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Math Help - polynomial rings

  1. #1
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    polynomial rings

    Let f(x) be a polynomial in \mathbb{Z}[x]. Prove that if f(x) has a root \alpha \in \mathbb{Z}, then f(x) has a linear factor in \mathbb{Z}[x].

    Suppose f(x) has a root \alpha \in \mathbb{Z}, then f(x) is reducible, and then f(x) = a(x)(x - \alpha) for some nonconstant polynomial a(x) \in \mathbb{Z}[x]. So x - \alpha is a linear factor in \mathbb{Z}[x].

    Is this correct? Did I miss anything important in this proof?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by dori1123 View Post
    Let f(x) be a polynomial in \mathbb{Z}[x]. Prove that if f(x) has a root \alpha \in \mathbb{Z}, then f(x) has a linear factor in \mathbb{Z}[x].

    Suppose f(x) has a root \alpha \in \mathbb{Z}, then f(x) is reducible, and then f(x) = a(x)(x - \alpha) for some nonconstant polynomial a(x) \in \mathbb{Z}[x]. So x - \alpha is a linear factor in \mathbb{Z}[x].

    Is this correct? Did I miss anything important in this proof?
    there is no assumption on the degree of f so you cannot say that a(x) is non constant..

    well, this is a direct consequence of the division algorithm. since you think that x-\alpha is the linear factor, then try writing f(x)=g(x)(x-\alpha) + R where R is a non zero constant or R=0.
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    Quote Originally Posted by kalagota View Post
    there is no assumption on the degree of f so you cannot say that a(x) is non constant..

    well, this is a direct consequence of the division algorithm. since you think that x-\alpha is the linear factor, then try writing f(x)=g(x)(x-\alpha) + R where R is a non zero constant or R=0.
    But \mathbb{Z}[x] is not a field, there's no Euclidean Algorithm...
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    Quote Originally Posted by dori1123 View Post
    But \mathbb{Z}[x] is not a field, there's no Euclidean Algorithm...
    The division algorithm does work for non-fields too, \mathbb{Z}[x] is an example.
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  5. #5
    MHF Contributor kalagota's Avatar
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    another example is \mathbb{Z} itself..
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    Suppose f(x) \in \mathbb{Z}[x] has a root \alpha \in \mathbb{Z}. By the division algorithm, f(x) = q(x)(x-\alpha)+r for some polynomial q(x) \in \mathbb{Z}[x] and for some r \in \mathbb{Z}. Since f(\alpha)=0, we have r=0. So f(x)=q(x)(x-\alpha) and so x-\alpha is a linear factor of f(x) \in \mathbb{Z}[x]. Is this correct?

    Also, for the converse to be true, if f(x) \in \mathbb{Z}[x] has a linear factor, then the linear factor must be a monic polynomial to guarantee that f(x) has a root in \mathbb{Z}. Right? That's the only condition I can think of.
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  7. #7
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    Quote Originally Posted by dori1123 View Post

    Also, for the converse to be true, if f(x) \in \mathbb{Z}[x] has a linear factor, then the linear factor must be a monic polynomial to guarantee that f(x) has a root in \mathbb{Z}. Right? That's the only condition I can think of.
    If f(x) = (x-a)g(x) then f(a) = 0, so a is a zero.
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    Quote Originally Posted by ThePerfectHacker View Post
    If f(x) = (x-a)g(x) then f(a) = 0, so a is a zero.
    I understand that the converse is true if f(a)=0 but why is a zero?
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    Quote Originally Posted by dori1123 View Post
    I understand that the converse is true if f(a)=0 but why is a zero?
    No, a is not zero, a is a zero of f. This means f(a) = 0.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    No, a is not zero, a is a zero of f. This means f(a) = 0.
    Oh, you mean a root. Sorry, I got used to say roots.
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