Let $C$ be a smooth projective curve over an algebraically closed field $k$, and let $k(C)$ be its function field. For elements $f_1,...,f_n \in k(C)$ and a nonconstant $t \in k(C) \setminus k$, we let the Wronskian $W_t(f_1,...,f_n)$ be the determinant of the $n \times n$ matric whose $j$-th row entries are the $(j-1)$-th derivatives of the $f_i$'s with respect to $t$.
Question 1. As stated in the title of the question, what does the last line mean?
Question 2. Let $z$ be another nonconstant element. How do we prove the equality $$W_z(f_1,...,f_n) = \left(\frac{dt}{dz}\right)^{{n \choose 2}}W_t(f_1,...,f_n)?$$
Question 1. As stated in the title of the question, what does the last line mean?
Question 2. Let $z$ be another nonconstant element. How do we prove the equality $$W_z(f_1,...,f_n) = \left(\frac{dt}{dz}\right)^{{n \choose 2}}W_t(f_1,...,f_n)?$$
Answer: This is not a complete answer but it illustrates how you may prove the formula in general. Let $K:=k(C)$ and let $t,z \in K$ be non constant elements. This is "by definition" the following: The elements $t,z$ are algebraically independent elements over the base field $k$ and they span polynomial sub rings $k[t], k[z] \subseteq K$. Let $d_1:=\partial_t, d_2:=\partial_z$ be partial derivatives wrto the variables $t,z$. It follows since $t,z \neq 0$ there is a non zero element $b\in K$ with $z=bt$. By the chain rule for derivation you get the formula
$$ d_2=\partial_z = \partial_z(t) \partial_t =d_2(t)d_1.$$
If $f,g \in K$ are two elements it follows
$$W_t(f,g)=fd_1(g)-gd_1(f).$$
You may also check that
$$ W_z(f,g)= fd_1(g)d_2(t)-gd_1(f)d_2(t)=d_2(t)W_t(f,g).$$
Hence the formula holds for $n=2$.