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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 $k$.

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    $\begingroup$ 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 $V\to k$.) $\endgroup$ Apr 26, 2020 at 2:43
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    $\begingroup$ Have you tried to see if your definition is natural with respect to vector space isomorphism? $\endgroup$ Apr 26, 2020 at 2:49
  • $\begingroup$ @EricWofsey you're right! I added the condition of the field $k$ to be infinite. Thanks! $\endgroup$
    – Albert
    Apr 26, 2020 at 3:05
  • $\begingroup$ @EthanDlugie I didn't, but I'll try! Thanks! $\endgroup$
    – Albert
    Apr 26, 2020 at 3:05

1 Answer 1

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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 $t=0$.)

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