I am currently reading the book "Linear Algebraic Group" by Springer, more precisely in chapter 4 where Lie algebras of linear algebraic groups are introduced. I would like to prove that the Lie algebra of $SL_n$ is given by the algebra $\mathfrak{sl}_n$ of $n\times n$ matrices with trace zero using the formalism developped in the book.
Let $k$ be an algebraically closed field and consider the linear algebraic group $G=GL_n$, whose affine algebra is given by $k[G]=k[T_{i,j},\Delta^{-1}]_{1\leq i,j \leq n}$ where $\Delta=\det(T_{i,j})$. Inside $G$, we have the subgroup $H = SL_n$ which is Zariski closed in $G$, hence it is again a linear algebraic group whose affine algebra is $k[H]=k[T_{i,j}]/(\Delta - 1)$.
I know already that the Lie algebra $L(G)$ of $G$ can be identified with the algebra $\mathfrak{gl}_n$ of $n\times n$ matrices with usual Lie brackets. That is because the $k$-derivations of $k[G]$ commuting with all left translations are exactly the $D_X$ for $X=(x_{i,j})\in \mathfrak{gl}_n$, defined by $$D_XT_{i,j}:=-\sum_{h=1}^nT_{ih}x_{hj}$$ To compute the Lie algebra $L(H)$ of $SL_n$, the book recommends to proceed in the following manner. I let $J=(\Delta-1)$ be the ideal of $k[G]$ defining $H$, and I consider $\mathcal D_{G,H}$ the set of all the $k$-derivations of $k[G]$ preserving $J$. Then, there is a natural isomorphism $$\mathcal D_{G,H}\cap L(G) \cong L(H)$$ Whence, to prove that $L(H)$ identifies with $\mathfrak{sl}_n$, I need to show the following:
We have $D_X(\Delta - 1) \in (\Delta - 1)$ if and only if $\operatorname{Trace}(X)=0$.
By argueing on dimension, it is actually enough to just prove that $D_X(\Delta - 1) \in (\Delta - 1)$ for $X = E_{i,j}$ or $E_{i,i}-E_{j,j}$ for $i\not = j$. I thought that it would be easy enough, but well... The computations are terrible and I can't reach the desired conclusion (even considering the case $n=2$).
Is there any "clever" way to tackle this problem ? Also, is there maybe another way to characterize the Lie algebra that would make the argument more direct ?
I believe I have found another way of argueing that $L(SL_n)$ is identified with $\mathfrak{sl}_n$ by computing the tangent space $T_{I_n}SL_n$ rather than by considering left-invariant derivations. Recall that $G = GL_n$ and $H = SL_n$.
By definition of the module of differentials $\Omega_H:=\Omega_{k[H]/k}$, I know that there are isomorphisms of $k$-vector spaces$$T_{I_n}H \cong \operatorname{Hom}_{k[H]}(\Omega_H,k_{I_n})\cong \operatorname{Hom}_k(\Omega_H(I_n),k)$$ where $k_{I_n}$ is $k$ seen as a $k[H]$-module via the identification $k[H]/M_{I_n}\cong k$, $M_{I_n}$ being the maximal ideal in $k[H]$ of regular functions vanishing at $I_n$ ; and then $\Omega_H(I_n) = \Omega_H/M_{I_n}\Omega_H$ (these notations are the ones used in Springer's book).
Now, because $k[H] = k[T_{i,j}]/(\Delta - 1)$, by denoting $t_{i,j}$ the image of $T_{i,j}$ in $k[H]$, I know that $\Omega_H$ is generated by the $dt_{i,j}$. If I consider the canonical basis $(e_{i,j})$ of $k[H]^{n^2}$ and define the map $k[H]^{n^2}\rightarrow \Omega_H$ sending $e_{i,j}$ to $dt_{i,j}$, then the kernel is the submodule generated by the element $\sum_{i,j=1}^n \frac{\partial \Delta}{\partial T_{i,j}}(t)e_{i,j}$ (the term "$-1$" disappears in the derivation). Thus I have a presentation of $\Omega_H$ as the quotient of the $n^2$-dimensional free module over $k[H]$ with the above kernel.
Now, I would like to understand the quotient $\Omega_H(I_n)$. This amounts to tensoring $\Omega_H$ with $k[H]/M_{I_n}\cong k$ over $k[H]$, the isomorphism with $k$ being given by evaluating in $I_n$. Thus $$\Omega_H(I_n)\cong k^{n^2}/(\sum_{i,j=1}^n \frac{\partial \Delta}{\partial T_{i,j}}(I_n)e_{i,j})$$
But now, if $X$ is the matrix whose $(i,j)-$coefficient is $T_{i,j}$, then $\frac{\partial \Delta}{\partial T_{i,j}}$ is its $(i,j)-$cofactor. Evaluating it at $I_n$ (that is, taking $T_{i,j}=\delta_{i,j}$), these partial derivatives are $1$ when $i=j$ and $0$ otherwise. Thus, $\Omega_H(I_n)\cong k^{n^2}/(\sum_{i}^n e_{i,i})$.
I need now to compute the dual of this vector space. That is the set of all linear form on $k^{n^2}\cong M_n(k)$ vanishing on the identity. Because every linear form of $M_n(k)$ has the form $M\mapsto \operatorname{Trace}(XM)$ for some $X\in M_n(k)$, I see now that $T_{I_n}H$ can naturally be identified with the space $\mathfrak sl_n$. It would remain to check that the Lie algebra structure of $T_{I_n}H$ coincide with that of $\mathfrak sl_n$. That is something I am still unsure, as the Lie algebra structure on the tangent space comes from that on the derivations of $k[H]$, so it's a little abstract.