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Basis of a Quotient Ring

  1. Jan 7, 2012 #1

    BVM

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    In my Abstract Algebra course, it was said that if
    [tex] E := \frac{\mathbb{Z}_{3}[X]}{(X^2 + X + 2)}.
    [/tex]
    The basis of E over [itex]\mathbb{Z}_{3}[/itex] is equal to [itex][1,\bar{X}][/itex].
    But this, honestly, doesn't really make sense to me. Why should [itex]\bar{X}[/itex] be in the basis without it containing any other [itex]\bar{X}^n[/itex]? How did they arrive at that exact basis?
     
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  3. Jan 7, 2012 #2

    micromass

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    Can you write [itex]\overline{X}^2[/itex] in terms of [itex]\overline{X}[/itex] and 1??
     
  4. Jan 8, 2012 #3

    BVM

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    Thank you for replying.

    I've solved the problem. Whereas I previously thought that I couldn't write any [itex]\bar{X}^n[/itex] in terms of 1 and [itex]\bar{X}[/itex] I've since realised that for instance: [itex]\bar{X}^2 = -\bar{X}-\bar{2} = 2\bar{X}+\bar{1}[/itex].

    My initial mistake as to think that in [itex]\mathbb{Z}_3[/itex] we can't define the negativity resulting in subtracting that polynomal, but obviously you can just add any 3n to it.
     
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