# What is the formal definition of polynomial ring of several variables?

Let's consider a polynomial ring of single variable.

One can define them informally by saying $$P(X)=\sum_{i=1}^n a_n X^n$$ while $$X$$ is an indeterminate variable.

However, since mathematics is based on first-order logic, one can only talk about something that actually exists (as a set). So as its name says, indeterminate variable is not a sentence in $$ZFC$$.

Nevertheless, we can formally define a polynomial ring $$R[X]$$ as a subset of $$R^\omega$$ whose support is finite. ($$R$$ is a commutative ring with unity)

Just like a single variable, I want to know what would be the formal definition of a polynomial ring of several variables. What would be a formal definition?

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I have figured out a candidate. That is,

Candidate for a "Definition of a polynomial ring $$R[X_1,...,X_n]$$". ($$R$$ is a commutative ring with unity)

Let $$R[X_1,...,X_n]$$ be the set of all functions $$f:\omega^n \rightarrow R$$ whose support is finite.

Define $$(f+g)(a,b,c)=f(a,b,c)+g(a,b,c)$$ and $$(f•g)(a'',b'',c'')=\sum_{a+a'=a''} f(a,b,c)•g(a',b',c')$$ (Note that in this way, $$+,•$$ are well defined.)

Then, it can be checked that $$(R[X_1,...,X_n],+,•)$$ is a ring.

Is this a formal definition of a polynomial ring?

• A possible formal definition is: $R[X_1,\dots,X_n]$ is the free object on $\{X_1,\dots,X_n\}$ in the category of commutative $R$-algebras. Commented Sep 5, 2014 at 7:43
• @egreg Isn't there a definition in ZFC, not category? Commented Sep 5, 2014 at 7:55
• Mathematics is not based on first-order logic. People were doing mathematics for thousands of years before anyone invented first-order logic, or sets for that matter. Commented Sep 5, 2014 at 8:04
• Commutative $R$-algebras form a variety in the class of $R$-algebras, so they have free objects. Just translate the property to ZFC. Commented Sep 5, 2014 at 8:05
• @Qioachu You are absolutely right. I meant my mathematics. I want my mathematics to be settled in one logical system.. Commented Sep 5, 2014 at 8:14

Your attempt is in the right direction, but it's unclear what you mean by $f(a,b,c)$.
Define an operation $+$ on $\omega^n$ in the obvious way, that is, componentwise. A function $f\colon \omega^n\to R$ is a polynomial in $n$ indeterminates if (and only if) $f(t)=0$ for all but a finite number of elements $t\in\omega^n$ (has finite support, in other words). For two polynomials $f$ and $g$ define $$f+g\colon t\mapsto f(t)+g(t)$$ and $$fg\colon t\mapsto \sum_{u+v=t}f(u)g(v)$$ After verifying that these operations are well defined in the set of polynomials, it's just a tedious verification showing that this set is a ring. The embedding of $R$ in this ring is the map sending $r$ into the function $\bar{r}$ that is $r$ on $(0,0,\dots,0)\in\omega^n$ and zero elsewhere: $$\bar{r}(t)=\begin{cases} r&\text{if t=(0,\dots,0)}\\ 0&\text{otherwise} \end{cases}$$ Note that the set of polynomials is, under addition, the free $R$-module on $\omega^n$.
• Thank you very much! I really appreciate :) And $(a,b,c)$ is a typo, i meant $(a_1,...,a_n)$. Commented Sep 5, 2014 at 8:32