ahmet

This is okay. ✔️ but what about the theorem appearing above. In Regard to the multiplication operation, Z2 - {0z2} should satisfy group criteria.(Should be a group for multiplication operation) is this also being satisfied? Here H=Z2 maybe you might be right, not sure. Because I am sure that my hodja had made me note that theorem. But already galois thorem is not specified for specific prime number (to be exempted),therefore you might be right.

no, this is not a group for multiplication operation.To be honest Galois's theorem does not specify the case for 2 ,which is prime number. But this prime number is unique prime number that is even number. There is one a very basic theorem that will not accept any set which has had just two elements to be a field. That basic theorem is given as below: Theorem: a ring having just two elements cannot be a group for/across multiplication operation.(Thus <H,+,.> cannot be a field) Proof: <H,+,.> is a ring. if <H,.> is a group, we should have at least x element in H such that for 0H≠ b , a.x=b should be satisfied. when we care one other theorem ,which states 0H.x=0H , we have b=0H and this is a contradiction. thus, as any ring that has just two elements cannot be a group in regard to multiplication operation, it will also not be accepted as a field. I think Galois theorem is invalid specifically for 2. meanwhile, I think wikipedia might be not suitable for mathematical sciences ,where the issue is upper level than MSc or equivalent. (for Z2 there 0 is not invertible. )

what is this ?

would you like to share your opinions in convenience of title? let us learn which engineering book you mostly enjoyed,pleased to read and why.

I think there should be three element at least. lets see 1+(-1)=0 but -1 is not element of {1,0} set. Thus we can say that additive inverse does not exist for "1" element in {1,0} in fact, the second operation should meet/satisfy group criteria to be field, and to be ring; first operation should also satisfy group criteria.

simply, I see the use of "de morgan" rules.

this is an ideal (also it is one of trivial ideal) but not a field. Could you explain how you have defined . and + operations here? what is more ,I can additioally say that there was one theorem claiming that a matehmatical structure containing just 2 element also would not be field.(i.e. not only just one element but just two element is also insufficient to be field.)

in normal or direct reflection,no. but...I mean or claim and also underline "coincidences" , I think that this was high potential. because it is always probable to find a new thing when you look for something. but these days ...I am sure that searches are dense. here,in turkey I see some associations (e.g. TUBITAK) also turkish authority ease some conditions For instance in previous times ethical board was mandatory prior to some researches but now this conditon has been eased. not only turkey, I predict many countries have high desire to find a cure or vaccine for covid-19. but what if by coincidence ...?

the cure of AIDS or vaccine of HIV be found by coincidence simultaneously while the studies across covid-19 are being performed ?, what is your opinion?

ahahahha really right!

as I remember Z and Q are not in the same category. One of them should be different. Menwhile, to you; which type of fields do you claim? there are many types of fields (E.g. P.F.F. , Euclid field or completeness field etc.) Anyway, It seems disccussing such contexts under another thread will be better.

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welcome @wtf, .I think I have done a mistake. (But some terms changing between languages, so I am not sure whether I would express all terms correctly (or equiavlently) Description: if <H,+,.> has 1H and has no zero divisors, and is commutative, then we call this ring with " completeness field" (I translated from turkish into english that term) Description: if H is commutative and has 1H and for every x element H-{0H}, at least y element H-{0H} such that x.y=1H conditions are being satisfied then we call this completeness field specifically with "object" (but not sure about the correctness of this term)( or equivalently this ring should satisfy group criteria for the second operation (multiplication)) All Q,R,C are objects. also, all finite completeness fields are also objects. Q,R and C should have no more than their trivial ideals

heyy ! I was just joking , but generally my jokes have potentiality to be also real. The thing I know I do not lie. ok,as they do not prefer to reply in the current status,I think I can postpone some operations. meanwhile,I was to use their materials about html+css ..(also maybe js) but I am still not sure whether interferring their specific samples would also provide me another copyright (?) (my own copyright) Thanks (Could you also respond: assume please I have contacted to the author of that or any else book. Also assume that the author kindly gives me permission to use the content. Should I also take the permisson from the publisher??)

