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?
[Math] Ring of Polynomials Commutative
abstract-algebrapolynomialsring-theory
Related Solutions
Notice the first construction you gave can also be described as the semigroup ring $R[\Bbb N ^k]$, and makes a polynomial ring with $k$ commuting indeterminates. Moreover, the coefficients commute with indeterminates, but not always with each other.
Of course, you can let go of both these properties and generate a ring of "generalized polynomials," which look like finite sums of finite products of coefficients interleaved with indeterminates.
I'm not sure that these rings have significant usage (definitely nothing compared to polynomials over commutative rings); however, its key advantage is that evaluation maps work properly.
For example, in the ordinary polynomial ring $\Bbb H[x]$ over quaternions, $xi-ix=0$, since coefficients commute with indeterminates. But this equation doesn't remain true if you try to evaluate it at quaternion values: $j$ would yield $0\neq 2ij$. But in the ring of generalized polynomials, evaluation works just fine.
Naturally, things are messier than normal for solutions to such generalized polynomials. For example, $ix-xi +1=0$ doesn't have any solutions over $\Bbb H$, but $x^2+1=0$ has uncountably many solutions.
For polynomials over any commutative coefficient ring, the (high-school) polynomial long division algorithm shows how to divide with remainder by any monic polynomial, i.e any polynomial $\rm\:f\:$ whose leading coefficient $\rm\:a=1\:$ (or a unit, i.e. $\,\rm a\mid 1),\,$ since this implies the leading monomial $\rm\,ax^n\,$ of $\rm\:f\:$ divides all higher degree monomials $\rm\:x^k,\,$ so the division algorithm works to kill all higher degree terms in (successive) dividends, leaving a remainder of degree $\rm < n = deg\ f.$
But this generally fails if $\rm\:f\:$ is $\rm\color{#c00}{not\ monic}$, e.g. $\rm\: x = ax\:q + r,\ \color{#c00}{a\nmid 1}\,$ has no solution since evaluating at $\rm\:x=0\:$ $\Rightarrow$ $\rm\:r=0,\:$ then eval at $\rm\:x=1\:$ $\Rightarrow$ $\rm\,1 = aq\,\Rightarrow\,\color{#c00}{a\mid 1\Rightarrow\!\Leftarrow}\,$ Conversely, if $\rm\,a\,$ is a unit, then $\rm\,ab = 1$ for some $\rm\,b\,$ so the division is possible: $\rm\: x = ax\cdot b + 0.$
However, it is possible to generalize division with remainder to non-monic polynomials by scaling the dividend by a sufficiently large power $\,\color{#0a0}{a^i}$ of the leading coefficient of the divisor, i.e.
Theorem (nonmonic Polynomial Division Algorithm) $\ $ Let $\,0\neq F,G\in A[x]\,$ be polynomials over a commutative ring $A,$ with $\,a\,$ = lead coef of $\,F,\,$ and $\, i \ge \max\{0,\,1+\deg G-\deg F\}.\,$ Then
$\qquad\qquad \phantom{1^{1^{1^{1}}}}\color{#0a0}{a^{i}} G\, =\, Q F + R\ $ $\ {\rm for\ some}\ \ Q,R\in A[x],\ \deg R < \deg F$
Proof $\ $ See here for a few proofs.
Best Answer
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.)