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The formal construction of the polynomial ring in one variable is briefly the following:

We take a ring $(R,+,\cdot)$ with $1_R$. We define $R^{ \mathbb{N}}$ be the set of all the sequences $(a_0,a_1,a_2,a_3,...)$, $a_i \in R^{ \mathbb{N}},\forall i\in \mathbb{N}$ and we define the following operations: $$+: R^{ \mathbb{N}} \times R^{ \mathbb{N}} \longrightarrow R^{ \mathbb{N}},\ ((a_0,a_1,a_2,...),(b_0,b_1,b_2,...))\mapsto (a_0,a_1,a_2,...)+(b_0,b_1,b_2,...):= (a_0+b_0,a_1+b_1,a_2+b_2,...)$$ and $$\cdot: R^{ \mathbb{N}} \times R^{ \mathbb{N}} \longrightarrow R^{ \mathbb{N}},\ ((a_0,a_1,a_2,...),(b_0,b_1,b_2,...))\mapsto (a_0,a_1,a_2,...) \cdot (b_0,b_1,b_2,...):=(c_0,c_1,c_2,...)$$ with $c_n=a_0b_n+a_1b_{n-1}+...+a_{n-1}b_1+a_nb_0,\forall n\in \mathbb{N}=\{0,1,...\}$.
Furthermore we define the equality $(a_0,a_1,a_2,...)=(b_0,b_1,b_2,...) \iff a_i=b_i, \forall i\in \mathbb{N}$. With these two binary operations we have that $(R^{ \mathbb{N}},+\cdot )$ is a ring with $1_{R^{ \mathbb{N}}}=(1_R,0_R,0_R,...)$. Now, polynomial is every element of the last ring of the form $(a_0,a_1,a_2,...,a_n,0_R,0_R,...)$. If $R[X]$ is the set of all the polynomials then $R[X]$ is a subring of $R^{ \mathbb{N}}$ and the mapping $f:R\longrightarrow R[X]$, $a\mapsto (a,0_R,0_R,...) $ is a monomorphism. So, we can say that $R$ is a subring of $R[X]$, and after this we have all the usual theorems.

My question is:

How we can do exactly the same construction in the polynomial ring in several variables with the same procedure?

PS 1: I apologize for my English. If you don't understand something ask me please.

PS 2: I know that a similar question already exists, but I think I have a different procedure.

Thank you in advance.

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    $\begingroup$ Typically, when considering the formal construction, I think one defines polynomial rings in several variables via induction, i.e. $R[X_{1}, \ldots, X_{n}] := (R[X_{1}, \ldots, X_{n-1}])[X_{n}]$. I suppose you could define them using sequences as above. For two variables, e.g., you could have $a_{0}$ be the degree $0$ monomial terms, $a_{1}, a_{2}$ be the degree $1$ monomial terms, $a_{3}, a_{4}, a_{5}$ be the degree $2$ monomial terms, etc. I think it should be possible, but very annoying, to write down the associated convolutions (multiplications) then; I wouldn't bother with this though. :) $\endgroup$ Commented Dec 28, 2016 at 3:27
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    $\begingroup$ Special case of a monoid ring, viz. $R[\Bbb N^{\large k}]\ \ $ $\endgroup$ Commented Dec 28, 2016 at 3:59
  • $\begingroup$ Thank you for you answers. Could anybody write me down the complete proof? $\endgroup$
    – Chris
    Commented Dec 28, 2016 at 17:36
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    $\begingroup$ Sources that do it "right": 1) Serge Lang, Algebra, Revised Third Edition, 2002 ( link.springer.com/book/10.1007/978-1-4613-0041-0 ), Chapter II, §3. 2) Thomas W. Hungerford, Algebra, 12th printing, 2003 ( link.springer.com/book/10.1007/978-1-4612-6101-8 ), Chapter III, §5. 3) Herbert Amann, Joachim Escher, Analysis I, 2005 ( springer.com/us/book/9783764377557 ), §8. The idea is either to use monoid rings as @BillDubuque suggested, or to follow the univariate construction but replace ... $\endgroup$ Commented Dec 29, 2016 at 10:47
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    $\begingroup$ ... sequences by "multisequences", i.e., maps from $\mathbb{N}^n$ to $R$. The recursive approach mentioned by @AlexWertheim has several drawbacks -- it fails generalizing to infinitely many variables or even to finitely many variables without a canonical ordering. $\endgroup$ Commented Dec 29, 2016 at 10:48

3 Answers 3

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First, there is a little mistake in your definition : a polynomial has only finitely many non-zero coefficients. Instead of $R^\mathbb{N}$, you should hence consider $S :=\{(a_0,\dots) \in R^\mathbb{N} \big| \exists d \in \mathbb{N}, \, \forall e>d,\, a_e=0\}$

To do the same construction for n variables, you just need to change the index set :

define $S_n :=\{(a_t) \in R^{\mathbb{N}^n} \big| \exists d \in \mathbb{N}^n, \, \forall t,\,sum(t)>sum(d) \Rightarrow a_t=0\}$, where $sum((t_1,\dots,t_n))$ denotes $t_1 + \dots + t_n$.

the idea : for $t = (t_1, \dots, t_n) \in \mathbb{N}^n$, $a_t$ is the coefficient of the monomial $X_1^{t_1}\dots X_n^{t_n}$.

