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Complete the table for the finite field

  1. Jun 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##({a, b, c}, *,+)## be a finite field. Complete the field table for the operations ##*## and ##+##
    ##\begin{array}{|c|c|c|c|}
    \hline * & a & b & c \\
    \hline a & ? & ? & ? \\
    \hline b & ? & ? & ? \\
    \hline c & ? & ? & b \\
    \hline
    \end{array}##

    ##\begin{array}{|c|c|c|c|}
    \hline + & a & b & c \\
    \hline a & a & ? & ? \\
    \hline b & a & c & ? \\
    \hline c & a & ? & ? \\
    \hline
    \end{array}##

    2. Relevant equations
    3. The attempt at a solution

    I figured that since it is a finite field the operation ##+## must be commutative and must have a symmetric table along the main diagonal.
    ##\begin{array}{|c|c|c|c|}
    \hline + & a & b & c \\
    \hline a & a & a & a \\
    \hline b & a & c & ? \\
    \hline c & a & ? & ? \\
    \hline
    \end{array}##
    Now i have to use the information from the first table that ##c*c = b##. I can use that by seeing that ##b+b = c## and than i can say:
    ##c*c = b##
    ##b+b = c##
    ##c*c + c*c = c##
    ##c*(c+c) = c##
    and i cant figure what this tells me. Other than this i don't see a clue in solving this. Could you provide a helpfull clue?
     
    Last edited: Jun 8, 2017
  2. jcsd
  3. Jun 8, 2017 #2

    Dick

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    I think there must be some errors in the question. The addition table is telling you that ##a+b=a=a+a##. Adding the additive inverse of ##a## to both sides tells you ##a=b##. ???
     
  4. Jun 8, 2017 #3
    Doesn't that also tell that ##a,b,c## all are additive identity ? so does not that means you put whatever you want in any box ?
     
  5. Jun 8, 2017 #4

    SammyS

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    What you are given:
    This is a finite FIELD.
    Three elements: a, b, c .
    Two binary operations: *, + .

    In my opinion:
    1. You can't assume that the * operation acts like multiplication.
    2. You can't assume that the + operation acts like addition.​

    In fact, it looks to be like the traditional roles are switched, with ##\ a\ ## being the zero element.
     
    Last edited: Jun 9, 2017
  6. Jun 9, 2017 #5

    rcgldr

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    I'm wondering if + and * have been mistakenly swapped for this question.
     
  7. Jun 9, 2017 #6

    Dick

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    Now that's a good point. They just say '+', not 'addition'. Sneaky.
     
    Last edited: Jun 9, 2017
  8. Jun 9, 2017 #7

    rcgldr

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  9. Jun 9, 2017 #8

    SammyS

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    As an Abstract Algebra exercise, it's not wise to make any assumptions regarding the roles of the two binary operations, without exploring what the given tables imply.

    It's clear that the ' + ' operation is not consistent with being the "addition" operation. This is apparent from the given fragment of the ' + ' table.

    Beyond this, OP (@diredragon) has demonstrated that the distributive law does not hold for ' * ' distributed over ' + ' .

    The problem can be solved by thinking more abstractly about the two operations. Thinking this way will be most helpful for OP and we are here to help OP.
     
    Last edited: Jun 9, 2017
  10. Jun 10, 2017 #9

    rcgldr

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    My impression is that the symbols have been swapped in error. It's not normal for such problems to change the accepted meaning of mathematical symbols. The problem could have used different symbols that don't have an implied meaning, or noted that the symbols used do not conform to the symbols as defined in mathematics, including abstract algebra. For example, what if a problem asked for the solution to a = 5 - 3, but swapped the meaning of = and - ?
     
