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Multiplication tables of rings

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data
    construct a multiplication table for the ring Z[tex]_{3}[/tex][[tex]\alpha[/tex]], [tex]\alpha[/tex][tex]^{2}[/tex] + 1(bar) = 0(bar)


    2. Relevant equations



    3. The attempt at a solution
    I'm actually confused on how to find the elements of the ring. My book and notes have thrown me off a bit and I can't find them. Hint: there are 9 elements
     
  2. jcsd
  3. Oct 16, 2008 #2
    I think for advanced courses like yours it might help me and others if you remind us of what things like (bar) mean

    and do you mean by Z^3 actually Z_3 or ZxZxZ or something completely different outside my knowledge?
     
  4. Oct 16, 2008 #3
    yea sorry the text editor on here sucks....it is supposed to be Z_3 (which is what i put) and (bar) refers to being part of that equivalence class so 1 bar is ...-2,1,4,7... and 0 bar is ...-6,-3,0,3,6... for the integer set Z_3 so the congruent mod 3.
     
  5. Oct 17, 2008 #4

    HallsofIvy

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    Then, strictly speaking, each member of Z3[/sup] is an equivalence class, each class containing exactly one non-negative integer less than 3. That is, there is an equivalence class containing 0, and equivalence class containing 1, and an equivalence class containing 2. That is, I presume, what you mean by [itex]\bar{0}[/itex], [itex]\bar{1}[/itex], and [itex]\bar{2}[/itex]. Since that does NOT have 9 members. Am I to assume that you mean the set of all numbers of the form [itex]a\alpha+ b[/itex] where a and b are in Z3 and [itex]\alpha[/itex] satisfies the equation above? That has nine members: 0, 1, 2, [itex]\alpha[/itex], [itex]\alpha+ 1[/itex], [itex]\alpha+ 2[/itex], [itex]2\alpha[/itex], [itex]2\alpha+ 1[/itex], and [itex]2\alpha+ 2[/itex].

    Set up your table so it has those both along the top and verticaclly on the left. Multiply each of the 81 pairs and reduce to one of those 9 forms by using the equation [itex]\alpha[/itex] satisfies. For example, [itex]\left(\alpha+ 1\right)\left(2\alpha+ 1\right)= 2\alpha^2+ 3\alpha+ 1[/itex]. Since [itex]\alpha^2+ 1= 0[/itex], [itex]\alpha^2= -1[/itex]. Of course, 3 is equivalent to 0 mod 3 so this reduces to [itex]-2+ 1= -1[/itex] which is 2 mod 3:[itex]\left(\alpha+ 1\right)\left(2\alpha+ 1\right)= 2[/itex].

    No, the LaTex editor here does not "suck" but it is a bad idea to try to combine both LaTex and non-LaTex in the same formula: use LaTex for the entire formula.
     
  6. Oct 17, 2008 #5
    awesome thanks
     
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