# Polynomial math question. **URL:** \ **Category:** Factual Questions\ **Created:** [May 18, 2005, 5:44am UTC](https://boards.straightdope.com/t/polynomial-math-question/304365 "2005-05-18T05:44:49Z")\ **Posts on this page:** 7\ **Page:** 2
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [May 22, 2005, 7:39am UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/21 "2005-05-22T07:39:16Z")
> [@RM Mentock](#): > > Actually, you said, “but the reasoning from earlier posts is all wrong” > > If we weren’t working over R, which obviously we weren’t since not all polynomials can be completely factored like that over R. I apologize for being unclear. It’s wrong in the sense that it’s unfounded. Once the foundation is properly relaid _then_ analogous arguments can go through. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [May 22, 2005, 7:54am UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/22 "2005-05-22T07:54:07Z")
> [@RM Mentock](#): > > Do you have a cite? Not offhand, though it’s one of those things that tends to exist more ephemerally than in the literature. Basically, the people who need the proper view learn it by reading a book or taking a class that assumes that viewpoint and bang their heads against the wall until finally they realize that their earlier classes had been lying in the name of simplicity of exposition. It’s definitely needed in differential and algebraic geometry, and I really don’t see that explicitly pointing out the choice (even if you don’t ever use the other choice) slows up the students in any way. Semi-analogous cases: In multivariable calculus classes, the space of tangent vectors to **R** [sup]3[/sup] at any point p is erroneously confounded with **R** [sup]3[/sup] itself. When the student tries studying differential geometry they must unlearn the naïve notion of tangent vectors. Some books do a better job than others in making this explicit. Many terms are around for historical reasons, being named before they were as well-understood as they are now. The homology “groups” in topology are more properly homology **Z** -modules, since that’s the concept that generalizes to homology with coefficients (homology R-modules) and additional algebraic structures (homology algebras). One book I’m sure is written in a style that makes the choice more explicit is Griffiths’ and Harris’ _Principles of Algebraic Geometry_. In fact, I seem to notice that it’s usually exactly this leap from the style of a typical complex analysis course that torpedoes most people who try to get beyond chapter 0. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [May 22, 2005, 9:20pm UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/23 "2005-05-22T21:20:38Z")
Z-modules are groups, abelian groups. But that’s an interesting point, that students might be hindered from moving up by a changing nomenclature. But what about the connection down, what about bridging to the student’s previous knowledge of groups? And now I’m curious. What would you say is the derivative of f(z) = (z + {\bar z})/2 ? ( the Re(z) ) ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [May 22, 2005, 9:41pm UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/24 "2005-05-22T21:41:26Z")
> [@RM Mentock](#): > > Z-modules are groups, abelian groups. But that’s an interesting point, that students might be hindered from moving up by a changing nomenclature. But what about the connection down, what about bridging to the student’s previous knowledge of groups? They’re called “groups” because they were first thought of as abelian groups. Ideally, a student would be taught that abelian group = **Z** -module, but just in case I’d define the homology modules and remind students that a **Z** -module is an abelian group. > [@](#): > > And now I’m curious. What would you say is the derivative of f(z) = (z + {\bar z})/2 ? ( the Re(z) ) It doesn’t have one. It has partial derivatives with respect to z and {\bar z}, but since neither of them vanish neither holds all the information. Alternatively, if you specify a curve in the plane, which specifies a relationship between z and {\bar z}, you can write a total derivative using the chain rule, which is then the total derivative along the curve. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [May 23, 2005, 12:52am UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/25 "2005-05-23T00:52:27Z")
> [@Mathochist](#): > > They’re called “groups” because they were first thought of as abelian groups. Ideally, a student would be taught that abelian group = **Z** -module, but just in case I’d define the homology modules and remind students that a **Z** -module is an abelian group. Doesn’t seem that hard 🙂 > [@](#): > > It doesn’t have one. It has partial derivatives with respect to z and {\bar z}, but since neither of them vanish neither holds all the information. And so, the derivative of f(z) = {\bar z} ? ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [May 23, 2005, 1:39am UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/26 "2005-05-23T01:39:26Z")
> [@RM Mentock](#): > > Doesn’t seem that hard 🙂 > > And so, the derivative of f(z) = {\bar z} ? The holomorphic derivative is zero. The anholomorphic derivative is 1. Again, “the derivative” is only defined when you restrict to a specific choice of complex structure and to either the sheaf of holomorphic or of anholomorphic functions with respect to that complex structure. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [May 28, 2005, 4:27pm UTC](https://boards.straightdope.com/t/polynomial-math-question/304365/27 "2005-05-28T16:27:22Z")
> [@Mathochist](#): > > The holomorphic derivative is zero. The anholomorphic derivative is 1. Again, “the derivative” is only defined when you restrict to a specific choice of complex structure and to either the sheaf of holomorphic or of anholomorphic functions with respect to that complex structure. But, once you’ve chosen the particular structure, the holomorphic derivative of F(z) = z is always 1, just as the holomorphic derivative of f(z) = {\bar z} is zero, as you say. So, for complex polynomial functions of z as in the OP, yes, there seems to be a canonical derivative. I disagree with your objections. [Previous page](https://boards.straightdope.com/t/polynomial-math-question/304365.md?page=1)