Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

How do you interpret the indeterminate "x" in ring theory from the set theory viewpoint? How do you write down R[x] as a set? Is it appropriate/correct to just say that

$R[x] = \{ f: R \to R | \exists n \in \mathbb{N}, \mbox{ and } c_0, \dots, c_n \in R \mbox{ such that } f(x) = c_0 + c_1 x + \dots + c_n x^n \}$

This appears to be a very analytic definition. Is there a better definition that highlights the algebraic aspect of the set of polynomials? $$ $$ EDIT: There is a thread on this at http://tea.mathoverflow.net/discussion/568/inconsistent-and-closedminded-question-closing/#Item_0 There is a single user who is composing a lengthy response to this question. Anyway, take a look.

share|improve this question
14  
See the first discussion about polynomials in Lang's Undergraduate Algebra and and in his graduate book called just Algebra. He explains two ways of how to give a precise meaning to "x". You can't define R[x] as polynomial functions R --> R in general because of the problem that, for example, when R = Z/p the polynomial function x^p - x is 0 everywhere but we do not want to consider x^p - x to be the polynomial 0. Abstract polynomials can be interpreted as polynomial functions, but there can be some loss of information, so from a general point of view we need another device to define R[x]. –  KConrad Jul 29 '10 at 23:23
3  
One definition is using sequences that are eventually 0. In your notation, f = (c_0,c_1,...,c_n,0,0,0,...). Then x = (0,1,0,0,0,...). Admittedly nobody actually thinks like that, just as nobody actually thinks of real numbers as Cauchy sequences of rationals modulo an equivalence relation, but if you need a definition of R[x] then the sequence approach is one starting point. –  KConrad Jul 29 '10 at 23:27
5  
Why was this question closed? It doesn't see that the OP is a troll. IT seems the question was a genuine confusion and satisfactory answers are given below. It was very mean to close it without at least directing the OP to math.stackexchange.com or sites of similar level. –  Anweshi Jul 31 '10 at 13:44
3  
I have voted to close because of inappropriate level: this is a basic undergraduate definitions issue. In addition to any abstract algebra textbook, this is explained in the Wikipedia article on polynomial rings, which the author should have consulted first. I disagree that MO should be used to answer popular questions "from other math newsgroups": according to the FAQ, "MathOverflow's primary goal is for users to ask and answer research level math questions, the sorts of questions you come across when you're writing or reading articles or graduate level books". –  Victor Protsak Jul 31 '10 at 17:35
3  
@Victor: What the OP asked is not a trivial matter. So thinks Andre Weil too. In his autobiography he writes that "what is 'x'?" is a profound question. –  Anweshi Jul 31 '10 at 19:43
show 18 more comments

closed as not a real question by Robin Chapman, Andrew Stacey, Wadim Zudilin, Harry Gindi, Pete L. Clark Aug 3 '10 at 22:57

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

5 Answers

Among infinitely many other places, this issue is discussed in Sections 4.3 and 4.4 of my notes on commutative algebra:

http://math.uga.edu/~pete/integral.pdf

In Section 4.3 I give Mariano's definition, with some commentary. A slight drawback to this definition is that it makes the associativity of the product look mysterious. In Section 4.4, I mention that this may be viewed as a special case of the semigroup algebra construction, namely we may define $R[x]$ to be the set of all finitely nonzero functions $t: \mathbb{N} \rightarrow R$ -- this is algebra, so $\mathbb{N} = \{{\bf 0},1,2,\ldots\}$ -- with pointwise addition and convolution product. The associativity still has to be checked, but it is relatively satisfying to do this once and for all in this level of generality. (And this will probably come in handy elsewhere, e.g. the associativity of the convolution product is precisely the content of the Möbius Inversion Formula.)

On the other hand -- when $R$ is commutative, as I assume from now on -- of course it does make sense to plug in a polynomial at any element of $R$: in other words, a polynomial determines a function from $R$ to $R$. Indeed, the evaluation map gives a homomorphism of rings from $R[t]$ to the ring of all functions from $R$ to $R$ under pointwise addition and pointwise multiplication. As I mention in my notes, when $R$ is an infinite integral domain, this evaluation map is injective and one can use this to deduce the associativity of the multiplication in $R[t]$ for free.

