When $k$ is algebraically closed a regular point is smooth I would like to show that in the case of a $k$-scheme of locally finite type with $k$ algebraically closed, a regular point is smooth. For smooth implies regular, the proof in Görtz and Wedhorn is pretty clear. 
Does anyone know a nice argument for the converse? I am using the definitions of GW, i.e. 
$x\in X$ is smooth if there exists an open neighborhood $U$ of $x$ and an open immersion $$j: U\hookrightarrow Spec \text{ }k[T_1,\dots , T_n]/ (f_1,\dots , f_{n-d})$$ for suitable $n$ and $f_i$, such that the Jacobian matrix $$J_{f_1,\dots , f_{n-d}}(x)=\left( \frac{\partial f_i}{\partial T_j}(x)\right)_{i,j}\in M_{(n-d)\times n}(k)$$ has rank $n-d$.   
$x\in X$ is regular in $x$ if the stalk $\mathcal O_{X,x}$ is a regular ring. 
 A: If $\mathcal O_{X,x}$ is a regular local ring with Krull dimension $d$, its maximal ideal $\mathfrak m_x\subset \mathcal O_{X,x}$ has a minimal set of generators consisting of $d$ elements. This is in fact equivalent to $$\dim_kT_xX=d,$$ where $T_xX=(\mathfrak m_x/\mathfrak m_x^2)^\vee$ is the Zariski tangent space of $X$ at $x$. This is a vector space over $k(x)=k$. Now since $X$ is locally of finite type over $k$, $x$ has an open neighborhood contained in an affine scheme $$\textrm{Spec }k[T_1,\dots,T_n]/(f_1,\dots,f_h)\subset \mathbb A^n_k$$ for some $f_1,\dots,f_h\in k[T_1,\dots,T_n]$ and $n\geq d$. The condition that $x$ is regular means, by the Jacobian criterion, that the matrix $$J_{f_1,\dots,f_h}(x)=\Bigl(\frac{\partial f_i}{\partial T_j}(x)\Bigr)\in M_{h\times n}(k)$$ has rank $n-d$.* In particular $h\geq n-d.$ Now in case $h>n-d$, we are entitled to remove some rows until we end up with just those generating the row space of $J_{f_1,\dots,f_h}(x)$. After relabelling them from $1$ to $n-d$, we see that the open neighborhood $$x\in U\subset \textrm{Spec }k[T_1,\dots,T_n]/(f_1,\dots,f_{n-d})$$ we are after can be chosen to be the intersection $$U=V\cap \textrm{Spec }k[T_1,\dots,T_n]/(f_1,\dots,f_{n-d}),$$ where $V$ is the open subset of $\mathbb A^n_k$ consisting of those $y\in\mathbb A^n_k$ such that $\textrm{rank }J_{f_1,\dots,f_{n-d}}(y)=n-d$.

*Note here that we may find other equations $g_1,\dots,g_\ell\in k[U_1,\dots,U_m]$ such that $x$ has an open neighborhood contained in $V(g_1,\dots,g_\ell)\subset\mathbb A^m_k$, and we may now construct the new matrix $J_{g_1,\dots,g_\ell}(x)$. But we would still have $n-\textrm{rank }J_{f_1,\dots,f_h}(x)=m-\textrm{rank }J_{g_1,\dots,g_\ell}(x)=d.$ This difference stays constant! It is kind of nontrivial that the notion of regularity in terms of the Jacobian criterion, which seems dependent upon a choice of equations vanishing around $x$, is in fact equivalent to the notion of $\mathcal O_{X,x}$ being a regular local ring, which is completely intrinsic to $x$.
