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Let $k$ be a field and $F=k \langle x_{1},\ldots,x_{n}\rangle $ be a finitely generated free associative algebra. Let $g = (g_{ij})$ be any matrix of the group $GL_n(k)$ and define the algebra automorphism $\rho_g:F\to F$ such that $x_i\mapsto \sum_j g_{ij} x_j$.

Then, I am looking for an algebra $A = F/I$ where the two-sided ideal $I\subset F$ is $GL_n(k)$-invariant, that is, for any $g \in GL_n(k)$, $\rho_{g}(I)\subset I$.

Please note that, the algebra $F/T(R)$ corresponding to an associative algebra $R$, where the two-sided ideal $T(R) = \{f\in F\mid f(r_1,\ldots,r_n) = 0,\ \mbox{for all}\ r_1,\ldots,r_n \in R\},$ is $GL_n(k)$-invariant. By the definition $F/T(R)$ is a relative free algebra in $n$ variables defined by $R$. But I want some other examples.

kumar
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    This is really stupid, especially since I don't think I've taken the time to properly understand your question, but -- isn't $GL_n(k)$ a $GL_n(k)$-invariant $k$-algebra? -- never mind, this isn't a finitely-generated algebra $F/I$ – Chill2Macht Apr 10 '17 at 10:14
  • $GL_n(k)$ is a group not an algebra. – kumar Apr 10 '17 at 11:42
  • If $I \subset F$ is $Aut(F/k)$ invariant, doesn't $I$ have to be the trivial ideal ? – reuns Apr 10 '17 at 12:32
  • You're right I was confusing it with the endomorphism algebra, of which it is a subset. – Chill2Macht Apr 10 '17 at 13:40

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