However, when $R$ is finite it is important to distinguish between polynomials in the formal sense and polynomial functions. In particular, your definition of a polynomial is not correct when $R$ is e.g. the finite field $\mathbb{Z}/p\mathbb{Z}$, because it does not distinguish between the polynomial $x^p - x$ and the zero polynomial: both induce the zero function. (That multiple polynomials may determine the same function has some positive aspects as well; it can be used to give a proof of the Chevalley-Warning theorem.)

share|improve this answer
3  
+1 I like this approach. I first saw it in Lang's Algebra when I was looking for a rigorous treatment of group rings. I had never realized that group rings and polynomial rings were both special cases of this general semigroup algebra construction. –  Keenan Kidwell Jul 30 '10 at 0:25
    
I like the discussion in the linked notes - perhaps you could add some of that discussion here? Your answer to the OP's question then is: 'x' has not much set-theoretical status - it just needs to be anything not in $R$. –  Jacques Carette Aug 1 '10 at 13:00
add comment

$R[x]$ is simply the free left $R$-module on the set of symbols $\{x^k:k\geq0\}$, on which a certain multiplication operation is defined. There is nothing analytic about that!

In particular, the elements of $R[x]$ are not functions. When $R$ is commutative, there is a somewhat canonical ring homomorphism $\phi:R[x]\to \mathcal F$, where $\mathcal F$ is the ring of all functions $R\to R$ with pointwise operations. But in general $\phi$ is not injective.

NB This should have been a comment to the question...

share|improve this answer
    
What does "somewhat canonical" mean here, I wonder? The map seems pretty canonical to me: e.g. it is functorial in $R$. –  Pete L. Clark Jul 29 '10 at 23:45
    
Well, the map $\phi':p\in R[x]\mapsto(a\in R\mapsto p(a+1)\in R)\in\mathcal F$ is also a (slightly unnatural) natural morphism... so there are a few of them: I'd not say any of them is canonical then :) –  Mariano Suárez-Alvarez Jul 29 '10 at 23:52
    
(By the way, $\mathcal F$ is not really a functor of $R$) –  Mariano Suárez-Alvarez Jul 29 '10 at 23:57
    
Yes, you're right. –  Pete L. Clark Jul 30 '10 at 0:03
add comment

I feel duty-bound to add an answer since I had desired that this question be re-opened. There is nothing much I am able to contribute in addition to the well-known things.

The functorial definition of the polynomial algebra $R[x]$ over a ring $R$ is the following. It is a ring $R[x]$ together with a homomorphism $R \to R[x]$ such that given any ring homomorphism $\phi \colon R \to S$ and any fixed element $a \in S$, there is a unique homomorphism $\phi^\prime \colon R[x] \to S$ such that $\phi^\prime (x) =s$.

This is a proffered way of saying that the indeterminate "x" should be free to vary without any restriction whatsoever except that it is an element of a ring. A concrete manifestation is given by the free $R$-module on the set of symbols $x^k$ where $k \geq 0$ together with a certain multiplication operation. Since we do not a priori know what is "x", we make this precise with a machinery of sequences; and when we are done, we call a particular element of the resulting ring to be "x".

I contend that the ring $\mathbb Z$ and the polynomial rings over it are very important objects in the category of commutative rings with identity. For one, any commutative ring with identity $R$ admits a unique homomorphism $\mathbb Z \rightarrow R$. Moreover, if we let $r$ run through the elements of $R$, then there is an obvious surjective map from $\mathbb Z[(X_r)_{r\in R}]$ to $R$ and this realization of every comm. unital ring as a quotient of some polynomial ring over $\mathbb Z$ can be used to construct the co-product in this category.

I should mention that fixing "x" in the polynomial algebra is in a sense fixing some "co-ordinate". The Symmetric Algebra over a vector space is interesting in this sense.

share|improve this answer
add comment

This began as a comment on Anweshi's answer, but it got too long.

I consider the functorial definition (as in Anweshi's answer but slightly modified --- see below) to be the best one, because it tells us what is really important about $R[x]$. Other definitions (using sequences, modules with products, symbols, etc.) are better viewed as constructions rather than definitions; they provide proofs that the functorial definition can be satisfied. The situation is analogous to the case of the real numbers, whose important property is (and whose definition therefore should be) that they form a complete Archimedean ordered field; Cauchy sequences, Dedekind cuts, etc. provide constructions.

