Be $k$ a field. I'm trying to define a function on $k[x_1, ..., x_n]$. However, I know of no way to write an arbitrary element of this ring efficiently. I read somewhere about using the $S_n$-orbit, but that seemed cumbersome too. Maybe I'll just have to accept it is cumbersome, but any ideas are appreciated.

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    $\begingroup$ If $I : \{1, \ldots, n\} \to \Bbb{N}$ is a sequence of $n$ natural numbers $\langle i_1, \ldots, i_n\rangle$, write $x^I$ for $x_1^{i_1} \cdots x_n^{i_n}$. The $x^I$ are a basis for $k[x_1, \ldots, x_n]$ as a vector space over $k$ and any element of $k[x_1, \ldots, x_n]$ can be written uniquely as a sum $\sum_I a_I x^I$ where only finitely many of the coefficients $a_I \in k$ are non-zero. Does that help? $\endgroup$
    – Rob Arthan
    Jun 13 '19 at 20:25
  • $\begingroup$ Hmm I see, but for my purposes not really... Given $n$ $k$-linear maps $f_1, \ldots f_n : V \rightarrow V$, I want to define a homomorphism $\rho : k[x_1, \ldots, x_n] \rightarrow \text{Aut}_k(V)$ by setting $\rho(x_i)v := f_i(v)$. I think I have found a way now, which I'll post in a minute. $\endgroup$ Jun 13 '19 at 20:39
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    $\begingroup$ $\sum_{i_1, \ldots, i_k} \lambda_{i_1i_2\ldots i_k}x_1^{i_1}x_2^{i_2}\ldots x_k^{i_k}$, for $0 \leq i_j \leq n_j$ and $1 \leq j \leq k$. (Here $n_j$ is the degree of the $j-$th variable.) $\endgroup$ Jun 13 '19 at 20:42
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    $\begingroup$ @JosvanNieuwman Be careful, such a function $\rho$ will only be defined if the $f_i$ all commute with each other. Otherwise you'll want the ring $k\langle x_1,\ldots,x_n\rangle$, the free noncommutative algebra. $\endgroup$
    – jgon
    Jun 14 '19 at 2:28
  • $\begingroup$ You're absolutely right. In my situation pairwise commutativity of the $f_i$ is given. $\endgroup$ Jun 14 '19 at 2:55

What $k[x_1,\dots,x_n]$ actually means is the free commutative $k$-algebra generated from the set $\{x_1,\dots,x_n\}$. We can generalize this to $k[I]$ for any set $I$. Of course, just writing $k[I]$ without comment will definitely lead to confusion. If I were writing a sizable article that leverages this notation a bit, I may well define the notation for an arbitrary set, and then state that $k[x_1,\dots,x_n]$ is shorthand for $k[\{x_1,\dots,x_n\}]$ (assuming all $x_i$ are distinct).1

For your purposes, a more practical compromise would be to write $k[\{x_i\}_{i\in I}]$ where you also have $\{f_i\}_{i\in I}$ and $\rho$ becomes simply $\rho(x_i)(v)=f_i(v)$. Often $[n]$ (or just $n$ itself) is used for notation for the $n$-element set $\{0,\dots,n-1\}$ (or $\{1,\dots,n\}$ if you're uncouth), so you could say $k[\{x_i\}_{i\in[n]}]$ and $\{f_i\}_{i\in[n]}$.

Notation like $\{a_i\}_{i\in I}$ is commonly used to represent the $I$-indexed family of objects $a_i$. If all the $a_i$ are drawn from some encompassing set, $A$, this can be identified with a function $I\to A$. In set theory, such a function is a set of pairs, so if you wanted to be really precise, you should have something like $\rho((i,x_i))(v)=f_i(v)$.

1 Actually, knowing me, I would not do this, and would only mention this to explain the connection to more typical notation if I mentioned it at all.


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