# Ring of Polynomials Commutative?

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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No, by definition. – Zhen Lin Feb 12 '14 at 21:01
No, it cannot be non-commutative. – Kevin Carlson Feb 12 '14 at 21:02
It is commutative. – Stefan Hamcke Feb 12 '14 at 21:02
Ah, a much better solution to the 15-character problem, @ZhenLin – Kevin Carlson Feb 12 '14 at 21:02
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. – Ayman Hourieh Feb 12 '14 at 21:03

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.)