I'm reading Humphreys' Linear Algebraic Groups, trying to understand the following argument found on the top of pg. 76.
Let $G$ be an algebraic group over some field $k$, with $x\in G$. Let $\gamma_x:G\to G$ be the commutator map sending $y$ to $yxy^{-1}x^{-1}$. Note that the fiber over $e\in G$ is exactly the centralizer of $x$, $C_G(x)$. Earlier it is shown that the differential $(d\gamma_x)_e:\mathfrak{g}\to\mathfrak{g}$ is $1-\mathrm{Ad}\;x$, where $\mathfrak{g}$ is the Zariski tangent space of $G$ at the identity. Notice then that the kernel of $(d\gamma_x)_e$ is $\mathfrak{c}_{\mathfrak{g}}(x)=\{a\in\mathfrak{g}\mid a=\mathrm{Ad}\;x(a)\}$.
I understand all of this just fine. What I don't understand is the next claim. Humphreys states that from the above setup, we can conclude that $\mathscr{L}(C_G(x))\subset\mathfrak{c}_{\mathfrak{g}}(x)$. How can we conclude this? I must not have a firm grasp of how to think about the Lie algebra associated to an algebraic group. The two definitions I'm familiar with are the space of left invariant derivations of $k[G]$, or the Zariski tangent space of $G$ at $e$. Neither one helps me see why the above containment should be true.
Let $f:X\to Y$ be a morphism of affine varieties. Let $y\in Y$ and denote by $X_y=f^{-1}(y)$ the fiber. It is a closed subvariety of $X$. Let $x\in X_y$ be any point and $I=I(X_y)$. Then we have $I\subseteq \mathfrak m_x$. Denote by $T_x:=(\mathfrak{m}_x/\mathfrak{m}_x^2)^\ast$ the Zariski tangent space of $x$ and similarly, define $T_y$. Let $T'_x$ be the tangent space of $X_y$ at $x$. We have $T'_x\subseteq T_x$ in a way which I will explain in the following. The goal is to show that $T'_x$, as a subspace of $T_x$, lies in the kernel of the differential map $T_x\to T_y$.
The morphism $f$ induces a ring homomorphism $\phi:\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ with the property that $\mathcal O_{X,x}\cdot \phi(\mathfrak m_y)=I\subseteq \mathfrak m_x$. The dual map $\phi^\ast:\mathfrak m_x^\ast \to \mathfrak m_y^\ast$ has the kernel $K:=\{ \ell\in\mathfrak m_x^\ast \mid \ell(I)=0 \}$, because $0=\phi^\ast(\ell)=\ell\circ\phi$ if and only if $I=\operatorname{im}(\phi)\subseteq\ker(\ell)$. The tangent space $T'_x$ of $X_y$ at $x$ is the dual of $(\mathfrak m_x/I)/(\mathfrak m_x/I)^2$ and the dual map of $\rho:\mathfrak m_x \to \mathfrak m_x/I$ will satisfy $\operatorname{im}(\rho)\subseteq K$. Therefore, when we pass to quotients, the inclusion $T'_x\hookrightarrow T_x$ will map into the kernel of $T_x\to T_y$.
This is precisely what we wanted.