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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?

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  • $\begingroup$ 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. $\endgroup$
    – egreg
    Commented Sep 5, 2014 at 7:43
  • $\begingroup$ @egreg Isn't there a definition in ZFC, not category? $\endgroup$
    – Rubertos
    Commented Sep 5, 2014 at 7:55
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    $\begingroup$ 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. $\endgroup$ Commented Sep 5, 2014 at 8:04
  • $\begingroup$ Commutative $R$-algebras form a variety in the class of $R$-algebras, so they have free objects. Just translate the property to ZFC. $\endgroup$
    – egreg
    Commented Sep 5, 2014 at 8:05
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    $\begingroup$ @Qioachu You are absolutely right. I meant my mathematics. I want my mathematics to be settled in one logical system.. $\endgroup$
    – Rubertos
    Commented Sep 5, 2014 at 8:14

1 Answer 1

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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 $\omega^n$.

In any case, thinking to polynomials from a higher point of view (that is, free objects in a suitable category) is more rewarding.

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  • $\begingroup$ Thank you very much! I really appreciate :) And $(a,b,c)$ is a typo, i meant $(a_1,...,a_n)$. $\endgroup$
    – Rubertos
    Commented Sep 5, 2014 at 8:32

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