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Some basic question about a quotient ring

  1. May 10, 2012 #1
    It's been awhile since I studied ring theory but here's a question about it:

    Let R = C[x1, x2, x3, x4, x5, x6, x7, x8] be a complex polynomial ring in 8 variables.

    Let

    f1 = x1 x3 +x5 x7 and
    f2 = x2 x4 +x6 x8.

    How do [itex]\bar{f}[/itex]1, [itex]\bar{f}[/itex]2 in (f1,f2)R/(x1,x2)R look like?


    Is it
    [itex]\bar{f}[/itex]1 = x5 x7 + (x1,x2)

    [itex]\bar{f}[/itex]2 = x6 x8 + (x1,x2)

    or is it

    [itex]\bar{f}[/itex]1 = x1 x3 + x5 x7 +(x1,x2)

    [itex]\bar{f}[/itex]2 = x2 x4 + x6 x8 + (x1,x2)?
     
    Last edited: May 10, 2012
  2. jcsd
  3. May 10, 2012 #2
    Since

    x1*x3 [itex]\in [/itex] (x1, x2),

    x1 x3+ x5 x7 ~ x1 x3 mod (x1, x2).

    So the coset x1 x3+ x5 x7 + (x1, x2) = x5 x7 + (x1, x2)?
     
  4. May 10, 2012 #3

    What is that?? What do you mean by [itex]\,\,(f_1,f_2)R/(x_1,x_2)R\,\,[/itex]? The denominator clearly is

    an ideal, but what is the numerator?





    DonAntonio
     
  5. May 10, 2012 #4
    > (f1,f2)R/(x1,x2)R?

    The numerator (f1, f2) is an ideal in R generated by f1 and f2 while the denominator (x1, x2) is an ideal in R generated by x1 and x2.
     
  6. May 10, 2012 #5
    So what I'm interested in is: how do the images of f1 and f2 look like in (f1,f2)R/(x1,x2)R?
     
  7. May 10, 2012 #6
    It seems to me that in (f1,f2)R/(x1,x2)R, we are killing off all polynomial combination of x1 and x2 (all polys of the form ax1 + bx2 where a and b are in R), or are we killing off something smaller than that?


    Thus, do

    [itex]\bar{f}[/itex]1 = x5 x7


    [itex]\bar{f}[/itex]2 = x6 x8

    in (f1,f2)R/(x1,x2)R ?
     
  8. May 10, 2012 #7

    mathwonk

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    Science Advisor
    Homework Helper

    yes. that quotient ring just cancels all terms containing x1 or x2.
     
  9. May 12, 2012 #8
    So (f1,f2)R/(x1,x2)R is generated by x5 x7 and x6 x8, right?

    I.e., (f1,f2)R/(x1,x2)R = (x5 x7, x6 x8)
     
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