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Can $k[x_1,...,x_n]$, the ring of polynomials with coefficients $\in k$ where $k$ is a field, ever be a non-commutative ring?

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    $\begingroup$ No, by definition. $\endgroup$
    – Zhen Lin
    Feb 12, 2014 at 21:01
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    $\begingroup$ No, it cannot be non-commutative. $\endgroup$ Feb 12, 2014 at 21:02
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    $\begingroup$ It is commutative. $\endgroup$ Feb 12, 2014 at 21:02
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    $\begingroup$ Ah, a much better solution to the 15-character problem, @ZhenLin $\endgroup$ Feb 12, 2014 at 21:02
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    $\begingroup$ In addition to what others have said, one can define a non-commutative polynomial ring by declaring the variables $x_1, \ldots, x_n$ to be non-commutative. $\endgroup$ Feb 12, 2014 at 21:03

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The notation $k[x_1,\dots,x_n]$ is used to denote the ring of polynomials in $n$ indeterminates over $k$ and that is, by definition, commutative.

There is a thing that could be seen as a non-commutative polynomial ring: the free algebra on $n$ symbols $x_1, \dots, x_n$ over $k$. It can be constructed in a way analoguously to an ordinary polynomial ring: elements are finite sums of (non-commutative) monomials of the form $\alpha \vec x$, where $\alpha \in k$ and $\vec x$ is a (finite, possibly empty, repetitions allowed) sequence of elements of the set $\{x_1,\dots,x_n\}$. I've seen it denoted by $k\langle x_1,\dots,x_n\rangle$ and by $k\{x_1,\dots,x_n\}$. (Note that this boils down to exactly is said in the comments: you declare the variables to be non-commuting; or, maybe more accurately formulated, you do not declare them to be commuting.)

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  • $\begingroup$ when we say "...polynomials in n indeterminates over $k$" do we mean the values that $x_n$ can take come from $k$? $\endgroup$ Nov 17, 2020 at 17:17
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    $\begingroup$ It is slightly more subtle. Polynomials are not functions, so there is - at least not by definition - no such thing as 'values that the $x_i$ can take on'. Given a $k$-algebra $A$, you can interpret a polynomials in $k[x_1,\dots,x_n]$ as functions from $A^n$ to $A$, so in that context you can say that the $x_i$ can take on values in $A$. And you can apply this to $A=k$, so a polynomial indeed can be interpreted as a function $k^n \to k$. $\endgroup$ Nov 17, 2020 at 17:23
  • $\begingroup$ That makes sense. So if we study $p = x + 1 \in \mathbb{Z}_{2}[x]$, and select $x = 6$ when $6 \notin \mathbb{Z}_{2}[x]$ we can still get something meaningful: $p = 1$? $\endgroup$ Nov 17, 2020 at 17:28
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    $\begingroup$ I would say $6 = 0 \in {\mathbb Z}_2$ (although formally it is of course that 'the residue class of $6$' = 'the residual class of $0$' ${}\in {\mathbb Z}_2$). $\endgroup$ Nov 17, 2020 at 18:07
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    $\begingroup$ But, yes, evaluating $x + 1 \in {\mathbb Z}_2[x]$ at $x = 6 (= 0)$ gives $1$. $\endgroup$ Nov 17, 2020 at 18:14

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