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Math Help - Irreducibilty in F[x]

  1. #1
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    Irreducibilty in F[x]

    Let f:[0,1]→R be defined by f(x)=x, show that the polynomial x^2+f in T[x] is irreducible, where T is the ring of continuous real valued function on [0,1].

    For irreduciblity do i just show that this does not have any roots. by evaluating the equation x^2 + f, at 0 and 1?
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  2. #2
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    Quote Originally Posted by empressA88 View Post
    Let f:[0,1]→R be defined by f(x)=x, show that the polynomial x^2+f in T[x] is irreducible, where T is the ring of continuous real valued function on [0,1].

    For irreduciblity do i just show that this does not have any roots. by evaluating the equation x^2 + f, at 0 and 1?
    the question is kind of strange because the continiuity condition is useless here. you need to prove that no real-valued (continuous) function g on the interval [0,1] exists such that

    (g(x))^2+x=0, for all x \in [0,1], which is trivially true.
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    (g(x))^2+x=0, for all x \in [0,1],

    so once i prove this equation, it should cover it?
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    Quote Originally Posted by empressA88 View Post
    (g(x))^2+x=0, for all x \in [0,1],

    so once i prove this equation, it should cover it?

    Of course: since the polynomial x^2+f has degree \leq 3 , it reducible iff it has a root in T\iff \,\,\exists g\in T s.t. g^2+f=0\in T[x]\iff g^2(t)+t=0\,\,\,\forall\,t\in [0,1] , since the zero element in the ring T is the zero map...and this is what NCA told you is trivially true.

    Tonio
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    Quote Originally Posted by empressA88 View Post
    (g(x))^2+x=0, for all x \in [0,1],

    so once i prove this equation, it should cover it?
    yes, and there's nothing to prove! just put x = 1, or any positive number, to see that the equality cannot hold.
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  6. #6
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    ok..thank you
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