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Isomorphic polynomial rings

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data

    I am required to prove that [tex] F_5[x]/(x^2 + 2) isomorphic to F_5[x]/(x^2 + 3) [/tex]

    now I have the solution in front of me so I more or less know what's going on, however there are some points of confusion......

    ....the solution states that [tex] x \rightarrow 2x [/tex] will define the desired isomorphism. The next line asserts that

    [tex] x^2 \rightarrow (2x)^2 + 2, 4x^2 + 2= -(x^2 -2) = -(x^2 +3) [/tex]

    [tex] x^2 = -2=3 [/tex] ............?????:confused:

    ..what is going on here surely [tex] x^2 \rightarrow 4x^2 [/tex]

    since [tex] U(x^2) = U(x)U(x)= 2x2x = 4x^2 [\tex]

    anyway this is not my only problem with the solution I have, it then goes on to assert that indeed the two rings are isomorphic and further that,

    U(a + bx) = a + 2bx is such an isomorphism

    The proof of this says

    U((a+bx)(c+dx)) = U(ac + adx bcx +bdx^2)= U(ac + 3bd +(ad + bc)x) = ac +3bd +2(ad+bc)x = *1

    U(a +bx)U(c+dx) = (a +2bx)(c+2dx) = ac + 2adx + 2bcx + 4bdx^2
    = ac+ 3bd + 2(ad + bc)x = *1

    fantastic! execept look at term 4 2 lines up, 4bdx^2 = 12bd=2bdmod 5 and not =3mod5.....:confused:

    so what am i missing here?
    Last edited: Mar 22, 2007
  2. jcsd
  3. Mar 24, 2007 #2
    I don't quite understand your first problem, but for your second problem you should set [tex]x^2=-3=2(mod 5)[/tex].
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