Let $k$ be a field of $\text{char}\,k \neq 2$. For any $n \ge 1$, define a $k$-algebra $A_n(k)$, on $2n$ generators, as a quotient of the free algebra $k\langle x_1, \ldots, x_n, y_1, \ldots, y_n\rangle$ by the two-sided ideal generated by the set$$\{x_iy_j - y_jx_i - \delta_{ij},\,x_i x_j - x_jx_i, \,y_iy_j - y_jy_i, \, i , j = 1, \ldots, n\}, \quad \delta_{ij} := \begin{cases} 1 & \text{if }i = j \\ 0 & \text{if }i \neq j.\end{cases}$$We define an $A_n(k)$-action on the vector space $M := k[t_1, \ldots, t_n]$ as follows:$$x_i \text{ acts as }{\partial\over{\partial t_i}}, \quad \text{resp. }y_i\text{ acts as multiplication by }t_i, \quad \forall i = 1, \ldots, n.$$I have two questions.
- How do I see that this action makes $M$ a cyclic $A_n(k)$-module?
- How do I see write this module in the form $M = A_n(k)/I$ for an appropriate left ideal $I \subset A_n(k)$?
For 1, just note that acting with the $y_i$ already generates $M$ since, given $f\in M$, we have $$f(y_1,\ldots,y_n).1=f(t_1,\ldots,t_n).$$ For 2, the ideal in question is the ideal generated by $(x_1,\ldots,x_n)$. To see this, let $$\phi: A_n(k)\to M$$ be the map defined by $\phi(X)=X.1$ where $X\in A_n(k)$ (so the map is "act on $1$"). Well, it is easy to check that $\phi$ is a module homomorphism since $$\phi(X_1X_2)=(X_1X_2).1=X_1.(X_2.1)=X_1.\phi(X_2).$$ As this map is surjective (by 1) it follows that $M\cong A_n(k)/\ker\phi$ by the first isomorphism theorem. To prove that $\ker\phi=(x_1,\ldots,x_n)$ one first needs to develop a basis theorem: the set $$\{y_1^{r_1}\cdots y_n^{r_n}x_1^{s_1}\cdots x_n^{s_n}\mid r_i,s_j\in\mathbb{Z}_+\}$$ is a basis. Once you know this, you can see that the only basis elements that are not killed by $\phi$ are those for which $s_1=\cdots=s_n=0$.