Addition is defined componentwise : $(a_t) + (b_t) = (a_t + b_t)$

Multiplication is somewhat trickier : $(a_t) \cdot (b_t) := (c_t)$, where for each $t \in \mathbb{N}^n$, $$c_t := \sum_{u, v\in \mathbb{N}^n, \, u+v=t} a_u \cdot b_{v}$$

where $u + v = (u_1 + v_1, \dots, u_n + v_n)$.

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  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Chris
    Commented May 31, 2019 at 21:22
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Perhaps there is a way to do this as the "free-est" possible associative, unital $R$-algebra with unit (thus the algebra structures are defined by just morphisms of rings with unity from $R$ to the algebras) such that the set of variables commutes with the "copy" of $R$ and each other. That would define it even for an infinite set of variables $T$.

We could start with formal products in the set of variables, such that all but a finite number of exponents are $0$, where if all the exponents are $0$ we'll identify that with $1_{R[T]}$ then define the product of such unit coefficient monomials, then define formal multiplication with elements of $R$ to obtain the set of monomials and constants from $R$; then formal sums in the previous set such that all but a finite number of terms has $0_R$ as coefficient, where multiplication is extended linearly. Etc...

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($R$ has to be commutative here.)

Frankly, to my mind this is a tedious and unenlightening way to construct polynomial rings, and in particular it makes the multiplication look a lot more mysterious than it is. Its only virtue is that it is relatively concrete and doesn't require you to introduce more concepts compared to other constructions, but all the concepts you use in other constructions are extremely useful and worth introducing.

Here is the construction I prefer. The key concept we need is the concept of the free $R$-module $R[X]$ over a set $X$ (not to be confused with the polynomial ring in one variable). An element of $R[X]$ is a formal $R$-linear combination $\sum_{x \in X} r_x x$ of finitely many elements of $X$. The action of $R$ is given by multiplying the coefficients by $r \in R$, and addition is done by adding coefficients. In particular the notation is consistent in the sense that $\sum_x r_x x$ is obtained by adding $r_x$ mulplied by $x$ for all $x$.

You can think of a formal linear combination as a function $X \to R$ which vanishes except at finitely many points if you want to pin down precisely what is meant, but personally I think this should be avoided. It really is much better to think in terms of formal linear combinations.

The polynomial ring on $k$ variables over $R$ can then be constructed, as an $R$-module, as the free $R$-module $R[\mathbb{N}^k]$ over $\mathbb{N}^k$, where here $\mathbb{N}$ includes $0$. (This $\mathbb{N}^k$ is the free commutative monoid on $k$ generators but strictly speaking we don't need to introduce this concept.) We write the element $(n_1, \dots n_k) \in \mathbb{N}^k$ as the monomial $x_1^{n_1} \dots x_k^{n_k}$. Multiplication is the obvious thing implied by this notation, namely

$$\left( \sum r_{n_1, \dots n_k} x_1^{n_1} \dots x_k^{n_k} \right) \left( \sum s_{m_1, \dots m_k} x_1^{m_1} \dots x_k^{m_k} \right) = \sum r_{n_1, \dots n_k} s_{m_1, \dots m_k} x_1^{n_1 + m_1} \dots x_k^{n_k + m_k}.$$

It is uniquely determined by the condition that it is $R$-bilinear and that exponents add, there is no need to talk about convolution, and associativity follows from the associativity of addition. You can check that $x_1^{n_1} \dots x_k^{n_k}$, with respect to this multiplication, is in fact the product of $n_1$ copies of $x_1$, then $n_2$ copies of $x_2$, etc. which means the notation is consistent.

The only thing you might worry about here is that our use of sum and monomial notation is circular somehow, but it's not, although it would perhaps be logically more scrupulous to use slightly different notation for the "formal" and the "actual" sum resp. product of monomials since at the beginning we haven't yet checked that they're the same.

This construction generalizes smoothly to the case of infinitely many variables, say indexed by a set $I$, where we replace $\mathbb{N}^k$ by the free commutative monoid $\bigoplus_{i \in I} \mathbb{N}$ on $I$ (this is smaller than $\mathbb{N}^I$ in general). It also generalizes smoothly to noncommutative polynomials, where we replace $\mathbb{N}^k$ by the free monoid on the set of variables $I$.

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  • $\begingroup$ Note that this standard monoid ring construction was already mentioned in the comments 10 years ago. $\endgroup$ Commented Jun 7 at 19:37

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