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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 $R$.

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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
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

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.

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