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As the title says, given that $\theta^3 + 11\theta - 4 = 0$, I'm trying to prove that $\frac{-\theta + \theta^2}{2}$ is an algebraic integer in $K = \mathbb{Q}(\theta)$.

I know that $x^3 + 11x -4$ is irreducible in $\mathbb{Q}[x]$ since it's irreducible in $\mathbb{F}_3[x]$. I also know that the set of algebraic integers forms an integral domain and thus I know that $-\theta + \theta^2$ is an algebraic integer, unfortunately that's the best I can do with that method since $\frac{1}{2}$ is specifically not an algebraic integer.

Clearly I need to somehow use the polynomial to solve this, but I can't see how, can anyone point me in the right direction? Thanks.

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3 Answers 3

up vote 7 down vote accepted

The minimal polynomial of $(\theta^2-\theta)/2$ is ${z}^{3}+11\,{z}^{2}+36\,z+4$.

One way to get this is: if $t = (\theta^2-\theta)/2$, express $t^3 + b t^2 + c t + d$ as a rational linear combination of $1$, $\theta$ and $\theta^2$, and solve the system of equations that say that the coefficients of $1$, $\theta$ and $\theta^2$ are all $0$.

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Thank you for the help Robert. However I would like to check my understanding: The reason its min poly couldn't be a quartic or higher is because $[\mathbb{Q}(\theta):\mathbb{Q}]=3$, and since $\theta$ and $\frac{\theta^2 -\theta}{2}$ can each be constructed from the other in $\mathbb{Q}(\theta)$, this means it couldn't be quadratic or lower. Thus its min poly had to be a cubic. Thanks. –  Blood Pudding Jul 27 '12 at 22:38
    
It is a cubic, but I don't think you need to know that in advance. If it was a quadratic, you'd have a solution where $d=0$. –  Robert Israel Jul 27 '12 at 23:43

This demonstrates what Robert Israel suggests.

Suppose that $$ \theta^3+11\theta-4=0 $$ and $$ \alpha=\frac{\theta^2-\theta}{2} $$ Then $$ \begin{align} \alpha^0&=\frac22\\ \alpha^1&=\frac{\theta^2-\theta}{2}\\ \alpha^2&=\frac{-5\theta^2+13\theta-4}{2}\\ \alpha^3&=\frac{19\theta^2-107\theta+36}{2} \end{align} $$ and $$ \begin{bmatrix}36&-107&19\end{bmatrix} \begin{bmatrix} 2&0&0\\ 0&-1&1\\ -4&13&-5 \end{bmatrix}^{-1} =\begin{bmatrix}-4&-36&-11\end{bmatrix} $$ Therefore, $$ \alpha^3+11\alpha^2+36\alpha+4=0 $$ and $\alpha$ is an algebraic integer.

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Thanks for the elaboration. –  Blood Pudding Jul 27 '12 at 22:30

Also, since $\theta$ is an algebraic integer, to show $\theta(\theta-1)/2$ is an algebraic integer, it suffices to show that it is $2$-adically integral.

Thus, the plan is to prove that either $2|\theta$ or $2|(\theta-1)$, in $\mathbb Z_2$.

Hensel's lemma shows that $x^3+11x-4=0$ has solutions $1,3$ mod $4$, and that both these give solutions in $\mathbb Z_2$. Thus, since the thing is a cubic, the third root is also in $\mathbb Z_2$, and (by looking at the constant term) is divisible by $4$, in fact. The solutions $\theta_1,\theta_2$ in $\mathbb Z_2$ congruent to $1,3$ mod $4$ both have the property that $2|(\theta_j-1)$, as desired.

True, this did not determine the minimal polynomial of $\theta(\theta-1)/2$.

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