Commutative Algebra - Science topic

e.g., see

for a

"Timetable scheduling problem"

"mixed-integer (linear) programming model"

which is "usually" solved by a genetic algorithm method

Good Luck

A charged particle whizes past a very massive body, therefore its trajectory bends somehow

and radiates (for Classical reasons). Therefore there might be some radiation surrounding a

Black hole. So who is responsible? Not QM. Perhaps only Einstein or Newton?

Let K={|z|<=1} be the closed unit circle (a connected compact) in C, R=O(K) be the ring of holomorphic functions f on K with d(f)=#{zeroes of f in K}, R is a ED.

Let a=a(z)=z^2+2z+1, b=b(z)=z (hence d(a)=2, d(b)=1). Let h=h(z)=sin(z)/z, h(0)=1.

Then

1) z^2+2z+1=(z+2)z+1, q=z+2, r=1 (hence d(r)=0)

2) z^2+2z+1=(z+2+h)z+1-zh, q'=z+2+h, r'=1-zh=1-sin(z) (hence d(r')=0,

since sin(z)=1 iff z=\pi/2 +2\pi*k, k\in Z).

I think no need to include closure property, if it is given that 'o' is a binary operation on the set G.

it is not clear for me

Dear Ahmadreza Khazaeipoul

For Cryptographic purposes.

I hope you find the attached article is helpful.

Best regards

Dear Raihan Dwianto

I hope you find the following are useful

Regards

Have a look at https://core.ac.uk/download/pdf/82015258.pdf

You may follow : https://www.tandfonline.com/doi/abs/10.1080/00927870601041458?scroll=top&needAccess=true&journalCode=lagb20 which could a bit helpful for understanding your problem.

Dear professor ,

In the set M_n of nxn square matrices, the Zero matrix commutes with all other matrices, and in particular Zero matrix commutes with any non- diagonalizable matrix. Obviously, commute does not imply diagonalizable.

If we consider the set D_n of the square diagonalizable matrices,

the statement: Commutative matrices are simultaneously diagonalizable

is true over the set D_n. Mathematically, this is what I exactly meant, because the other case is trivial and well-known.

So, your suggestion is to make a general statement:

Diagonalizable commute matrices are simultaneously diagonalizable.

This will exclude all trivial cases. :-)

Best wishes

@ Yasir,

Google your request to reach

(1) https://link.springer.com/content/pdf/10.1007/BF02364730.pdf

(2) https://www.researchgate.net/.../240933988_Endomorphism_rings_power_series_rings_...

(3)

Best regards

Some questions can be solved in a non-commutative graded case:

Generally, this number is not unique.  For example, {x} and {x²,x+x²} are both minimal sets of generators of k[x]. For the subalgebra k of k[x₁,...,x_n] every minimal set of generators has 0 elements (duh). I suspect that this is the only example where the number of elements in a minimal set of generators is unique. 

The situation is different in the graded case. The polynomial ring k[x₁,...,x_n] has a grading. If R is a graded sub-algebra then every minimal set of homogeneous generators of R has d elements, where d is the dimension of the k-vector space m/m² and m is the ideal generated by all homogeneous polynomials in R of positive degree. If R is not finitely generated then d is infinite. Every minimal set of (not necessarily homogeneous) generators of R must have at least d elements. 

Dear Kelvin,

I think you meant: B=A\otimes _K A' (i.e. tensor product over the field K). Next, we should prove this by definition: if a\in A and deg(a)=r\in R, a'\in A' and deg(a')=r'\in R are two R-homogeneous elements then we assign deg(a\otimes a')=(r,r')\in R². It remains to show that this grading  is compatible with multiplication in B=A\otimes _K A' (which is not hard).

Best regards, Yuri

"if P(x) takes value 0 for n+1 different values of x",  "then all its coefficients are actually zeros."

1) Then p(x) is of degree >=n+1

2) No reason , that all coefficients then are zero.

3) A polynomial can has zeros in some domain without be identically zero.  Please reformulate your additional information in the question.

Thank you so much, Ahmed Munir.

I did ask this question in mathstackexchange recently, but this is my first question on ResearchGate.

The condition you mentioned in your first paragraph is not the minimality I'm talking about, if I understood you correctly.  The one I'm talking about is that there is no set with fewer elements that generates the ideal.  In fact the minimality you mentioned (if I understood it) is never satisfied:  For example, if we have an ideal J = (f, g), then we have for example (g + 1)f + (-f)g = f.  For example in Z[x], J = (2, x) is minimally generated by 2 and x even though we can get both 2 and x as polynomial linear combinations of 2 and x.

