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Thread: The Fields

  1. #1 The Fields 
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    Excuse me ... .


    A field is structure of algebra which and is abelian group ... .

    Until now, in abstract algebra, we have known three field : , , and (read respectively: point, real line, and complex plane) under usual additive opetration and usual multiplicative operation ... .

    Topologically,

    the point is seem like the ... ,

    the real line is seem like the ... , and

    the complex plane is seem like the ... .



    Are there the fields which like topologically for ... ?


    Thank you all ... .


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  3. #2  
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    I believe that it only works for dimension 4 and 8, called the quaternions and octonions respectively. The reasons are still above my head so I can't really give a good explanation. There are other fields though, for example the finite fields with a finite number of integers under modular arithmetic.


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  4. #3  
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    @ TheObserver

    A set of quarternion can form a ring ... , but not a field ... because the semigroup is not commutative ... . In simple example, , , ... . To form a field, we have to add the requirement of commutation ... .

    I'm sorry if I do mistakes ... . Thank you ... .
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  5. #4  
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    No that sounds right, my bad.
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  6. #5  
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    TheObserver: I applaud your humility, we need more of that round these parts. But as I am fond of saying, notation is arbitrary. is usually used for the rational numbers. (is this field? tell us). The quaternions are usually denoted as after their discoverer W. R. Hamilton. Octonians? No idea! But I think that trfrm is about right

    But this caught my interest.....
    Quote Originally Posted by TheObserver View Post
    There are other fields though, for example the finite fields with a finite number of integers under modular arithmetic.
    You need to be a bit more careful how you express yourself with pedants like me around.

    You are mostly right though.

    Consider the integers. This is clearly a commutative ring but not a field (tell us why).

    Now consider the ring , those integers exactly divisible by some . Then one can form something called the quotient ring by the rule that if any members of differ by the integer , which is denoted as , in shorthand as . A simple example is worth more that a thousand words.....

    Suppose we say that specifies the set , then the only other element in the quotient ring will be the set . These sets comprise what are called "equivalence classes", and it is customary to elect a class representative, which in this case we may call as 0 and 1. So there is a ring isomorphism between and .

    Any is a ring (prove it!), but is only a field if is a prime number.

    The proof of this assertion is not easy, and rather messy, but may try if you want. But I am unsure why you state that this is a "finite field", truly I don't know the answer
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    Right, I forgot that its only a field for primes, we talked about that a lot too. My algebra teacher in first year called them finite fields, thats the only reason I use that name. I took a great year long honours linear algebra course in my first year but I never really followed through with ring theory or field theory or anything. The only algebra Ive done since then was a group theory course last year so Ive lost a lot of those details. Thanks for the refresh.
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  8. #7  
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    OK, good. It might just be worth pointing out to those members who visit both the Math and the Physics section - and why wouldn't they - that mathematicians and physicists use the term "field" in different ways.

    In mathematics, a field is defined as a set for which two operations are defined - these are by convention called "addition" and "multiplication" - each of which has an inverse and an identity. Queerly enough, as trfrm detected, there seems not to be all that many fields in this sense.

    Even more queer (possibly) that, up to isomorphism, there is only one ordered field, and that is (where "ordered" has the usual and obvious meaning)

    In physics, by (possible) contrast a field is the assignment to each and every point in some manifold a quantity, scalar, vector or tensor, for example.

    There is a really, really neat way to do this, using the theory of "fibre bundles". Maybe I will bore you all again with a new thread on this. We'll see......

    PS I have thought long and hard about how to bring these two different usages of the term into register - and failed! Anyone any ideas?
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  9. #8  
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    Quote Originally Posted by trfrm View Post
    Excuse me ... .


    A field is structure of algebra which and is abelian group ... .

    Until now, in abstract algebra, we have known three field : , , and (read respectively: point, real line, and complex plane) under usual additive opetration and usual multiplicative operation ... .

    Topologically,

    the point is seem like the ... ,

    the real line is seem like the ... , and

    the complex plane is seem like the ... .



    Are there the fields which like topologically for ... ?


    Thank you all ... .
    "opetration"? If the mistakes your doing is all of this kind...Then dont worry...nobody will notice.
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  10. #9  
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    Quote Originally Posted by Guitarist View Post
    Any is a ring (prove it!), but is only a field if is a prime number.
    I am disappointed nobody attempted a proof of this.

    So try this - my assertion may be viewed as a corollary to the following more general theorem:

    Suppose a commutative ring equipped with a multiplicative identity, and let be an ideal in . Then is a maximal ideal if and only if the quotient is a field.

