Assume that I have a finitely presented algebra $A$ over the complex numbers (by which I mean that $A$ is generated over $\mathbb{C}$ by finitely many elements $x_1,...,x_n$ subject to finitely many relations). Let now $y_1,\ldots, y_m$ be a finite subset of elements of $A$. Is there an algorithm which gives a presentation of the subalgebra $B$ of $A$ generated by $y_1,...y_m$ ? In particular, I am interested in the following example: Let $A = \mathbb{C}B_3 = \mathbb{C}\langle \sigma_1,\sigma_2 | \sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2\rangle$ be the group algebra of the Braid group on 3 braids. Is there any nice presentation of the subalgebra generated by $\sigma_1$ and $\sigma_2 + \sigma_2^{-1}$ ?
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$\begingroup$ Just to make sure I'm understanding the question correctly: The presentation that you seek for the subalgebra $B$ need not use the known $y_1,\dots,y_m$ as its generators but rather can use completely different elements to generate $B$, right? $\endgroup$– Andreas BlassCommented Jul 30, 2015 at 16:03
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$\begingroup$ I would prefer to know the relations between the given generators, but also a presentation with respect to another set of generators will be good. $\endgroup$– Ehud MeirCommented Jul 31, 2015 at 15:14
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If we modify your first question only slightly, then the answer to the question is no, there is no algorithm. By Theorem 1 of the paper G. Baumslag, W. W. Boone and B. H. Neumann, Some unsolvable problems about elements and subgroups of groups, Math. Scand. 7, 191-201 (1959), let $G$ be a finitely presented group for which is it undecidable whether elements have finite order. Let $A=\mathbb{C}G$. For each element $g\in G$, we can take $B_g=\mathbb{C}[g]$ to be the subalgebra generated by $g$. Any non-trivial relation in $B_g$ would tell us $g$ has finite order, and so there can be no algorithm (which is independent of $g$) to yield a presentation of $B_g$.
answered Aug 4, 2015 at 13:07