Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show that $1,\alpha,\alpha^2/2$ is an integral basis of $R=\mathcal{O}\cap \mathbb{Q}[\alpha]$. ($\mathcal{O}$ is ring of algebraic integers)

$\alpha$ is a root of $f(x)=x^3+2x^2+4$ which is irreducible in $\mathbb{Q}(x).$ $\alpha^2$ is a root of $g(x)=x^3-4x^2-16x-16$ which is irreducible in $\mathbb{Q}(x).$

I found the discriminant as $disc(\alpha)=-16.5.7$

Also, I know $\alpha^2/2 \in R$ but $\alpha^2/4 \notin R$ and if $\frac{a+b\alpha}{2}\in R$ with $a,b\in \mathbb{Z}$ then a and b are both even.

But, then how to show that $1,\alpha,\alpha^2/2$ is an integral basis of $R$.

share|improve this question
    
Well, that is basis of the vector space $\;\Bbb Q(\alpha)/\Bbb Q\;$ , and all its elements are integral, so...unless you're expecting something else. –  DonAntonio Dec 21 '13 at 17:33
    
could you please explain : that is basis of the vector space $\mathbb{Q}(\alpha)/\mathbb{Q}$ –  104078 Dec 21 '13 at 17:37
1  
If $\;\alpha\;$ is a root of a rational irreducible polynomial of degree $\;n\;$, the vector space (field extension) $\;\Bbb Q(\alpha)\;$ over $\;\Bbb Q\;$ has a basis $\;\{1,\alpha,\alpha^2,...,\alpha^{n-1}\}\;$ , and in your case this happens with $\;\{1,\alpha,\alpha^2\}\;$ . Now just check that multiplying the third element by $\;\frac12\;$ doesn't change their being linear independent... –  DonAntonio Dec 21 '13 at 17:40
    
as in this case $1,\alpha,\alpha^2$ is a $\mathbb{Q}$-basis. I know that $\alpha^2/2\in R$, also as you mentioned multiplying the third element by $\frac{1}{2}$ keeps the lin-independcy. so, $1,\alpha,\alpha^2/2$ is an integral basis. I mean the fact that $\alpha^2/2\in R$ is necessary, right? –  104078 Dec 21 '13 at 17:46
    
@DonAntonio I am sorry, but I don't see how this is a proof for $\{1,\alpha,\alpha^2/2\}$ being an integral basis. The fact that it is a $\Bbb Q$-Basis of $\Bbb Q(\alpha)$ does not show that it is a $\Bbb Z$-Basis of $R$. For example $\{1,\sqrt{5}\}$ is a $\Bbb Q$-basis of $K=\Bbb Q(\sqrt{5})$ but it is not a $\Bbb Z$-basis for $\mathcal O_K = \Bbb Z[\frac{1+\sqrt{5}}{2}]$ as required for being an integral basis. –  benh Dec 21 '13 at 23:08

1 Answer 1

Let $K/\Bbb Q$ be a finite field extension and $R$ be the ring of integers of $K$. In order to show that a given $\Bbb Q$-Basis $(\alpha_1, ...,\alpha_n)$ of $K$ whose elements are already in $R$ is an integral basis, there are a couple of useful tricks, saving us from exhausting calculation of traces: Let $d_K$ be the discriminant of $K/\Bbb Q$ Then:

  • Lemma 1: $d(\alpha_1,...,\alpha_n)$ = $c^2 d_K$ for some $c\in \Bbb Z$.

  • Lemma 2: if $d(\alpha_1,...,\alpha_n)=d_K$, then $(\alpha_1,...,\alpha_n)$ is an integral basis.

  • Lemma 3: $d_K \equiv 0,1 \mod 4$

  • Lemma 4: If $\sigma_1,...\sigma_n$ are the homomorphisms $K\rightarrow \Bbb C$, then $d(\alpha_1,...,\alpha_n) = \det((\sigma_i \alpha_j)_{i,j})^2$

So let's consider the basis you gave: $(1,\alpha,\alpha^2/2)$ is a $\Bbb Q$-Basis that is already in $R$. Fine. We know that $d(1,\alpha,\alpha^2)$ coincides with the discriminant $\Delta f=2^4\cdot 5\cdot 7$ of $f$. Also, by Lemma 4 the discriminant has some linearity in its arguments, more precisely: $$d(1,\alpha,\frac{\alpha^2}{2})=\frac{1}{4}d(1,\alpha,\alpha^2)=\frac{1}{4}\Delta f=2^2\cdot 5 \cdot 7.$$ By Lemma 1 we have $d_K=2^2 \cdot 5\cdot 7$ or $d_K = 5\cdot 7$. But the last is a contradiction to Lemma 3, so $$d_K=2^2 \cdot 5\cdot 7 = d(1,\alpha,\frac{\alpha^2}{2}).$$ By Lemma 2, we conclude that $(1,\alpha,\alpha^2/2)$ is an integral basis, which completes the proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.