How can I show that for a field $K$, in the free algebra on $2$ generators $K\langle x,y\rangle$, the two-sided ideal $$\big\langle\!\big\langle xy^ix\;\big|\;i\in\mathbb{N}\big\rangle\!\big\rangle =\bigg\{\sum_{i}g_ixy^ixh_i \;\bigg|\; i\in\mathbb{N},\, g_i,h_i\in K\langle x,y\rangle\bigg\}$$ is not generated by any finitely many $f_1,\ldots,f_k\!\in\!K\langle x,y\rangle$, if this is even true?
If not, what other generators of the ideal must I take?
Is there a simpler set of generators of an ideal that is not finitely generated?
Let $I$ be the two sided ideal generated by $xy^ix$ for all $i \in \mathbb N$ and let $I_n$ be the ideal generated by $xy^ix$ with $i < n$.
First prove that $I_n$ does not contain $xy^nx$. Do this by arguing that every monomial contained in every polynomial of this ideal contains a string of the form $xy^ix$ with $i < n$ whereas $xy^nx$ does not. This gives $I_n \neq I$ for all $n$.
Next argue that if there was a finite set of generators for the ideal $I$ then we would have $I = I_n$ for some $n$. Do this by observing that for each of the generators $f$ we must have $f = \sum_ig_ixy^{a_i}xh_i$ where the sum is over only finitely many terms. Hence taking $n$ large enough we must eventually generate $f$.