Timeline for What is the theory of polynomials?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2012 at 14:55 | vote | accept | Jacques Carette | ||
| Aug 27, 2011 at 16:14 | answer | added | beroal | timeline score: 3 | |
| Aug 27, 2011 at 12:26 | comment | added | Jacques Carette | Todd and Emil both have given me excellent answers. I would be hard-pressed to choose one (i.e. 'accept') over the other. Is there a common protocol in this situation? | |
| Aug 26, 2011 at 15:18 | answer | added | Emil Jeřábek | timeline score: 7 | |
| Aug 26, 2011 at 12:54 | comment | added | Jacques Carette | @Emil: thank you for taking the time to expand on your answer, I now understand it much better. You should make this into an actual answer! I will probably implement both solutions, under different names. | |
| Aug 26, 2011 at 10:53 | comment | added | Emil Jeřábek | ... and all axioms of the form $a+b=c$ or $a\cdot b=c$ that are true in $R$, where $a,b,c\in R$. If you also want composition, you include it as a binary operation $\circ$ in the signature, and you add the axioms $(u+v)\circ w=(u\circ w)+(v\circ w)$, $(u\cdot v)\circ w=(u\circ w)\cdot(v\circ w)$, $a\circ u=a$, $x\cdot u=u$, where $a\in R$, and $u,v,w$ are universally quantified variables. (You can also add $u\circ(v\circ w)=(u\circ v)\circ w$ and $u\circ x=u$ for good measure, but this is not necessary to get the property of $R[x]$ being the initial algebra in the variety.) | |
| Aug 26, 2011 at 10:49 | comment | added | Emil Jeřábek | Actually, I suggested elements of $R$ to represent constants, not unary functions (I thought it was obvious from the definition of atomic diagram, but apparently this concept, coming from model theory, is unknown to participants in this discussion). That is, the signature of $V$ consists of nullary functions $\{x\}\cup R$ and binary functions $\{+,\cdot\}$, and its axioms (which I will state now in the not-necessarily-commutative version) consist of the axioms of rings (with (−1)⋅x taking place of −x), axioms $a\cdot x=x\cdot a$ for $a\in R$, ... | |
| Aug 25, 2011 at 20:44 | comment | added | Andrew Stacey | Jacques: Emil is suggesting that it be defined by the zeroary operations $\lbrace 0,1,x\rbrace$, unary operations $R$ (acting by scalar multiplication), and binary operations $\lbrace +,\times\rbrace$, together with the obvious identities. In that description, it is the initial algebra in a variety of algebras in the sense of universal algebra. The models of this theory are the unital $R$-algebras with a distinguished element (the image of $x$). So $R[x]$ is the initial "$R$-algebra with distinguished element". To go further, one needs to use co-operations. | |
| Aug 25, 2011 at 20:04 | history | edited | Jacques Carette | CC BY-SA 3.0 |
clarification
|
| Aug 25, 2011 at 20:01 | comment | added | Jacques Carette | @Emil: My whole problem comes in at the level of giving a sufficiently explicit 'signature' to the 'theory of polynomials' so that it can be accepted by a mechanized mathematics system. I can do that for a 'plethory' from Todd's information, but not from your comemnts -- even though, as a mathematician, I understand that your answer is correct, just insufficiently precise for me to translate into a "theory presentation". My problem might be that I don't know how to unpack 'free V-algebra over one generator' into a syntactic definition. I should add that aspect to my question, to clarify. | |
| Aug 25, 2011 at 19:55 | comment | added | Jacques Carette | @Andrew: Definitely addition and multiplication. I can forgo composition, that is 'extra' (but will want it later). I had not thought about the coproducts, probably because they are rarely explicitly pointed out that way in the classical literature on computations with polynomials. See my next comment to Emil too. | |
| Aug 25, 2011 at 19:25 | comment | added | Gerhard Paseman | Jacques, indeed the question proper is quite answerable. My comment was inspired by the title and the universal-algebra tag. Using just those, I can think of several possible answers. None of them apply to the question proper unfortunately. So the reader I mention is actually someone who shares a similar reaction, one who expected something else of the question proper from the title and the tag. It looks like your specific question is getting a good amount of attention. Gerhard "Ask Me About System Design" Paseman, 2011.08.25 | |
| Aug 25, 2011 at 18:57 | comment | added | Andrew Stacey | Exactly what answer you want depends on what structure polynomials have for you. I would guess that you allow yourself to multiply and add them. Do you let yourself compose them as well? How about a coproduct? How about two coproducts? ($x \mapsto x \otimes x$ and $x \mapsto 1 \otimes x + x \otimes 1$) Depending on how much you allow, the "answer" will range between what Emil says in the comments above and what Todd says in his answer below. | |
| Aug 25, 2011 at 16:54 | comment | added | Emil Jeřábek | @Jacques: The signature of the free algebra has nothing to do with its set of generators. Having said that, you can formally add the generator to the signature as a new constant, and in the expanded signature, the algebra will be free over an empty set of generators, as I wrote in my second comment. Is it unclear? | |
| Aug 25, 2011 at 16:42 | comment | added | Jacques Carette | @Gerhard: I am aware that the literature is vast. But the question, I believe, is quite narrow and should be answerable. | |
| Aug 25, 2011 at 16:41 | comment | added | Jacques Carette | @Michael: yes. @Emil: What is 'a generator'? I know what they are mathematically! I mean, what is the signature which corresponds to 'over one generator'? | |
| Aug 25, 2011 at 16:10 | comment | added | Gerhard Paseman | I am tempted to respond with clone theory, tame congruence theory, the theory of using terms to approximate arbitrary maps, spectra, and other research found within general algebra. However, that addresses only the title, not the question. Let me thus assure the reader that there is rich literature outside of category theory and within general algebra that deals with polynomials, even of one variable. Gerhard "Ask Me About General Algebra" Paseman, 2011.08.25 | |
| Aug 25, 2011 at 15:08 | answer | added | Todd Trimble | timeline score: 11 | |
| Aug 25, 2011 at 14:58 | comment | added | Michael Bächtold | @Jacques: by polynomials you mean any ring isomorphic to $R[x]$ without providing the additional information of the generator "x"? | |
| Aug 25, 2011 at 14:51 | comment | added | Emil Jeřábek | Actually, you can include $x$ in the signature as yet another constant, with no further axioms. Then $R[x]$ becomes the free $V$-algebra over the empty set of generators, aka the initial algebra in $V$ as a category. | |
| Aug 25, 2011 at 14:29 | comment | added | Emil Jeřábek |
Assuming $R$ is commutative: let $V$ be the variety in the signature $\{+,-,0,\cdot,1\}\cup R$ whose axioms consist of the axioms of commutative rings plus the atomic diagram of $R$. Then $R[x]$ is the free $V$-algebra over one generator. This is a universal algebra characterization. However, you apparently mean category theory rather than universal algebra; I suppose you can characterize free algebras as initial in a suitable category (I’m not sure what to do about the generator).
|
|
| Aug 25, 2011 at 14:19 | history | asked | Jacques Carette | CC BY-SA 3.0 |