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I let $M$ denote the free commutative monoid generated by some elements $x_1,\dots, x_n$. Suppose too that $R$ is a commutative ring. Then I can construct the free algebra of $M$ over $R$, denote it $R[M]$.

Recall that $R[M]$ is then the set of mappings $f\colon M\to R$ such that $f(m)\neq 0$ for only finitely many $m\in M$. Then $+$ and $\cdot$ are given by $(f+g)(m)=f(m)+g(m)$, $(fg)(m)=\sum_{pq=m}f(p)q(p)$, $0(m)=0$, and $1(1)=1$ and $1(m)=0$ if $m\neq 1$. Moreover, $R[M]$ has a subring isomorphic to $R$ by associating $r\in R$ with $r'\in R[M]$ such that $r'(1)=r$ and $r'(m)=0$ otherwise. It also has a submonoid of the multiplicative monoid isomorphic to $M$ by associating $m\in M$ with $m'\in R[M]$ given by $m'(m)=1$ and $m'(n)=0$ otherwise.

Viewing $M$ and $R$ as being in $R[M]$, I know that any element of $R[M]$ can be written as $\sum r_im_i$. So I'm curious, is $R[M]$ isomorphic to $R[x_1,\dots,x_n]$? To me it appears that the elements of the sets look more or less the same, as does the $+$ operation, but I'm not sure if they actually are isomorphic or not. Thanks for any explanation.

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Yes; the polynomial ring in $n$ variables with coefficients in $R$ is isomorphic to the semigroup ring $R[\mathbb{N}^n]$. For example, this is the way in which the polynomial ring is defined in Hungerford. –  Arturo Magidin Jul 9 '12 at 5:02
    
@ArturoMagidin Great, that was exactly what I was looking for, thanks. –  Buble Jul 9 '12 at 6:06
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up vote 2 down vote accepted

As Arturo Magidin said in comments, the monoid ring $R[\mathbb N_0^n]$ and the ring of polynomials in $n$ variables $R[x_1,\dots,x_n]$ are indeed naturally isomorphic. (I write $\mathbb N_0$ to indicate that $0$ is included, as it's not always considered an element of $\mathbb N$.) Wikipedia article Monoid Ring gives this as a primary example.

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