Question about the definition of the Jacobian ideal

Let $R=\frac{k[[x_1,\ldots,x_n]]}{(f_1,\ldots,f_c)}$ (or $\frac{k[x_1,\ldots,x_n]}{(f_1,\ldots,f_c)}$), where $k$ is a field with characteristic $0$ and $0\ne(f_1,\ldots,f_c)\subseteq (x_1,\ldots,x_c)$.

In the paper On the fitting ideal in free resolution (see also Jacobian ideal reference), there is a definition of the Jacobian ideal of $R$. Assume that the height of $(f_1,\ldots,f_c)$ in $k[[x_1,\ldots,x_n]]$ is $h$, i.e. $h=\operatorname{inf}\{\operatorname{height}(p)\mid p\in \operatorname{Spec}(k[[x_1,\ldots,x_n]]), (f_1,\ldots,f_c)\subseteq p\}$. Then the Jacobian ideal of $R$ is defined to be the ideal of $R$ generated by $h\times h$ minors of the Jacobian matrix $\frac{\partial(f_1,\ldots,f_c)}{\partial(x_1,\ldots,x_n)}$.

I have two questions about the definition.

Must the Jacobian ideal of $R$ be non-zero? I am confused that why the definition considers $h\times h$ minors of the Jacobian matrix. Why not higher minors?

Consider the example $R=\frac{k[[x,y,z]]}{(xy,xz)}$, the height of $(xy,xz)$ in $k[[x,y,z]]$ is $1$. We can check easily that there is a $2\times 2$ minor of the Jacobian matrix that is non-zero in $R$. So it seems that it is reasonable to consider higher minors.

Thank you in advance.

The ideal can be zero, but only if $R$ is non-reduced: take for example $(x^2,xy,y^2)$ which has Jacobian ideal $(2x^2,2y^2,4xy)=0$. If $R$ is reduced, then $\operatorname{Spec} R$ is either a reduced variety (in the case you're taking $k[x_1,\cdots,x_n]$) or the germ of a reduced variety (in the case you're taking $k[[x_1,\cdots,x_n]]$) and such things are generically smooth if the characteristic of $k$ is zero. This is covered in most introductory algebraic geometry texts if you're looking for a reference.

The reason for considering $h\times h$ minors is that you're trying to look at nonsingularity of $V(f_1,\cdots,f_c)$, which happens when the Jacobian matrix is of rank $n-\dim_p V(f_1,\cdots,f_c)$ at each point of $p\in V(f_1,\cdots,f_c)$.