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Math Help - Values that make a field (need answer checked)

  1. #1
    Member elninio's Avatar
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    Values that make a field (need answer checked)

    I just need my answer checked for this particular problem.

    The question is:
    "For which values of a=1,2,3,4 is Z_5[x]/<x^2+a> a field? Show your work.

    My answer:
    We need to know when the polynomial is irreducible in Z_5.
    For this, we check to see if a is a perfect square in Z_5. The squares are:
    0=0 1=1 2=4 3=4 4=1
    So the values of a which arnt squares are 2 and 3, these being the values that give a field.

    Correct?
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  2. #2
    Member Haven's Avatar
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    Since x^2 +a is a quadratic polynomial, x^2 +a either has two linear factors or it's irreducible. If a polynomial f(x) has a linear factor (x-b) then f(b) = 0.

    So we check for every value of a, if and when x^2 + a = 0

    e.g. f(x) = x^2 + 1
    f(0) = 1, f(1) = 2, f(2) = 5 = 0. So f(x) is not irreducible.
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  3. #3
    Member elninio's Avatar
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    I see!
    So my method was wrong, but after rechecking, it seems 2 and 3 are indeed the correct values.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Your method was correct. You are both essentially doing the same thing - eliminating the values of a for which the polynomial has a root.
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