ok. this is already rational idea. but contacting over the phone,will not this be non sense? because they will be able to refute later? I think I need written (material)/ permission (like an e mail) ok,now I decide to postpone somethings. furthermore,as far as I know copyright is not like a patent , and is valid only in the published country. But I do not really dominate all the relevant contents

maybe,I will try that but this will be a bit time later. Furthermore, do you have information about the questions on this issue ,for instance , what does require me to request a permission and not. (Specifically: I do not know whether I have to take permission on constructing a website or websites. I intend to use almost all of the codes they provided in some samples, but I also know that I would definitely interfere to that codes, for instance I shall change css part, and will add some new codes. however,I shall not use any extra (for instance I shall not copy paste any else section of the book and will not use the information provided in example (appearing on that sample of website) but I shall use just the codes succintly.

and what to do now,should I send a complaint form? inquiring: why don't you reply?!

I have tried to contact wiley online library customer care and requested permission to use some parts in some of the books published by this publication but they still have not responded although two mails forwarded.(I mean,the same request two times sent) So, can I decide that they have already allowed me to use as they do not care or are in silent mode?

hi, once again; I would ask a new question on this issue: DO YOU BELIEVE THAT I WOULD LOSE THE SMARTNESS OF MY VOICE IF I GET OLDER? (if we compare the case with young ages) now I am 31 (of course proudly feel myself as 18 , but I do not know what would happen at my for instance 45 or 60) I am sure I am trying to calculate rather non mathematical probability ahahaha all in all it seems normal, because we have emotions ..

hi, I do not know what exactly caused you to think so. but no problem for now. meanwhile I do not think that that description was available here. a reminding: with trivial ideals I meant for instance in <H,+,.> ring, H and {0H} are trivial ideals. (There should be no more ideals for this ring)

ahahaha @studiot ,you seem like you criticize the sets pahaha rather than criticizing people ,well ,very well. I really appreciate the tone of your tongue. haha. (like saying; Z is not a successfull set for multiplicative inverses and N already seems like just a stupid set, he even cannot contain additive inverses ) Meanwhile, I saw @wtf ' s explanations (provided in the link) but I also thought that there would be many extension fields. mmm may I ask a question , with what are you expresing a ring satisfying all the conditions given below in english? (in fact bourbaki has provided that definition but I could not immediately /quickly find) ---->> with no zero divisor. ---->> commutative --->> contains unitary element of multiplication one more question: what do you mathematically call a specific ring with, that has not had any else ideals except its trivial ideals?

hi, algebra is not as easy as it appears/stands. Some notions even could be stricter than you might imagine. I have not come across with any well described such expresion. But of course some new groups might be defined with some expressions (like you provided). Also ,gauss ring may be assessed in this regard.but if you provide more contexts, maybe I and some other mathematicians might help you. mm,now,among the contexts and when I would overview the big picture under this thread, studiot's this sentence seems like so much pretty good or cute (quoted below).

hi again, I commonly believe that I was not good in algebra.But I shall try to represent basic contexts as well as I can. first, studiot's comment seems like a solution for some problems. because to my knowledge , we are allowed to say I < R and Q < R (also,even R<C (here C shows complex numbers)) but in spite of all of the details given above, in existing methods I have not seen at anywhere or have not come across any expression ,claiming Q < I , I was thinking this would be correct from the notes I taken in the university. but generally as; there is no doubt or contradictions between mathematical thorems in general or it is not a common action, I taught or said to 12nd grade students that any rational number will never be an irrational number. But how to make a clear comparison between I and Q sets. Or how are the mathematical characters of such sets ? thus, studiot's comment seems like a good solution and may provide a new set (under assumption; all (of (my) ) the claims appearing above was right) between I and R ,which is bigger than I and Q and also includes both of these sets. (some contexts that might potentially be relevant to this subject. 1) yes again under assumption that we would be able to say Q was a subset of I, studiot's solution shows an ideal of R, but he presumably created a new set. (if so, congratulations) 2) gauss integer numbers ring 3) some other algebraic contexts (E.g. H/K where K is a prime or maximal ideal of H,not sure)