an anti-automorphism which reverses the strings.
Bonus question: are there any nontrivial two-sided ideals? I can't think of any right now.
Consider the infinite set of prefixes: $$ P=\{\ \underbrace{00\dots0}_{n\text{ times}}\underbrace{11\dots1}_{n\text{ times}}\ :\ n\in \{1, 2, 3, \dots\}=\Bbb Z_{>0}\ \}\ . $$ Let $L_p$ be the "ideal" starting with some prefix $p\in P$. Let $L$ be the union of all $L_p$ for $p$ running in $P$.
Then $L$ is an ideal, since taking some $\lambda \in L$, there is some $p\in P$ with $\lambda \in L_p$, and then any word of the shape $\lambda \sigma$ is in $L_p$, thus also in $L$.
The ideal $L$ does not accept a finite list $Q$ of prefixes. Let us show this. Assume there is such a finite list. In particular, there is a maximal length $N$ for these finitely many prefixes. Consider $Q'$, the list of all words of length $=N$ that may occur as a prefix for some word in $L$. Then $Q'$ is a list of prefixes of length $N$ for $L$. So for each $q'$ in $Q'$, any word of the shape $q'*$ is also in $L$. (Here, $*$ is a place holder for any finite word. The words $q'*$ build a "cylinder" inside the language.) Now we get a contradiction. The "zero" word $z_N=00\dots0$ of length $N$ is a prefix, since there is some word in $L$ starting with $z_N$, but the "cylinder" $z_N*$ is not included in $L$. Just consider the word $z_N10\dots$ . If this is in $L$, then $N=1$, but of course, $L$ has no generating set of prefixes of length one.
For the bonus, we no longer use prefixes / suffixes, but subwords. Consider the simple subword $0$. Define the subset $B$ as the subset of words containing at least one $0$ letter / subword. Then $B$ is not the full language, some finite sequence of ones is never in $B$. To get more complicated bilateral "ideals" use a (finite or infinite) list of generating subwords.