4
$\begingroup$

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.

$\endgroup$
5
  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ $\sum_{i_1, \ldots, i_k} \lambda_{i_1i_2\ldots i_k}x_1^{i_1}x_2^{i_2}\ldots x_k^{i_k}$, 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.