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    did some research and i understand almost all of it wew!

    so a field is a commutative ring which each non 0 element is invertible but i don't understand the part where google says "by definition any fields is a commutative ring"
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    (Original post by will'o'wisp2)
    did some research and i understand almost all of it wew!

    so a field is a commutative ring which each non 0 element is invertible but i don't understand the part where google says "by definition any fields is a commutative ring"
    It is just saying the same thing again. What is a field? The first part of your sentence says that it is a commutative ring with an extra condition - so, by definition, any field is a commutative ring (it must be, we've just said that a field is a special kind of commutative ring).
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    (Original post by Pangol)
    It is just saying the same thing again. What is a field? The first part of your sentence says that it is a commutative ring with an extra condition - so, by definition, any field is a commutative ring (it must be, we've just said that a field is a special kind of commutative ring).
    oh right i see ._____. thanks man
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    (Original post by will'o'wisp2)
    did some research and i understand almost all of it wew!

    so a field is a commutative ring which each non 0 element is invertible but i don't understand the part where google says "by definition any fields is a commutative ring"
    Done it just by comparing the required axioms for both definitions from Wikipedia:

    The requirement of a commutative ring is that it must be a set A with two operations a+b and a \cdot b, and these must have the following for all a,b,c in the set:

    a+b=b+a is also in A
    (a+b)+c=a+(b+c)
    There is an identity element for addition, 0, such that a+0=0+a=a
    There is an inverse element for addition, -a, such that a+(-a)=0
    (a\cdot b) \cdot c=a\cdot (b \cdot c)
    There exists an identity elements for multiplication, 1, such that a \cdot 1 = 1\cdot a = a
    Multiplication distributes over addition: a \cdot (b+c)=a\cdot b + a\cdot c
    AND multiplication is commutative a \cdot b = b \cdot a



    Then a field is a set B with two binary operations a+b and a\cdot b, and must have the following for all a,b,c in B:

    a+(b+c)=(a+b)+c
    a \cdot (b \cdot c) = (a \cdot b) \cdot c
    a+b=b+a
    a \cdot b = b \cdot a
    There is an identity element for addition, 0, such that a+0=0+a=a
    There is an inverse element for addition, -a, such that a+(-a)=0
    Multiplication distributes over addition: a \cdot (b+c)=a\cdot b + a\cdot c
    There are multiplicative inverses a^{-1} (for a being nonzero) in B such that a \cdot a^{-1}=1

    Now all the conditions line up except the multiplicative inverses which are missing from the commutative ring. In fact, IF the commutative ring has the property with the multiplicative inverses, then it automatically satisfies the definition of a field. This is going from the commutative ring to a field, but now looking at it the other way, every field must be some commutative ring.

    In other words, NOT EVERY commutative ring is a field, but every field is some commutative ring.


    EDIT: tfw your internet cuts off and someone beats u 2 it
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    (Original post by RDKGames)
    Done it just by comparing the required axioms for both definitions from Wikipedia:

    The requirement of a commutative ring is that it must be a set A with two operations a+b and a \cdot b, and these must have the following for all a,b,c in the set:

    a+b=b+a is also in A
    (a+b)+c=a+(b+c)
    There is an identity element for addition, 0, such that a+0=0+a=a
    There is an inverse element for addition, -a, such that a+(-a)=0
    (a\cdot b) \cdot c=a\cdot (b \cdot c)
    There exists an identity elements for multiplication, 1, such that a \cdot 1 = 1\cdot a = a
    Multiplication distributes over addition: a \cdot (b+c)=a\cdot b + a\cdot c
    AND multiplication is commutative a \cdot b = b \cdot a



    Then a field is a set B with two binary operations a+b and a\cdot b, and must have the following for all a,b,c in B:

    a+(b+c)=(a+b)+c
    a \cdot (b \cdot c) = (a \cdot b) \cdot c
    a+b=b+a
    a \cdot b = b \cdot a
    There is an identity element for addition, 0, such that a+0=0+a=a
    There is an inverse element for addition, -a, such that a+(-a)=0
    Multiplication distributes over addition: a \cdot (b+c)=a\cdot b + a\cdot c
    There are multiplicative inverses a^{-1} (for a being nonzero) in B such that a \cdot a^{-1}=1

    Now all the conditions line up except the multiplicative inverses which are missing from the commutative ring. In fact, IF the commutative ring has the property with the multiplicative inverses, then it automatically satisfies the definition of a field. This is going from the commutative ring to a field, but now looking at it the other way, every field must be some commutative ring.