Regarding your second paragraph, not every polynomial in J(p) is of the form g(p(x)) for some g(x) in J.  For example, xg(p(x)) is probably not of that form.  For example, if J = (x2) and p(x) = x3, then J(p) = (x6).  In this example, x7 is in J(p) but it is not of the form g(x3) for any g in J (or any g at all).

Yes, you can use the same definition for a non-commutative associative ring R. Note, however, that the resulting Nagata extension will typically be a non-associative ring!

For arbitrary (a,m), (b,n), (c,p) the equation ((a,m)*(b,n))*(c,p)=(a,m)*((b,n)*(c,p)) is equvalent to

n\sigma(a)c + mbc + p\sigma(ab) = nc\sigma(a) + mbc + p\sigma(ba).

If R is commutative, then the above equation is obviously satisfied, but if R is non-commutative then we can not immediately tell.

Dear colleague Kultchitsky,

Problem of the solving the Dirichlet's problem for the upper semi-plane is very interesting and usefull. I asked the same question, and generally gave the answer, missing only some details at the end, (but if it is a problem they can be added)..

Using well-known invariance of Laplace differential equation in regard to conformal mappings, we exceed the Laplace's equation                                                                 T_xx(x.y) + T_yy(x.y) = 0                                                                   to the Laplace's equations                                                                                                                F_uu + F_vv = 0,        

 can be found on the web-site

How to solve the Dirichlet's problem for the upper semi-plane ? - ResearchGate. Available from: https://www.researchgate.net/post/How_to_solve_the_Dirichlets_problem_for_the_upper_semi-plane [accessed May 18, 2016].

All best

Mirjana

wit

 (1) +(2) becomes the following Dirichlet's  problem

Dear Geng,

From a formal point of view, the construction of what you call ‘infinitely big’ and ‘infinitely small’ is different.

Cantor's infinite cardinal hierarchy arises from his attempt to generalise the idea of equinumerosity from the finite to the infinite. The limit construction in mathematics comes from the need to define in a mathematically concise manner a concept that was used in physical theorising (17th-18th  century Wallis, Leibniz, Mercator, Kepler, ) where 1/\infty, dx, partial derivatives and tendencies of series were used without proper definitions. (Note that the limit concept handled both things ‘series diverging to infinity’ and a proper use of the ‘infinitessimal dx’ in derivatives.) Finally the construction of hyperreals which provides another meaning to the notion of an infinitesimal, only arose in the context of first order logics and non-standard models of field theory.  

One should also note the difference between the context in which a concept was proposed, and what it was later argued to represent. (Eg. hyperreals were not introduced to represent infinitesimals, although they are now thought to allow such representation and maybe much better than the limit construction. Similarly Cantor’s hierarchy of cardinalities was not introduced to represent the ‘infinitely big’, but was a side product of the attempt to formalise equinumerosity in purely set-theoretic terms).

There is, in principle, no good or bad in ‘infinities’ as there is no good or bad in any definition, as long as it is correctly and well defined. The question is rather, what you need them for (pragmatics of mathematics), what they allow you to understand (epistemology of mathematics) and whether you can do without it (ontology of mathematics). Your normative question of what should or should not be defined will have other meanings from these different perspectives and different answers, depending on what kind of stance you take on pragmatics, epistemology and/or ontology.

I hope this helps,

Best,

Eric

I could not open your file. It says that it is corrupt. Can you please just give us one example or two of your suggestions?

Please see an example in the following useful link:

If I understand your question correctly, I think the answer is no.  For instance, 

almost complete intersections I (which define C-M quotients) are linked to Gorenstein

ideals J, and the canonical module of R/I is J/(regular seq). But Gorensteins can

have arbitrarily large numbers of generators. For explicit examples, try codim 3 Pfaffians.

Hello, Kelvin,

consider the (right) transversal T for H in G, i.e. G=disjoint union_{t \in T} H.t  of left cosets, |T|=|G:H|. It is known that FG=direct sum_{t \in T} FH.t, one of the summands being simply FH (for, say, t=1). Each summand is a free FH-module isomorphic to FH. If you take the quotient FG/FH you get direct sum_{t \in T, t \ne 1} FH.t of free FH-modules of the cardinality (if finite) |T|-1=|G:H|-1. For instance, consider the case G=H, the quotient FG/FG is trivially 0. Best wishes, Yuri

