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Math Help - Gauss' Lemma and algebraic integers

  1. #1
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    Gauss' Lemma and algebraic integers

    Let a be an algebraic integer and suppose that f(a)=0 for a polynomial f, with f in Q[x] (a polynomial with rational coefficients). Suppose f(x) is irreducible and monic. Show that f(x) is in Z[x].

    I feel kind of stupid... I know that Gauss' lemma reduces problems in Q[x] to the integers but I don't see exactly how to draw this conclusion.
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    Quote Originally Posted by robeuler View Post
    Let a be an algebraic integer and suppose that f(a)=0 for a polynomial f, with f in Q[x] (a polynomial with rational coefficients). Suppose f(x) is irreducible and monic. Show that f(x) is in Z[x].

    I feel kind of stupid... I know that Gauss' lemma reduces problems in Q[x] to the integers but I don't see exactly how to draw this conclusion.
    since f(x) is irreducible, every root a_j of f(x) (conjugates of a) is also an algebraic integer. now recall that we can write f(x)=x^n - s_1x^{n-1} + s_2x^{n-2} - \cdots + (-1)^n s_n, where

    s_1=\sum_{j=1}^n a_j, \ s_2 = \sum_{1 \leq i < j \leq n} a_ia_j , \ \cdots , s_n=a_1a_2 \cdots a_n. thus each s_j is an algebraic integer. we also have s_j \in \mathbb{Q} and so s_j \in \mathbb{Z}, because a rational algebraic integer is an integer.

    one thing, you should've posted this question in number theory subforum.
    Last edited by NonCommAlg; May 24th 2010 at 05:37 PM.
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    Quote Originally Posted by robeuler View Post
    Let a be an algebraic integer and suppose that f(a)=0 for a polynomial f, with f in Q[x] (a polynomial with rational coefficients). Suppose f(x) is irreducible and monic. Show that f(x) is in Z[x].

    I feel kind of stupid... I know that Gauss' lemma reduces problems in Q[x] to the integers but I don't see exactly how to draw this conclusion.
    This is another approach directly from the definition of an algebraic integer.

    By definition, an element \alpha \in K is called an algebraic integer if \alpha is the root of some monic polynomial with coefficients in \mathbb{Z}, where K is an extension field of \mathbb{Q}. Let g(x) \in \mathbb{Z}[x] be such a monic polynomial with g(\alpha)=0.

    If f(\alpha)=0, then f(x)=g(x)h(x), where f(x), h(x) \in \mathbb{Q}[x] and g(x) \in \mathbb{Z}[x]. If f(x) is irreducible in \mathbb{Q}[x], then g(x) \in \mathbb{Z}[x] should also be irreducible in \mathbb{Q}[x] and h(x) should be a unit in \mathbb{Q}[x], i.e., a unit in \mathbb{Q}. Further, if f(x) is monic, then f(x)=g(x). Thus f(x) \in \mathbb{Z}[x].


    ----------(Some additional remarks)-----------
    A corollary from the above definition is that the algebraic integers in \mathbb{Q} are integers \mathbb{Z}. For instance, if \beta is an algebraic integer in \mathbb{Q}, then the minimal polynomial of \beta=a/b \in \mathbb{Q}, where a and b are integers and b \neq 0, is bx - a . Since \beta is an algebraic integer by hypothesis, \beta should be the root of a monic polynomial with coefficients in \mathbb{Z}. Thus b is 1 and \beta \in \mathbb{Z}.
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    Quote Originally Posted by TheArtofSymmetry View Post

    If f(\alpha)=0, then f(x)=g(x)h(x), where f(x), h(x) \in \mathbb{Q}[x] and g(x) \in \mathbb{Z}[x].
    that is not true. what we can say is this: g(x)=f(x)h(x) and that's because f is irreducible and f(a)=g(a)=0.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    A corollary from the above definition is that the algebraic integers in \mathbb{Q} are integers \mathbb{Z}. For instance, if \beta is an algebraic integer in \mathbb{Q}, then the minimal polynomial of \beta=a/b \in \mathbb{Q}, where a and b are integers and b \neq 0, is bx - a . Since \beta is an algebraic integer by hypothesis, \beta should be the root of a monic polynomial with coefficients in \mathbb{Z}. Thus b is 1 and \beta \in \mathbb{Z}.
    It's true that \beta should be the root of some monic polynomial with integer coefficients, but why should bx-a have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having \beta as a root" is far from saying "all polynomials with integer coefficients having \beta as a root are monic". The second statement is not true!
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    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bruno J. View Post
    It's true that \beta should be the root of some monic polynomial with integer coefficients, but why should bx-a have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having \beta as a root" is far from saying "all polynomials with integer coefficients having \beta as a root are monic". The second statement is not true!
    I can't for the life of me remember why  \frac ab is not an algebraic integer for  b\geq2 . Can anyone say why?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    I can't for the life of me remember why  \frac ab is not an algebraic integer for  b\geq2 . Can anyone say why?
    Because of Gauss's lemma : we know that if f(x) = g(x)h(x) \in \mathbb{Z}[x] is monic, and g,h \in \mathbb{Q}[x], then in fact g,h \in \mathbb{Z}[x]. So if f has the rational root \alpha, we can write f(x)=(x-\alpha)q(x); we have x-\alpha, q(x) \in \mathbb{Q}[x] and therefore x-\alpha \in \mathbb{Z}[x] \Rightarrow \alpha \in \mathbb{Z}. (It might not be immediately obvious that q(x) has rational coefficients, but think about it!)
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    Quote Originally Posted by chiph588@ View Post
    I can't for the life of me remember why  \frac ab is not an algebraic integer for  b\geq2 . Can anyone say why?
    you can see it directly: suppose r = \frac{a}{b}, \ a \in \mathbb{Z}, \ b \in \mathbb{N}, \ \gcd(a,b)=1, is an algebraic integer. then, by definition, there exists p(x)=x^n + c_1x^{n-1} + \cdots + c_n \in \mathbb{Z}[x] such that p(r)=0.

    thus a^n + bc_1a^{n-1} + \cdots + b^na_n=b^n p(r)=0 and so b \mid a^n, which is impossible unless b=1, because \gcd(a,b)=1.
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    Quote Originally Posted by NonCommAlg View Post
    that is not true. what we can say is this: g(x)=f(x)h(x) and that's because f is irreducible and f(a)=g(a)=0.
    No, this is true.

    Irreducibility of f forces the choice of g to be irreducible in Q[x], and g(x) in Z[x] is a monic polynomial satisfying g(a)=0, where a is an algebraic integer.

    g should always exists for sure in this case, otherwise a is not an algebraic integer.

    Quote Originally Posted by Bruno J. View Post
    It's true that \beta should be the root of some monic polynomial with integer coefficients, but why should bx-a have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having \beta as a root" is far from saying "all polynomials with integer coefficients having \beta as a root are monic". The second statement is not true!
    Because bx-a is the minimal polynomial of \beta and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by TheArtofSymmetry View Post
    Because bx-a is the minimal polynomial of \beta and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure.
    It's true that the minimal polynomial over \mathbb{Q} of an algebraic integer has integer coefficients - but this is a consequence of Gauss's lemma. It's not trivial!
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