polynomial ring

Definition

Let $R$ be a commutative ring. The polynomial ring $R[x]$ is the set of expressions

$$r_0+r_{1}x+r_2{x}^2+\cdots +r_{m-1}{x}^{m-1}+r_m{x}^m$$

where $r_i\in R$ are the coefficients of the polynomial.

Addition and multiplication are defined using addition and multiplication of polynomials as would be expected.

Note the idea can be expanded to more variables, for example $R[x_1,x_2,\dots ,x_n]$ are polynomials with $n$ indeterminates.

Universal Property

Let $f:R\to S$ be a homomorphism of commutative rings. Fix elements $s_1,\dots ,s_n\in S$. There exists a unique ring homomorphism $F:R[x_1,\dots ,x_n]\to S$ with the following properties:

• $F(x_i)=s_i$ for $1\le i\le n$.
• $F(r)=f(r)$ for each constant polynomial $r\in R[x_1,\dots ,x_n]$ (interpreted as an element of $R$).

Proof

Existence:

Since $f\in R[x_1,\dots ,x_n]$ are $R$-linear combinations of products of $x_1$ through $x_n$, we can represent an element by the sum

$$f=\sum _{\alpha }{r}_{\alpha }{x}_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}$$

where $\alpha$ is an index on the degree of the polynomial.

Standard notation

This is the notation that I will be using for this proof, even though it is not standard. It is important to note that there may several different summands with the same degree. For example the $x_1^2{x}_2$ and $x_1{x}_2^2$ terms could have different coefficients. Note that the standard notation for a polynomial is

$$\sum _{\alpha \in Z_{\ge 0}}{r}_{\alpha }{x}^{\alpha },r_{\alpha }\in R,x^{\alpha }=\prod _{i=1}{x}_i^{{\alpha }_i}$$

This sum must be finite. Thus, consider the map

$$\begin{array}{rl}F:R[x_1,\dots ,x_n]& ⟶S\\ \sum _{\alpha }{r}_{\alpha }{x}_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}& ⟼\sum _{\alpha }f(r_{\alpha })s_1^{{\alpha }_1}\dots s_n^{{\alpha }_n}\end{array}$$

Note that $F(r)=f(r)s_1^0\dots s_n^0=f(r)$ for $r\in R$, and $F(x_i)=f(1_S)s_i=s_i$ since $f$ is a ring homomorphism.

$F$ satisfies the property that $F(a+b)=F(a)+F(b)$ trivially since it is defined for $R$-linear combinations.

Next, consider the product of two polynomials $a,b\in R[x_1,\dots ,x_n]$

$$\left(\sum _{\alpha }^n{r}_{\alpha }{x}_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}\right)\left(\sum _{\beta }^m{r}_{\beta }{x}_1^{\beta }\dots x_n^{\beta }\right)=\sum _{\alpha +\beta =i}^{n+m}(r_{{\alpha }_i}{r}_{{\beta }_i})x_1^{{\alpha }_{i_1}+{\beta }_{i_1}}\dots x_n^{{\alpha }_{i_n}+{\beta }_{i_n}}$$

where ${\alpha }_i$ takes in account for the possible many terms of the same degree again.

Then, looking at $F$ we see

$$\begin{array}{rl}F(a)F(b)& =\left(\sum _{\alpha }^{n}f(r_{\alpha })s_1^{{\alpha }_1}\dots s_n^{{\alpha }_n}\right)\left(\sum _{\beta }^{m}f(r_{\beta })s_1^{\beta }\dots s_n^{\beta }\right)\\ & =\sum _{\alpha +\beta =i}^{n+m}f(r_{{\alpha }_i}{r}_{{\beta }_i})s_1^{{\alpha }_{i_1}+{\beta }_{i_1}}\dots s_n^{{\alpha }_{i_n}+{\beta }_{i_n}}\\ & =F(ab)\end{array}$$

since $S$ is commutative, so we can collect like terms just indexing by powers of $s_i$.

Uniqueness:

Next, to prove that $F$ is the unique ring homomorphism with these properties, we can note that every element of $R[x_1,\dots ,x_n]$, each term in ${\sum }_{\alpha }{r}_{\alpha }{x}_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}$ can be expressed as $(r_{\alpha })(x_1{)}_1^{\alpha }\dots (x_n{)}^{{\alpha }_n}$, so since any function that satisfies the required properties would hold on the generators $R,x_1,\dots ,x_n$ then they will be the same for every element. Thus, the ring homomorphism is unique.