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Thread: 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 $\displaystyle f(x)$ is irreducible, every root $\displaystyle a_j$ of $\displaystyle f(x)$ (conjugates of $\displaystyle a$) is also an algebraic integer. now recall that we can write $\displaystyle f(x)=x^n - s_1x^{n-1} + s_2x^{n-2} - \cdots + (-1)^n s_n,$ where

    $\displaystyle s_1=\sum_{j=1}^n a_j, \ s_2 = \sum_{1 \leq i < j \leq n} a_ia_j , \ \cdots , s_n=a_1a_2 \cdots a_n.$ thus each $\displaystyle s_j$ is an algebraic integer. we also have $\displaystyle s_j \in \mathbb{Q}$ and so $\displaystyle s_j \in \mathbb{Z},$ 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 04:37 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 $\displaystyle \alpha \in K$ is called an algebraic integer if $\displaystyle \alpha$ is the root of some monic polynomial with coefficients in $\displaystyle \mathbb{Z}$, where K is an extension field of $\displaystyle \mathbb{Q}$. Let $\displaystyle g(x) \in \mathbb{Z}[x]$ be such a monic polynomial with $\displaystyle g(\alpha)=0$.

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


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

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

    thus $\displaystyle a^n + bc_1a^{n-1} + \cdots + b^na_n=b^n p(r)=0$ and so $\displaystyle b \mid a^n,$ which is impossible unless $\displaystyle b=1$, because $\displaystyle \gcd(a,b)=1.$
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    Quote Originally Posted by NonCommAlg View Post
    that is not true. what we can say is this: $\displaystyle g(x)=f(x)h(x)$ and that's because $\displaystyle f$ is irreducible and $\displaystyle f(a)=g(a)=0.$
    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 $\displaystyle \beta$ should be the root of some monic polynomial with integer coefficients, but why should $\displaystyle bx-a$ have to be that polynomial? Saying "there exists a monic polynomial with integer coefficients having $\displaystyle \beta$ as a root" is far from saying "all polynomials with integer coefficients having $\displaystyle \beta$ as a root are monic". The second statement is not true!
    Because bx-a is the minimal polynomial of $\displaystyle \beta$ 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 $\displaystyle \beta$ and the minimal polynomial of an algebraic integer has integer coefficients and monic for sure.
    It's true that the minimal polynomial over $\displaystyle \mathbb{Q}$ of an algebraic integer has integer coefficients - but this is a consequence of Gauss's lemma. It's not trivial!
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