The slight modification that I'd make in Anweshi's formulation of the functorial definition is that I regard this definition not as simply defining $R[x]$ but as defining the pair $(R[x],x)$. The most obvious reason for the change is that $x$ is explicitly used in the statement of the universal property. In fact, the definition simply says that the pair $(R[x],x)$ is the universal example of a ring with a homomorphism from R and a specified element. (I've been assuming that we're dealing with commutative rings; otherwise, I should either say "a specified central element" or work with fancier "polynomials" where the coefficients and $x$ can be interleaved in monomials, rather than just having a coefficient times a power of $x$.) A side benefit of this form of the definition is that it gives an immediate answer to the question "what is $x$": it is the specified element in the universal example.

share|improve this answer
    
Even more precision (or pedantry): Instead of just the pair $(R[x],x)$, include also the obvious homomorphism from $R$ to $R[x]$, since it is also used in the universal property. (That this homomorphism is one-to-one should not be part of the definition; it follows easily from the universal property.) –  Andreas Blass Aug 4 '10 at 15:33
    
Yes, your formulation is the correct one. Mine was too vague and without making precise in this way it does not make sense. –  Anweshi Aug 4 '10 at 18:02
add comment

The sequence definition above given in the comment by KConrad is how I first learned to rigorously define R[x] from both Nick Metas and Kenneth Kramer as an undergraduate. Niether worried much about the associativity of product multiplication since it wasn't stated in terms of a free left R-module.

Mario's subsequent definition gives an algebraic structure to R[x] which gives x a very explicit construction-it was also used implicitly by John Terilla in a subsequent ring and module theory course. To me,in mathematics,explicit is good. It seems to be that talking about "the set of symbols" is always something that makes me leery in mathematics because then you get into sticky metaphysics about what you mean by a symbol.(Not that that's strictly bad,mind you-just runs you off topic into quagmires that don't have anything to do with the original question.)

It seems to me Mariano's definition can be made more explicit by replacing "set of symbols" with KConrad's sequences.Essentially, x then becomes a function space of sequences whose range is {0,1} with R[x] then being taken as a free R-module of elements of x.Note this preserves the idea that the elements of R[x} are not necessarily themselves functions although they are built from a special class of functions i.e. Cauchy sequences of 0's and 1's.

Pietro's definition in the comments above is essentially a special case of this where copies of R are defined in "binary construction" by Cauchy sequences. It has the advantage of being able to give a simple construction of a graded algebra,which of course is quite useful in studying the overall algebra and as a dividend,the polynomial notation drops out trivially.

In fact,the whole thing can be simplified by just assuming the Cauchy construction of R and then using Pietro's definition straight away.

That's my 35 cents on it,anyway.

share|improve this answer
2  
"Set of symbols $\{x^n:n\geq0\}$" means simply "some set in bijection with $\mathbb N_0$". How can you get from that even a bit of sticky metaphysics? –  Mariano Suárez-Alvarez Jul 30 '10 at 2:30
1  
It's not clear in the original question if $R$ denotes an arbitrary ring, or the field of real numbers. I suspect the former is what was meant but the question does seem potentially ambiguous. –  Yemon Choi Jul 30 '10 at 2:48
4  
Given that the question has the phrase "in ring theory", I think we can assume that $R$ is supposed to be an arbitrary ring. In any case, this is not a situation in which restricting oneself to something like $\mathbb{R}$ makes things any simpler or clearer. –  Andy Putman Jul 30 '10 at 2:52
3  
By the way, did you know that Ernest Gowers (who prepared the second edition of Fowler) was the great-grandfather of the Fields' medalist? –  Andy Putman Jul 30 '10 at 4:50
3  
E. Gowers' "The Complete Plain Words" en.wikipedia.org/wiki/The_Complete_Plain_Words is in my view a little gem. (I think I first heard of it in an essay-of-sorts by a mathematician, who mentioned the family connection to W.T. Gowers Fields '98.) –  Yemon Choi Jul 30 '10 at 6:46
show 5 more comments

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