3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. :) $\endgroup$ – Alex Wertheim Dec 28 '16 at 3:27
  • 5
    $\begingroup$ Special case of a monoid ring, viz. $R[\Bbb N^{\large k}]\ \ $ $\endgroup$ – Bill Dubuque Dec 28 '16 at 3:59
  • $\begingroup$ Thank you for you answers. Could anybody write me down the complete proof? $\endgroup$ – Chris Dec 28 '16 at 17:36
  • 1
    $\begingroup$ 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 ... $\endgroup$ – darij grinberg Dec 29 '16 at 10:47
  • $\begingroup$ ... 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. $\endgroup$ – darij grinberg Dec 29 '16 at 10:48
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Chris May 31 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.