Basis-free definition of derivative of polynomial functions on a vector space

Let $$V$$ be a finite dimensional vector space over an infinite field $$k$$. The ring of polynomial functions on $$V$$ is the subalgebra of the $$k$$-algebra of all functions $$V\to k$$ generated by the dual space $$V^*$$, and is denoted by $$k[V]$$.

Let $$(e_1,\dots,e_n)$$ be an ordered basis of $$V$$ and let $$(f_1,\dots,f_n)$$ be its dual basis, then an element of $$k[V]$$ is a polynomial in $$f_1,\dots,f_n$$. We can then define a (formal) derivative as follows: First, fix $$i\in\{1,\dots,n\}$$ and define $$\partial_{e_i}(f_1^{r_1}\cdots f_{i-1}^{r_{i-1}}f_i^{r_i}f_{i+1}^{r_{i+1}}\cdots f_n^{r_n}) = r_i f_1^{r_1}\cdots f_{i-1}^{r_{i-1}}f_i^{r_i-1}f_{i+1}^{r_{i+1}}\cdots f_n^{r_n},$$ for all $$r_1,\dots,r_n\in \mathbb{Z}_{\geq 0}$$. Extending by linearity we obtain a well defined derivation $$\partial_{e_i}:k[V]\to k[V]$$. Then for $$v\in V$$, write $$v = \sum_{i=1}^n a_i e_i, \qquad a_1,\dots,a_n\in k$$ and define $$\partial_v(f) = \sum_{i=1}^n a_i \partial_{e_i}(f), \qquad \forall f\in k[V].$$

When we take $$V=k^n$$ and $$(e_1,\dots,e_n)$$ as the canonical ordered basis, the $$i$$-th vector in the dual basis is the coordinate function $$x_i:k^n\to k$$ given by $$x_i(a_1,\dots,a_n) = a_i$$, and $$k[V]$$ is precisely the polynomial ring $$k[x_1,\dots,x_n]$$ and the derivation $$\partial_v$$ coincides with the known formal directional derivative on that polynomial ring.

The main issue with this definition is that it depends on the chosen basis $$(e_1,\dots,e_n)$$. I would like to know if there is a basis-free definition of the derivative $$\partial_v$$ for a ring of polynomial functions $$k[V]$$ on a finite dimensional vector space $$V$$ over an infinite field $$k$$.

• You should really assume $k$ is infinite here or else the representation of an element of $k[V]$ as a polynomial in the $f_i$ is not unique. (Or, if you allow $k$ to be finite, you need to change your definition of $k[V]$ to be something that gives "formal" polynomials instead of functions $V\to k$.) Apr 26, 2020 at 2:43
• Have you tried to see if your definition is natural with respect to vector space isomorphism? Apr 26, 2020 at 2:49
• @EricWofsey you're right! I added the condition of the field $k$ to be infinite. Thanks! Apr 26, 2020 at 3:05
• @EthanDlugie I didn't, but I'll try! Thanks! Apr 26, 2020 at 3:05

Well, you could just define $$\partial_v$$ as the unique $$k$$-linear derivation on $$k[V]$$ such that $$\partial_v(f)=f(v)$$ for all $$f\in V^*$$. Of course, you have to prove that such a derivation actually exists (and is unique, but that part is easy), and for that you probably want to pick a basis, but the definition itself does not involve a basis.
Another possibility is to just adapt the classical calculus definition. First, note that if $$f\in k[V]$$ and $$v\in V$$, then the function $$x\mapsto f(x+v)$$ is also in $$k[V]$$ (this is clear if $$f\in V^*$$, and remains true if you take products and linear combinations). Now given $$f\in k[V]$$ and $$v\in V$$, you can define a function $$g:k\to k[V]$$ by $$g(t)=(x\mapsto f(x+tv))$$, and $$g$$ will actually be a polynomial function, i.e. a function of the form $$g(t)=\sum_{k=0}^m c_kt^k$$ for $$c_k\in k[V]$$ (again, this is clear if $$f\in V^*$$ and remains true if you take products and linear combinations). You can then define $$\partial_v f$$ to be the linear coefficient $$c_1$$ of this polynomial $$g$$. (Observing that $$c_0=g(0)=f$$, this linear coefficient is just what you get by taking the quotient $$\frac{f(x+tv)-f(x)}{t}$$ as a polynomial in $$t$$ and then plugging in $$t=0$$.)