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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. :) $\endgroup$ Dec 28, 2016 at 3:27
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    $\begingroup$ Special case of a monoid ring, viz. $R[\Bbb N^{\large k}]\ \ $ $\endgroup$ Dec 28, 2016 at 3:59
  • $\begingroup$ Thank you for you answers. Could anybody write me down the complete proof? $\endgroup$
    – Chris
    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 ... $\endgroup$ 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. $\endgroup$ Dec 29, 2016 at 10:48

2 Answers 2

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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
    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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