Indeterminate "x" in Abstract Algebra/Ring Theory - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-20T07:49:19Z http://mathoverflow.net/feeds/question/33865 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory Indeterminate "x" in Abstract Algebra/Ring Theory yac 2010-07-29T23:09:02Z 2010-08-03T22:34:48Z <p>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</p> <p>$R[x] = \{ f: R \to R | \exists n \in \mathbb{N}, \mbox{ and } c_0, \dots, c_n \in R \mbox{ such that } f(x) = c_0 + c_1 x + \dots + c_n x^n \}$</p> <p>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 <a href="http://meta.mathoverflow.net/discussion/568/inconsistent-and-closedminded-question-closing/#Item_0" rel="nofollow">http://meta.mathoverflow.net/discussion/568/inconsistent-and-closedminded-question-closing/#Item_0</a> There is a single user who is composing a lengthy response to this question. Anyway, take a look.</p> http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory/33866#33866 Answer by Mariano Suárez-Alvarez for Indeterminate "x" in Abstract Algebra/Ring Theory Mariano Suárez-Alvarez 2010-07-29T23:11:10Z 2010-07-29T23:16:29Z <p>$R[x]$ 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 <em>nothing</em> analytic about that!</p> <p>In particular, the elements of $R[x]$ are <em>not</em> 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.</p> <p><strong>NB</strong> This should have been a comment to the question...</p> http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory/33869#33869 Answer by Pete L. Clark for Indeterminate "x" in Abstract Algebra/Ring Theory Pete L. Clark 2010-07-29T23:32:13Z 2010-07-31T13:37:01Z <p>Among infinitely many other places, this issue is discussed in Sections 4.3 and 4.4 of my notes on commutative algebra:</p> <p><a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">http://math.uga.edu/~pete/integral.pdf</a></p> <p>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 <strong>semigroup algebra</strong> 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 <code>$\mathbb{N} = \{{\bf 0},1,2,\ldots\}$</code> -- with pointwise addition and <em>convolution product</em>. 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 <a href="http://en.wikipedia.org/wiki/Mobius_inversion_formula" rel="nofollow">Möbius Inversion Formula</a>.)</p> <p>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.</p> <p>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 <em>not</em> 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.) </p> http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory/33881#33881 Answer by Andrew L for Indeterminate "x" in Abstract Algebra/Ring Theory Andrew L 2010-07-30T02:17:01Z 2010-07-30T02:23:54Z <p>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.</p> <p>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 <em>strictly bad</em>,mind you-just runs you off topic into quagmires that don't have anything to do with the original question.) </p> <p>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. </p> <p>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. </p> <p>In fact,the whole thing can be simplified by just assuming the Cauchy construction of R and then using Pietro's definition straight away. </p> <p>That's my 35 cents on it,anyway.</p> http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory/34438#34438 Answer by Anweshi for Indeterminate "x" in Abstract Algebra/Ring Theory Anweshi 2010-08-03T20:36:59Z 2010-08-03T20:40:16Z <p>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. </p> <p>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 $\phi^\prime (x) =s$.</p> <p>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".</p> <p>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.</p> <p>I should mention that fixing "x" in the polynomial algebra is in a sense fixing some "co-ordinate". The <a href="http://en.wikipedia.org/wiki/Symmetric_algebra" rel="nofollow">Symmetric Algebra</a> over a vector space is interesting in this sense.</p> http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory/34447#34447 Answer by Andreas Blass for Indeterminate "x" in Abstract Algebra/Ring Theory Andreas Blass 2010-08-03T22:34:48Z 2010-08-03T22:34:48Z <p>This began as a comment on Anweshi's answer, but it got too long. </p> <p>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. </p> <p>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 <code>$(R[x],x)$</code>. 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 <code>$(R[x],x)$</code> 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.</p>