Gauss' Lemma and algebraic integers

Oct 2008
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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.
 

NonCommAlg

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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:
May 2010
95
38
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 \(\displaystyle \beta \in \mathbb{Z}\).
 

NonCommAlg

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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 \(\displaystyle f(a)=g(a)=0.\)
 

Bruno J.

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Jun 2009
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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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chiph588@

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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?
 

Bruno J.

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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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NonCommAlg

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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 \(\displaystyle \gcd(a,b)=1.\)
 
May 2010
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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.

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.
 

Bruno J.

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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!