The Lie algebra $\mathfrak{sl}(N, \mathbb{C}[t])$ can be thought of as matrix valued polynomials. As a vector space $$\mathfrak{sl}(N, \mathbb{C}[t]) = \bigoplus_{k =0}^{\infty}\mathfrak{sl}(N, \mathbb{C})t^k $$ and we have $$[At^m, Bt^n] = [A,B]t^{n+m}$$ Since $\mathfrak{sl}(N, \mathbb{C})$ is simple, it can be readily seen that $\mathfrak{sl}(N, \mathbb{C}[t])$ is generated by elements of $\mathfrak{sl}(N, \mathbb{C})$ and $\mathfrak{sl}(N, \mathbb{C})t$. I would like to present this Lie algebra in as compact way as possible in terms of generators and relations. I have the following conjecture. Consider the Lie algebra $\mathfrak{g}$ spanned by $x, J(x)\;| x \in \mathfrak{sl}(N, \mathbb{C})$ with the following relations \begin{cases}{} & [x, J(y)] = J([x, y]) \\ & [x, [J(y), J(z)]] = [J(x), [y, J(z)]] \end{cases} Then $$\mathfrak{g} \cong \mathfrak{sl}(N, \mathbb{C}[t])$$ More precisely, the following surjective homomorphism is in fact an isomorphism of the two algebra. \begin{align*} &\mathfrak{g} \twoheadrightarrow \mathfrak{sl}(N, \mathbb{C}[t])\\ &x \mapsto x\\ &J(x)\mapsto tx \end{align*} I have been trying to prove it directly, but to no avail. Is this true?
If it doesn't seem feasible to directly prove or disprove, is there any general theory to deal with these kind of questions?
I have the follosing proof for $\mathfrak{sl}_2(\mathbb{C})$ case. However, I am not sure it's correct, because I am making weaker assumptions then in the paper on which I am relying (Proposition 3.1). I would be very grateful if someone can point out any flaws in the argument or confirm it's sound.
Generators of $\mathfrak{sl}(2, \mathbb{C}[t])$
Consider the Lie algebra $\mathfrak{g}$ spanned by $e,f,h, J(e), J(f), J(h)$ with the following relations
\begin{cases}{} & e,f,h \text{ satisfy the usual } \mathfrak{sl}(2,\mathbb{C}) \text{ relations} \\ (1) \label{first_rel} & [x, J(y)] = J([x,y]) %\label{second_rel} %& $[[x, J(y)], [J(z), J(w)]] = [[J(x), J(y)], [z, J(w)]]$ \end{cases}
Denote
\begin{align*} & e_n := (\frac{1}{2}\text{ad}_{h_1})^n e\\ & f_n := (-\frac{1}{2}\text{ad}_{h_1})^n f\\ & h_n := [e, f_n] \end{align*}
Claim
$[x_k, y_l] = [x,y]_{k+l}$ where $x_n = (ae + bf + ch)_n := ae_n + bf_n+ ch_n$
Proof
We just need to check it for basis elements, more precisely that for ${k+l=n}$ \begin{align*} & [e_k, e_l] = 0\\ & [f_k, f_l] = 0\\ & [h_k, h_l] = 0\\ & [e_k, h_l] = -2e_n\\ & [f_k, h_l] = 2f_n\\ & [e_k, f_l] = h_n \end{align*} Base case $n=1$ follow from (1). Assume that it's true for all $k+l < n$. First let's check that $[e_k, e_l] = [e_1, e_{n-1}]$ when $k, l > 0\;|\; k+l=n$. Indeed $$0 = [h_s, [e_1, e_{n-s-1}]] = 2([e_{s+1}, e_{n-s-1}] + [e_1,e_{n-1}])$$ In particular $[e_{n-1}, e_1] = [e_1, e_{n-1}]$, which means that all the above are zero. Moreover $$[e, e_n] = [e, [h_1, e_{n-1}]] = -2[e_1, e_{n-1}] + [h_1, [e, e_{n-1}]]=0$$ Similarly $[f_k, f_l] = 0$.
Let's continue writing similar identities $$ [e_s, [h_1, h_{n-s-1}]] = 0 \Rightarrow [h_{n-s-1},e_{s+1}] = [h_1, e_{n-1}] = 2e_n $$ and $$[h_{n-s-1},f_{s+1}] = [h_1, f_{n-1}] = -2f_n$$ Now note that $[h, e_n] =2e_n$ since $h$ and $h_1$ commute. Similarly $[h, f_n] = -2f_n$. Hence we have also proved that $$ [f_k, h_l] = 2f_n \text{ and } [h_l, e_k] = 2 e_n $$ It follows from definition that $[e_l, f_k] = h_n$. Indeed $$ [ad_{h_1} e_{l-1}, f_k] = [e_{l-1}, -ad_{h_1} f_k] \Rightarrow [e_l, f_k] = h_n $$ It now only remains to prove that $[h_l, h_k] = 0$. But that easily follows from what we already derived $$ [h_l, [e, f_k]] = 2[e_l, f_k] -2[e, f_{l+k}] = 2h_n - 2h_n = 0 $$
Corollary
The Lie algebra spanned by $e, f, h, h_1$ satisfying relations \begin{align*} &[e,f] = h\\ & [h,e] = 2e\\ & [f, h] = 2f\\ & [h, h_1] = 0\\ & [e, f_1] = h_1 \end{align*} is isomorphic to $\mathfrak{sl}(2, \mathbb{C}[t])$, where $e_1, f_1$ denote $\frac{1}{2}[h_1,e]$ and $\frac{1}{2}[f,h_1]$ respectively
Indeed, by virtue of Jacobi identity, we can derive all the neccessary relations to apply the above claim, e.g. $$ 2[e, f_1] = [[e, f], h_1] + [f, [e, h_1]] = [f, [e, h_1]] = 2[e_1, f] $$ $$[h, e_1] = [[h,h_1], e] + [h_1, 2e] = 2e_1$$ The consequence of the claim is that there is an isomomorphism from the Lie algebra $\mathfrak{g}$ to the Lie algebra $\mathfrak{a}$ spanned by elements $$J_n(x), x \in \mathfrak{sl}(2, \mathbb{C})$$ with relations $$ [J_k(x), J_l(y)] = J_{k+l}([x,y]) $$ It can be easily seen that these relations are satisfied in $\mathfrak{sl}(2, \mathbb{C}[t])$ by elements $xt^k, yt^l$ Moreover, the elements $xt^n$ span $\mathfrak{sl}(2, \mathbb{C}[t])$, so the homomorphism ${\mathfrak{a}\to \mathfrak{sl}(2, \mathbb{C}[t])}$ is surjective, while comparing the dimensions of each degree (both algebras are clearly graded), we conclude that it is also injective.