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We got this definition of a polynomial ring during our last lecture in Commutative Algebra:

$\forall_{n\in \mathbb{N}_+}\:\forall _{R\:-\:ring}\:\:R\left[X_1,\:...,\:X_n\right]=\left\{f:\:\mathbb{N}^n\:\rightarrow \:R\::\:\left|supp\:f\right|\:<\:\infty \right\}$ where $supp\:f\:=\:\left\{\alpha \in \mathbb{N}^n\::\:f\left(\alpha \right)\:\ne \:0\right\}$

This doesn't really make (intuitive) sense to me, wouldn't it be better to say that it's a set of such functions with finitely many zero values (and not non-zero values)?

E.g. we know that $x^2$ has only one zero value, but infinitely many nonzero values (even in $\mathbb{N}$, given it contains 0 as we do in our lecture)

So I'd really like to understand why do we define polynomial rings in such a way, maybe on an example it would be better to understand why?

Also, why do we only care about $\mathbb{N}$? Why don't we define it e.g. $f:\:\mathbb{Z}^n\:\rightarrow \:R$ or something different

user
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    You misread it. It says that that coef map $f$ has only finitely many nonzero values, i.e. only finitely many monomials occur in the associated polynomial. – Bill Dubuque Mar 21 '21 at 13:34
  • No, I understood it properly, maybe I should edit that I mean nonzero value. It still does not make any sense to me, according to that definition, $x^2$ wouldn't be in any polynomial ring as it is the complete opposite, it has finitely many zero values, but infinitely many nonzero values – user Mar 21 '21 at 13:36
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    @user No, you didn't understand it properly, Bill is correct. – Rushabh Mehta Mar 21 '21 at 13:38
  • Ok, then I guess I just don't get it. We have the condition that #$supp:f:<\infty$, so #$\left{\alpha \in \mathbb{N}^n:::f\left(\alpha \right)\ne 0\right}<\infty$. This means that there are just a finite number of $\alpha$ such that $:f\left(\alpha \right)\ne 0$. If we take $x^2$, a polynomial, it does not fulfill this condition. – user Mar 21 '21 at 13:41
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    e.g. the polynomial $, a + b, x y^2 + c, x^3 y^4, $ is the map $,f,$ sending $,(0,0)\to a,\ \ (1,2)\to b,\ (3,4)\to c,\ $ and all else to $,0\ $ – Bill Dubuque Mar 21 '21 at 13:42
  • Welcome to Math.SE. Is there a reference you can cite? I think I have an idea what's behind the confusion, but it would be best to have the definition quoted or cited in the body of your Question. – hardmath Mar 21 '21 at 13:42
  • Ok I'll give the definition we got from our presentation during our last lecture – user Mar 21 '21 at 13:43
  • That def matches what I wrote. See the example I gave above. The domain $\Bbb N^n,$ is the expt vectors of the monomials in $f$. The map $f$ gives you the coef of each monomial. To say it has finite support means that that only finitely many monomials having nonzero coefs.occur in $f\ $ – Bill Dubuque Mar 21 '21 at 13:47
  • Yes, you are right, but wouldn't this mean that e.g. $x^2$ wouldn't be an element of a polynomial ring here? Also, is there a special reason why do we care about # supp f < infinity? – user Mar 21 '21 at 13:48
  • Actually, your example doesn't make sense to me. If I take e.g. (1,1), I get a+b+c and that's not 0 – user Mar 21 '21 at 13:53
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    This is a formal (set-theoretic) rigorous construction of polynomial rings that does not use informal notions like "variables". For infinitely many terms you can use power series or semigroup rings. – Bill Dubuque Mar 21 '21 at 13:54
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    The expt tuple $(j,k)$ corresponds to the monomial $x^j y^k$ or $,x_1^j x_2^k$ and the value of $f$ on the tuple tells you the coef of $f$ on the associated monomial. – Bill Dubuque Mar 21 '21 at 13:57
  • Ok now I get it, it's that f is a coefficient map, so that's why your example makes sense. But where is it stated in our given definition that it is a coefficient map? – user Mar 21 '21 at 14:14
  • Note that Note also that this is a definition / construction of a polynomial ring, not a polynomial (by itself). In this view polynomials exist only as elements of these types of rings, just as vectors exist only as elements of vectors spaces. Unlike in more informal contexts, there is no meaning assigned to "polynomial" (or "vector") in absence of such containing structures. – Bill Dubuque Mar 21 '21 at 14:28
  • Rather, we are concerned only with their "holistic" algebraic properties of the ements ("polynomials"), i.e. how they relate to each other under the operations of the structure (not their internal representation - which are artefacts of the particular construction). – Bill Dubuque Mar 21 '21 at 14:28
  • To understand the need for such rigorous definitions see this post which shows clearly the difficulties of informally describing polynomial rings and their quotients by showing Cauchy's attempt at doing such for thring theoretic construction $,\Bbb C = \Bbb R[x]/(x^2+1),,$ and the (well-deserved) scathing critique by Hankel. – Bill Dubuque Mar 21 '21 at 14:32
  • To motivate the coef map note that in the univariate case the map sending $\Bbb N$ to $R$ can be viewed as the coef sequence of the polynomial, then $ f = (c_0,c_1,c_2,\ldots)\leftrightarrow \sum c_i x^i.,$ Then additions is sequence ("vector") addition and the product is the Cauchy product. This should be explained in any good textbook. – Bill Dubuque Mar 21 '21 at 14:49
  • See also here on making rigorous informal "variable" or "indeterminate" notation, and see here on axiomatic approaches to polynomial rings. – Bill Dubuque Mar 21 '21 at 14:54
  • Thank you very much! – user Mar 21 '21 at 15:14

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