    In other words, NOT EVERY commutative ring is a field, but every field is some commutative ring.


    EDIT: tfw your internet cuts off and someone beats u 2 it
    the bold one isn't in the field part so is that a part of the field too? but thanks so much man, this is going to my notes, understood very well now
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    (Original post by will'o'wisp2)
    the bold one isn't in the field part so is that a part of the field too? but thanks so much man, this is going to my notes, understood very well now
    Yes the multiplicative identity is one of the requirements in the field that I missed out on, just add that in.
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    (Original post by RDKGames)
    Yes the multiplicative identity is one of the requirements in the field that I missed out on, just add that in.
    got a bit worried then, since it was going to my notes, still awesome job man
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    (Original post by RDKGames)
    Yes the multiplicative identity is one of the requirements in the field that I missed out on, just add that in.
    just quickly if the field is  \mathbb Z_6 then for the multiplcative inverse if i do [4]\cdot [4] then if it equals 1 then it's a multiplicative inverse?
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    (Original post by will'o'wisp2)
    just quickly if the field is  \mathbb Z_6 then for the multiplcative inverse if i do [4]\cdot [4] then if it equals 1 then it's a multiplicative inverse?
    In  \mathbb Z_6, [4]\cdot [4] = [4], but [5]\cdot [5] = [1] so [5] has a multiplicative inverse but [4] doesn't.
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    (Original post by atsruser)
    In  \mathbb Z_6, [4]\cdot [4] = [4], but [5]\cdot [5] = [1] so [5] has a multiplicative inverse but [4] doesn't.
    but for 5x5 isn't that 25 so

    0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3, 4,5,0,1,2,3,4,5, 0

    isn't it [0] ?

    but if it equals 1 then it has a multiplicative inverse got it cheers
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    (Original post by will'o'wisp2)
    but for 5x5 isn't that 25 so

    0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3, 4,5,0,1,2,3,4,5, 0

    isn't it [0] ?

    but if it equals 1 then it has a multiplicative inverse got it cheers
    It's [1].

    You do it mod 6, so since 24 is a mult of 6, one higher than that will leave remainder 1.
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    (Original post by RDKGames)
    It's [1].

    You do it mod 6, so since 24 is a mult of 6, one higher than that will leave remainder 1.
    so i don't start with 0? -__-

    ok then understood
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    (Original post by will'o'wisp2)
    so i don't start with 0? -__-

    ok then understood
    Do it from a table

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    (Original post by RDKGames)
    Do it from a table

    it's clearly undeniable that you're right but why can't i count it? why does it go wrong when i count it? .__.

    for z3 if i count it up to let's say 21 i get [2] when i should get [0]
    it's a problem when i count back to 0, it's like i have to miss it out for some reason
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    (Original post by will'o'wisp2)
    it's clearly undeniable that you're right but why can't i count it? why does it go wrong when i count it? .__.

    for z3 if i count it up to let's say 21 i get [2] when i should get [0]
    it's a problem when i count back to 0, it's like i have to miss it out for some reason
    Start counting the amount of digits you have from 0 instead of 1 then you'll find that you need 26 digits for the number 25. Your final digit is 24 mod 6 as it stands. If you start from 1 then you're saying 1 mod 6 = 0 which is not true for your first digit.
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    (Original post by RDKGames)
    Start counting the amount of digits you have from 0 instead of 1 then you'll find that you need 26 digits for the number 25. Your final digit is 24 mod 6 as it stands. If you start from 1 then you're saying 1 mod 6 = 0 which is not true for your first digit.
    i see thanks
 
 
 
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