# Non monogenic extension and computation of minimal polynomial

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
• Where did you take this exercise from? Jun 19, 2021 at 0:37
• My professor give to me the exercice, i don't know where he takes the exercices honestly. Why?
– 3m0o
Jun 19, 2021 at 11:57
• 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 $\alpha = \frac{1+ \sqrt{d}}{2}$.
– 3m0o
Jun 19, 2021 at 13:56

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 $$\text{disc } K=-503$$.