    Try to prove this.

    PS by edit - if you think this an unreasonable request, remember this is a MATHEMATICS forum, and attempting (not necessarily succeeding in) proofs is all part of the process. Go for it ......
    Last edited by Guitarist; December 19th, 2012 at 05:07 PM.
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  11. #10  
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    No takers, huh? You guys are no fun, but I guess it's mine own fault, as I put you off by admitting the proof is not easy. Ah well, silly me...

    Anyway lemme step through a proof which might educate some of you about rings, fields and ideals.

    But first note that if is a commutative ring, we must have that and . And, most specifically, that if then

    Second, an ideal is a subring such that, for some and any that and . If is a commutative ring, obviously these are equal.

    The ideal will be maximal in iff, for any other ideal then either OR . Note the following important properties of ideals: .

    The quotient ring has as elements and by the above (A is commutative by construction, I is so by definition). Note also by the above that is a zero element in

    So to the proof. Stay awake you at the back!!

    Consider the quotient ring , and let be ideals in such that . Let . Then, evidently, is not a zero element in .

    Now, if and only if is a field, I can always find some element such that ( since in any field , we must have, for any , some where ), so by the above and therefore .

    So if by the forgoing, we must have that using the cancellation law for addition in a field (strictly, it only needs to be an integral domain, but every field IS an integral domain)

    I will set , where, self evidently so I will have that , again using the cancellation law.

    But and and, further that by construction, so that , then I conclude that

    Thus, for any , I have that which is quite simply , thus , and therefore is a maximal ideal in .
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  12. #11  
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    @ Guitarist

    Thank you very much for the answer ... .
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  13. #12  
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    Quote Originally Posted by trfrm View Post
    A field is structure of algebra which and is abelian group ... .
    You forgot one more condition: multiplication must be distributive over addition. The two operations in a field are not independent of each other: they are related by the distributive law.



    Quote Originally Posted by trfrm View Post
    Until now, in abstract algebra, we have known three field : , , and
    When defining a field, mathematicians usually insist that , that is, the additive identity and the multiplicative identity must be distinct. Hence is not normally considered a field since a field must contain at least two elements.



    Quote Originally Posted by Guitarist View Post
    Even more queer (possibly) that, up to isomorphism, there is only one ordered field, and that is (where "ordered" has the usual and obvious meaning)
    Not true: is also an ordered field (and it’s not isomorphic to ). And it’s not the only other one either – so is . More examples: http://en.wikipedia.org/wiki/Ordered_field.



    Quote Originally Posted by Guitarist View Post
    Any is a ring (prove it!), but is only a field if is a prime number.

    The proof of this assertion is not easy, and rather messy, but may try if you want.
    The proof is actually quite straightforward. Since is a commutative ring with multiplicative identity, to prove that is a field where is prime it is enough to show that any nonzero element has a multiplicative inverse. Let be such an element, i.e. . Then since prime, and so by Fermat’s little theorem, , i.e. . Now consider . Since is prime, and so the exponent in is non-negative; hence is a valid integer. By the division algorithm, there exist integers such that and . (Note: cannot be 0 since does not divide .) It follows that , i.e. is the required multiplicative inverse of in .
    Last edited by Nehushtan; March 2nd, 2013 at 11:45 PM.
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  14. #13  
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    Quote Originally Posted by Nehushtan View Post
    You forgot one more condition: multiplication must be distributive over addition. The two operations in a field are not independent of each other: they are related by the distributive law.
    I’m sorry ... . I had forgotten to add this requirement for building a field ... .
    For , where is a field, then ... .

    Quote Originally Posted by Nehushtan View Post
    When defining a field, mathematicians usually insist that , that is, the additive identity and the multiplicative identity must be distinct. Hence is not normally considered a field since a field must contain at least two elements.
    Why do the additive identity and the multiplicative identity must be distinct ... ? Is the distinction one of all conditions of a field ... ?

    Isn’t , , and ... ?

    The Fields
    Last edited by trfrm; July 19th, 2018 at 11:22 PM.
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  15. #14  
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    Quote Originally Posted by trfrm View Post
    Why do the additive identity and the multiplicative identity must be distinct ... ? Is the distinction one of all conditions of a field ... ?
    They donít have to be distinct; itís just the way most mathematicians define a field. If you let then the only possible structure satisfying all the field axioms is Ė whereas the fields mathematicians normally work with are much less trivial than that.

    But thereís nothing wrong with considering to be a field if you want. Iím just saying that most mathematicians donít.
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