    Last edited: Jun 10, 2017
  11. Jun 10, 2017 #10
    Neither operation is defined as usual addition or multiplication as their symbols might suggest. From the definition of a field and already filled table places the problem asks for the rest of the table to be filled based on that which was given.
    ##c*c=b##
    ##b+b=c##
    ##(c*c)+b=c##
    ##(c+b)*(c+b)=c##
    Shouldn't this hold from the definition of a field. + is here a secondary operations and the general ##a+(b*c) = a+b * a+c##
     
  12. Jun 10, 2017 #11

    rcgldr

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    But the symbols are defined in mathematics, including fields in general, that also apply to finite fields and abstract algebra with + meaning "addition" and ⋅ (or * or x) meaning "multiplication". The question calls it a finite field, and the term finite field defines meanings for the symbols + and ⋅ (or * or x), which the question violates. The question should have used symbols without defined meanings if part of the problem was to determine which table was addition and which was multiplication.

    https://en.wikipedia.org/wiki/Field_(mathematics)#Definition

    There are also classic rules about a single element of the field being the additive identity 'a', where a + b = b, and another single element of the field being the multiplicative identity 'm', where m ⋅ b = b. Wiki words this a bit different:

    https://en.wikipedia.org/wiki/Field_(mathematics)#Classic_definition

    In the OP's question, the second table for + violates the rule about a single additive identity for the field.

    Consider this example of a field with 4 elements that follows the rules of the classic definition:

    https://en.wikipedia.org/wiki/Field_(mathematics)#A_field_with_four_elements
     
    Last edited: Jun 10, 2017
  13. Jun 10, 2017 #12
    The question intended only to test the knowledge of what fields are and how their properties relate to the tables. I do not know the exact name of the tables but in my language they are something like ,,Kaeily tables". I may have mispronaunced the name. If you are confused about + in the second table just use whatever symbol you wish and the question remains the same. It does not mean addition. It means some operation defined by symbol +. Properties are to be drawn from the fact that its a finite field and the given data.
     
  14. Jun 10, 2017 #13

    SammyS

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    We call it a Cayley table .

    Yes.

    You may want to include parentheses to avoid any ambiguity.
    ##a+(b*c) = (a+b) * (a+c)##​

    You have identified the + operation as being the "secondary operation". The table for + shows that ## ({a, b, c}, +) ## does not form a Group, because of the behavior of ##\ a \,.\ ## It's not the identity for + and it does not have an inverse for the + operation.

    You should be able to determine the (only possible) identity element for + .

    Can you ?

    Added in Edit:
    Actually, for the general case of the distributive property, you should use other symbols, rather than ##\ a,\, b,\, c,.##

    ##x+(y*z) = (x+y) * (x+z)##​
     
    Last edited: Jun 10, 2017
  15. Jun 11, 2017 #14
    The only possible identity element for + is ##c## since ##b+b=c## so the table for + looks like:
    ##\begin{array}{|c|c|c|c|}
    \hline + & a & b & c \\
    \hline a & a & a & a \\
    \hline b & a & c & b \\
    \hline c & a & b & c \\
    \hline
    \end{array}##
    How would i complete the * table? I know that ##c## can't be the identity so that leaves ##a## or ##b##.
     
  16. Jun 11, 2017 #15

    Dick

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    You know ##a## is the 'zero' for the multiplication ##+##. What does that tell you?
     
  17. Jun 11, 2017 #16
    That it is the identity element for the ##*##?
    So in general, for finite field ##(A, *, +)## if some element ##x## obeys ##a+x = x## and ##x + a = x## than that element ##x## is the identity for the operation ##*##?
     
  18. Jun 11, 2017 #17

    Dick

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    Yes, prove that using the distributive property.
     
  19. Jun 11, 2017 #18
    Let ##(A, +, *)## be a finite field. If: ##(\exists t)## ##(\forall x \in A)## ##x*t = t*x = t## than ##y## is an identity element for operation ##+##.
    Proof:
    Using the distributive law:
    ##t*(x+y) = t##
    ##t = t*x+t*y##
    ##t = t + t##
    This is true only is ##t## is neutral element for the ##+##.
    Let's suppose that it is in fact the neutral element. Then:
    ##(x+t)*y = x*y + t*y = x*y + t = x*y##
    Is this valid? And id the second part necessary or is the first one enough? I kinda thought that the first was lacking as the same result could come up if ##t## is also the zero element of ##+## as well it being of ##*##.
     
  20. Jun 11, 2017 #19

    Dick

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    That's looks correct, except I think you have reversed the ##*## and ##+## back to their conventional meanings.
     
  21. Jun 11, 2017 #20
    I meant for it to be general with + the primary and * the secondary operation. It's more relatable this way as well :)
     
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