Hi, if I correctly understood the problem,

you have a direct sum R[x] =R[x^2]+x*R[x^2] of R-modules, second summand consisting of sums of odd powers of x. Both summands are obviously isomorphic, and both are isomorphic to R[x] as R-modules (substitute t for x^2). The quotient R[x]/R[x^2] is isomorphic as R-module to

x*R[x^2], hence to R[x^2]  and consequently to R[x]. Best regards, Yuri 

Something you should have: http://www.ams.org/journals/tran/1966-121-01/S0002-9947-1966-0202782-5/S0002-9947-1966-0202782-5.pdf

Also I've found: Whether or not the new definition of simplicity is an actual restriction

is not known. That is, whether or not commutative powerassociative

nilalgebras without proper ideals exist (except of course

for the zero algebra of dimension 1) is still an open problem.

http://www.ams.org/journals/bull/1955-61-06/S0002-9904-1955-09974-9/S0002-9904-1955-09974-9.pdf.

So it's a very serious problem (If it haven't  been solved yet.)

Dear Enas,

I hhave cheeked.

your proof  and your examples are true.

Thank you Peter. What is the correct statement?

R = Z_4 is IF ring which is not von Neumann regular

One should also add CoCoA (see lin below) to the list of the freely available computer algebra systems specifically dedicated to commutative algebra (and non-free one could also add Magma). As written in the thread mentioned by Artur, there is not real point in asking which of the three is the "best" one, as this depends where much on what is important for you: breadth of the library, speed of computation, ease of programming. The next question is whether you have some specific problem in mind (for which perhaps only one of the three has some code available) or whether you are asking generally. But even for a very specific problem there is usually no simple answer. We developed recently a new algorithm for computing resolutions and implemented it in CoCoa. Then we run a lot of benchmarks comparing it with Singular and Macaulay2: each system had some examples where it was faster than the other one. Computational commutative algebra is a very complex field and there is usually not a "best" solution to a particular class of problems.

I have now made some progress; the program the helped me was Singular.

Here is an article that discusses the type of factorization you mention. (For what it's worth, I regard this as a very good article in this particular area.)

Mahdi Javadi and Michael Monagan.

On Factorization of Multivariate Polynomials over Algebraic Number and Function Fields.

Proceedings of ISSAC '09, 199–206, ACM Press, 2009.

It seems to me that a crucial point about your problem is that you don't just want to realize a given operator as a commutator, but as the commutator with a fixed (given) operator. While I don't know the answer to your question, you could try to attack it by studying the action on eigenfunctions for D. If [D,A]=iaL_z, then for any eigenfunction f of D, you get a differential equation on Af . Maybe this helps in solving the problem or spotting problems.

I see another possible approach, provided that one has a good understanding of [D,L_z] and of iterated commutators of this form. If this is the case, one could try to understand the Lie algebra generated by D and L_z .

Yes, the problem is still open. It is known as Kaplansky's conjecture.

Here are a couple of references:

Have a look at I. R. Shafarevich, Algebra I, paragraph 1, 2,3,4 at least, where there are many interesting examples of "coordinatisations". 

Thanks Olgur. But does still such an emaple exist if we suppose that R is unmixed of dimension 2?(Unmixed = every associated prime is of height zero)

Hi Rogier

Thank you for very much for your answer and book references are always appreciated! Yes I know these are implemented somehow in Singular. In fact Singular is one the programs that I mainly use. Unfortunately the cases we are dealing with are not always complete intersections, although in most cases they are and are generated by codim number of generators.

Also are you aware about the Hilbert-Samuel function/multiplicity number for singularities ? Do you remember any good books where the algorithms and def.s for these would be introduced in . I would really appreciate all book tips.

Also about a question about the 'intersection multiplicity': If two varieties intersect

transversally do you remember if there are kind of 'straight' or sequence of commands to verify this with Sage, Singular, Macaulay2 or CoCoa without coding the algorithms completely yourself ? Just to save time.

Also if you remember good books where these problems are tackled it

would also be good to know !

But thanks again for you response !

regards

Samuli Piipponen

Hi Tribid

Commutative algebra has various applications outside pure mathematics. You probably should look more computationally oriented books, where some chapters and exercises deals especially with applications. Look for example the two books of,

auths. Cox, Little, & O'Shea

1. Ideals, Varieties and algorithms

2. Using commutative algebra

Besides the the theory the cover a range of various applications.

You can also look at my publications list if you are interested :

seems that only few is -).

regards

- Samuli

Let R be a right Harada ring. Then every R-module is a direct sum of a small module and an injective R-module. Hence every R-module has C*-property.

as many as divisors of n. All ideals of Z/(n) are principal, by use of the euclidean algorithm...