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Math Help - Quotient Rings in Galois Theory - Help

  1. #1
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    Quotient Rings in Galois Theory - Help

    Hi

    I have the proof of this statement:

    'Let f(x) be a member of K[x] where f(x) is non-zero and K is a field.
    Suppose that K[x]/K[x].f(x) is a field.
    Then f(x) is irreducible.'

    The proof begins with:
    'Assue f(x) is not irreducible.

    If f(x) is a unit implies K[x].f(x) = K[x]'
    Is this becasue f(x) is a unit, so it has an inverse so multiplying f(x) by it's inverse multiplied by another element in K[x] can create any element in K[x]? (Due to K[x] being closed under multiplication)
    ie: a(x)b(x)g(x) = a(x) where b(x) = inverse f(x)
    a(x) a member of K[x]

    The next bit is that the previous line implies
    ' K[x]/K[x].f(x) = {0'} the equivalence class of zero'

    This is saying that every element is equivalent to zero. Is this because if we have:
    a' = b' (the equivalence classes)
    then a(x)-b(x) is a member of K[x].f(x) = K[x]
    This is true for all pairs of elements a(x) and b(x) in K[x] so all are equivalent to 0.
    ie: a(x) - 0 is a member of K[x]
    so a' = 0' for all a'

    Thanks for your help
    I know there is a lot of info here, but it's been bothering me for ages. I can't get my head around quotient rings.
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  2. #2
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    Quote Originally Posted by alanaj5 View Post
    Hi

    I have the proof of this statement:

    'Let f(x) be a member of K[x] where f(x) is non-zero and K is a field.
    Suppose that K[x]/K[x].f(x) is a field.
    Then f(x) is irreducible.'
    If f was reducible then we can write f=gh where \deg g,\deg h < \deg f. Then [g],[h]\not = [0] since f\not | g,f\not |h but [gh] = [f] = [0]. Thus, K[x]/K[x](f) is not an integral domain and so cannot be a field.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    If f was reducible then we can write f=gh where \deg g,\deg h < \deg f. Then [g],[h]\not = [0] since f\not | g,f\not |h but [gh] = [f] = [0]. Thus, K[x]/K[x](f) is not an integral domain and so cannot be a field.

    I get why f=gh and that neither g or h is a unit.
    But I don't understand why we know f\not | g,f\not |h and why this gives that [g],[h]\not = [0] and why [gh] = [f] = [0]
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  4. #4
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    Quote Originally Posted by alanaj5 View Post
    I get why f=gh and that neither g or h is a unit.
    But I don't understand why we know f\not | g,f\not |h and why this gives that [g],[h]\not = [0] and why [gh] = [f] = [0]
    Remember what [g] means it is the set \{ g + fq |q\in K[x] \}. Thus, if g_1\equiv g_2(\bmod f) then [g_1] = [g_2]. We cannot have that f|g because the degree of g is less than the degree of f. Therefore, f\not |g and so it means [g]\not =[0] because otherwise if [g]=[0] then g\equiv 0(\bmod f) \implies f|g - a contradiction.
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  5. #5
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    Thanks, I understand now. I hadn't thought about the quotient ring as modf before. It makes it so much easier.
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