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Let $ K= \mathbb{Q}[\theta] $, with $\theta$ a root of $p(x)=x^3-x^2-2x-8$.

  1. Check that the polynomial $p$ is irreducible over $ \mathbb{Q}[x]$
  2. Knowing that $ \operatorname{disc}(1,\theta,\theta^2) = -4 \cdot 503$ prove that $ [ \mathcal{O}_K : \mathbb{Z}[\theta] ] = 2 $ and conclude that the discriminant of $ K$ is equal to $-503$ Hint: Check that $ \beta := \frac{\theta+\theta^2}{2} $ is an integer
  3. Find the norm of $ \theta$ and $ \theta+1$. Use this information to prove that $2 $ totally splits in $K$
  4. Conclude that $K$ is not monogenic.

For some points i have trouble, in point 2

  1. We have that if it is not irreducible then it has a in his decomposition at least one linear factor, i.e. $x-a$, where $a \mid 8$ thus checking directily that $\pm d$ where $d$ is a divisor of $8$ are not root we conclude that polynomial $ p$ is in fact irreducible.
  2. I know that the minimal polynomial of $\beta$ is $x^3-3x^2-10-8$ and solutions say that it is sufficient to calculate the charactheristic polynomial of the $\mathbb{Q}$-linear map $[ \times \beta] : K \to K$ has coefficient in $ \mathbb{Z}$, and i agree with that since minimal polynomial and characterstic polynomial are equals in this case since the extension field is separable and $K=\mathbb{Q}[\theta]=\mathbb{Q}[\beta] $. But i don't understand how to compute the characterstici polynomial. In fact it is given by $ \det (x I_3 - [\times \beta] )$ now since $\theta^3 = \theta^2 +2\theta+8 $ and $ \theta^4 = 3\theta^2 + 10 \theta + 8$ i get that $ \beta (x + y \theta + z \theta^2 ) = \frac{1}{2} \left((8y + 16z) + (x+2y+12z)\theta + (x+2y+4z)\theta^2 \right)$ thus the multiplication matrix is given by $ [\times \beta] = \begin{pmatrix} 0& 4 & 8\\ 1/2 & 1 &6 \\ 1/2 & 1 & 4 \end{pmatrix} $. And the characteristic polynomial should be $ \det \begin{pmatrix} x& -4 & -8\\ -1/2 & x-1 &-6 \\ -1/2 & -1 & x-4 \end{pmatrix} $ but the determinant is $x^3-5x^2-8x-4$ so i don't understand where i get wrong...

-Question 1: Where I did it wrong? How to compute the minimal polynomial?

Taking the fact that $ \beta$ is an integer we get that $ \operatorname{disc}(1,\theta,\beta) = ( \det M)^2 \operatorname{disc}(1,\theta,\theta^2) $, where $ M$ is the matrix of changing the $\mathbb{Q}$-basis, thus we have that $ M = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 1/2 \\ 0 & 0 & 1/2 \end{pmatrix}$ and thus we have that $ \operatorname{disc}(1,\theta,\beta)= \frac{1}{4} \cdot 4 \cdot (-503) = -503$, it is square free thus it is a $\mathbb{Z}$-basis of $ \mathcal{O}_K$, and thus the discriminant of $K$, i.e. the discriminant of an integral basis of $ \mathcal{O}_K$, it is actually $-503$, then since $(1,\theta,\theta^2) $ are $ \mathbb{Q}$-linear independent it's follow that

$ \operatorname{disc}(1,\theta,\theta^2) = [\mathcal{O}_K : \mathbb{Z}[\theta]]^2 \cdot \operatorname{disc}(K)$ and thus we have that $ [\mathcal{O}_K : \mathbb{Z}[\theta]]=2$

-Question 2: I deduced from the fact that $(1,\theta,\beta)$ is a $\mathbb{Z}$-basis of $ \mathcal{O}_K$ that the discriminant is $-503$ then from this that $[\mathcal{O}_K: \mathbb{Z}[\theta]]=2 $, it is correct? Since the exercice say to do the contrary but i don't see how to do it.

-Question 3: why it is not sufficient to say that $ K$ is not monogenic by the fact that $ [ \mathcal{O}_K : \mathbb{Z}[\theta]] =2 $ ?? Maybe because it could exsits another $ \alpha$ different from $\theta$ such that $ \mathcal{O}_K = \mathbb{Z}[\alpha] $ and $ K=\mathbb{Q}(\alpha) $ ?

