How to represent polynomial rings in multiple variables.

Question

If $R$ is a ring, we can form the polynomial ring in $x$ as $$R[x]=\{\sum_{i=1}^nax^i|a \in R \wedge n\in \mathbb{N} \cup\{0\}\} $$ where the $\sum_{i=1}^nax^i$ are formal sums. Every source I look at deals with polynomials with just one variable, and only a select few go as far as mentioning $R([x])[y] \cong R([y])[x]$ and just call it $R[x,y]$. None mention explicitly $R[x_1, ... ,x_n]$. How could one explicitly and rigorously define it in the manner above with $R[x]$? What do elements of $R[x_1, ... ,x_n]$ look like explicitly? What do their operations formally look like?

Answer 1

$R[x_1,\dots, x_n]$ is by definition the free associative and commutative $R$-algebra, $\;R^{(\mathbf N^n)}$, built on the monoid $\;\mathbf N^n$.

In one indeterminate, it's the algebra $\; R^{(\mathbf N)}$ of all eventually $0$ sequences of elements of $R$. Addition is defined componentwise, and multiplication of $(c_n)_{n\ge 0}$ and $(c'_n)_{n\ge0}$ is $(d_n)_{n\ge0}$, defined by $$d_n=\sum_{i+j=n}c_ic'_j$$

In this context, $X$ denotes the sequence $(0,1,0,\dots,0,\dots)$. One can check that $$X^2=(0,0,1,0\dots),\quad X^3=(0,0,0,1,0,\dots),\quad\text{&c.}$$

With several indeterminates, it goes along the same lines, replacing natural numbers in indices by $n$-tuples of natural numbers. For instance, $R[X,Y]$ is the set $R^{(\mathbf N^2)}$, and the indeterminates are \begin{align*}X&=\bigl(0_{(0,0)}, 1_{(1,0)}, 0_{(i,j)},\dots\bigr)\quad\text{for}\enspace(i,j)\neq(1,0),\\ Y&=\bigl(0_{(0,0)}, 1_{(0,1)}, 0_{(i,j)},\dots\bigr)\quad\text{for}\enspace(i,j)\neq(0,1). \end{align*}

One may even consider polynomials with indeterminates indexed by an arbitrary set $I$, which, as a set, is simply $$R^{(\mathbf N^{(I)})}.$$

Answer 2

You define it recursively.

$$ R[x_1,x_2, \dots, x_{n+1}] = \left\{ \sum_{j=0}^k a_j x_{n+1}^j \mid a_j \in R[x_1, x_2, \dots, x_n], k \in \Bbb{Z}_{\geq 0} \right\} $$

So a generic element is a polynomial in $x_{n+1}$ with coefficients in $R[x_1, x_2, \dots, x_n]$.

An example of an element of $R[x,y,z]$ is $45 + ((3x) + ((4x)y^{12})z + ((x^{12})y)z^3$, where the recursion is indicated by parentheses (although normally you would not write these parentheses).

The algebraic operations are unchanged from the one variable case. For addition and subtraction proceed degree-by-degree in the last variable adding/subtracting the coefficients as you go. Since this definition is recursive, the phrase "adding/subtracting coefficients" describes a recursive process. Multiplication also proceeds by degrees in the output, with the coefficients multiplied by the obvious recursive technique.

Answer 3

Bernard's answer is very good, but here are two more way of looking at a polynomial ring. It is always helpful to have multiple perspectives.

Let $K$ be a commutative ring, and $X$ some set of variables. A polynomial ring $K[X]$ on a set of variables $X$ is the free commutative $K$-algebra on $X$, i.e the most general commutative $K$-algebra we can construct with $X$. It is determined up to unique isomorphism by its universal property.

Here's the first method for making such a polynomial ring. Let's denote by $K\{X\}$ the free $K$-module on $X$, i.e. with basis given by the elements of $X$. Then form the symmetric algebra $\operatorname{Sym}(K\{X\})$. Using the universal property of the free $K$-module and the universal property of the symmetric algebra, it is easy to check that this satisfies the correct universal property for a polynomial ring, and hence it is the polynomial ring $K[X]$ (up to unique isomorphism).

Here is a different, but very enlightening construction that mimics the construction of monoid and group rings. Notice that the free $K$-module functor turns products into tensor products (i.e. is strong monoidal). That is, $K\{X \times Y \} \cong K\{X \} \otimes_K K\{Y \}$. Thus, it preserves monoid objects, commutative monoid objects, Hopf monoid objects etc. That is, if $X$ is such an algebraic object in $\mathsf{Sets}$, then so is its image in $K\mathsf{Mod}$. Let $\operatorname{Free}_\mathsf{CMon}: \mathsf{Sets} \to \mathsf{CMon}$ denote the free commutative monoid functor. Take the free commutative monoid on $X$, and compose with the free $K$-module functor to get $K\{ \operatorname{Free}_{\mathsf{CMon}} (X)\}$. This is a commutative monoid object in $K\mathsf{Mod}$, but that is precisely the definition of a commutative $K$-algebra! Again, just check that this construction satisfies the right universal property, and you're done.