  1. Since $p$ does not divide the discriminat it's follow that it is not ramiefied, thus since the extension degree of field is $3$ we get three possible case, since $ N(2 \mathcal{O}_K) = 8$ then we have:

a) $ 2 \mathcal{O}_K $ is prime

b)$ 2 \mathcal{O}_K = \mathfrak{P}_1 \mathfrak{P}_2$, with $ \mathfrak{P}_i $ primes

c) $ 2 \mathcal{O}_K = \mathfrak{P}_1 \mathfrak{P}_2 \mathfrak{P}_3$, with $ \mathfrak{P}_i $ primes.

We prove we are in case c). Suppose we are n case 1, then we calculate the norm $ N(\theta) = \left| \theta \mathcal{O}_K / \mathcal{O}_K \right| = \left| \mathbb{Z} / N_{K/\mathbb{Q}}( \theta) \mathbb{Z} \right| =\det ( [ \times \theta]) = 8 $ and similarly $ N(\theta-1) = \det( [ \times \theta])= 10$ thus we have that in the decomposition of $ (\theta -1)\mathcal{O}_K$ there is a prime of norm $2$, since $2 \mid 10$ we get that $2 \mathcal{O}_K $ is not prime, otherwise there is not a prime of norm $2$. Suppose that we are in case b), we get that since $ \theta - \theta +1 = 1$ then the two prime ideals are coprime thus wlogwma $ \mathfrak{P}_1 \mid (\theta-1) $, we have that $ (\theta) = \mathfrak{P}_2^e $, since $N(\theta)=8$ and thus there is only prime above $2$ in his prime decomposition. But we have also that $ N(\mathfrak{P}_2) N(\mathfrak{P}_1) = N(2 \mathcal{O}_K)= 8$ and $N(\mathfrak{P}_1) = 2$ so $ N(\mathfrak{P}_2)=4$ and this is not possible since we must have $ 8 = 4^e$ that has no solution. So we have that $2 $ totally splits in $ \mathcal{O}_K$.

  1. We have that $ K$ is of degree 3 and $2 < 3$ totally splits in $ \mathcal{O}_K$ thus since $ 2 \mid [ \mathcal{O}_K : \mathbb{Z}[\alpha] ] $ for any $ \alpha \in \mathcal{O}_K $ of degree $3$ the results follows
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    $\begingroup$ Where did you take this exercise from? $\endgroup$
    – russoo
    Jun 19, 2021 at 0:37
  • $\begingroup$ My professor give to me the exercice, i don't know where he takes the exercices honestly. Why? $\endgroup$
    – 3m0o
    Jun 19, 2021 at 11:57
  • $\begingroup$ Answer to question 1: I did a stupid error forgetting to multipliying by $1/2$ one of the term inside the matrix. Answer to question 3: Yes for example quadratic extension with $ d \equiv 1 \mod 4$ is monogenic but the $\alpha \neq \sqrt{d} $ in fact $ \alpha = \frac{1+ \sqrt{d}}{2} $. $\endgroup$
    – 3m0o
    Jun 19, 2021 at 13:56

1 Answer 1

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If I understand your comments in a correct way, you found the answers to question 1 and 3 by yourself. So, only question 2 is open. Here is an answer:

Your argument using the fact that $1, \theta, \beta$ is a $\mathbb{Z}$-basis is correct. However, there is also a way to solve the problem without knowing that $1, \theta, \beta$ is a $\mathbb{Z}$-basis and this is probably what the creator of the exercise had in mind. It goes like this: We already know that $$-4 \cdot 503=[\mathcal{O}_K:\mathbb{Z}[\theta]]^2 \cdot \text{disc }K$$ It follows that $[\mathcal{O}_K:\mathbb{Z}[\theta]]=1$ or $2$. From the hint, we know that $\beta \in \mathcal{O}_K$. At the same time, $\beta \notin \mathbb{Z}[\theta]$. Hence, $\mathcal{O}_K \neq \mathbb{Z}[\theta]$ and therefore $[\mathcal{O}_K:\mathbb{Z}[\theta]]=2$. It follows that $\text{disc } K=-503$.

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