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This question arose from the need to see the splitting behaviour of primes over an extension of number fields. One criterion is the Kummer's theorem, which gives the decomposition of the base prime in an extension, depending on a chosen integer that generates the ring of integers.

But my question is,

"How does one show in general that for finite extensions $L / K$, there exists an algebraic integer $\alpha \in L$, so that $K[\alpha] = \mathcal O_L$? "

A related question is,

If $L$ is the compositum of subfields $K_1$ and $K_2$ over $K$, then can $\mathcal O_L$ be described in terms of $\mathcal O_{K_1}$ and $\mathcal O_{K_2}$? You may assume any additional conditions or even special cases like quadratic and quartic extensions.

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2  
One doesn't show that in general because it's false in general (e.g. take $K = \mathbb{Q}, \alpha = \sqrt{-3}$.) –  Qiaochu Yuan Feb 23 '13 at 1:23
    
Qiaochu, I have edited the question. –  Abhishek Parab Feb 23 '13 at 1:27
    
That is also false in general. (I don't have a counterexample ready off the top of my head though.) –  Qiaochu Yuan Feb 23 '13 at 1:27
5  
Example 8.1.6 in Stein's A Brief Introduction to Classical and Adelic Number Theory gives a counterexample: the ring of integers in $\mathbb{Q}(\alpha)$ where $\alpha^3 + \alpha^2 - 2 \alpha + 8 = 0$ is not of the form $\mathbb{Z}[\beta]$ for any $\beta$. –  Qiaochu Yuan Feb 23 '13 at 1:30
3  
Dear Abhishek, you may find a partial answer to your second question in my answer here. –  Bruno Joyal Feb 23 '13 at 1:31
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1 Answer

If $K$ is a number field, and there exists $\alpha$ such that $\mathcal{O}_K=\mathbb{Z}[\alpha]$, then $K$ is called monogenic. The first example of a non-monogenic number field is due to Dedekind, who showed that for $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of the cubic polynomial $x^3-x^2-2x-8$, the ring of integers does not satisfy $\mathcal{O}_K=\mathbb{Z}[\delta]$ for any $\delta$. In what follows, I will provide a full proof of this claim.

Proof that $\mathcal{O}_K\neq \mathbb{Z}[\delta]$ for any $\delta$: First, we verify that $f$ is indeed irreducible over $\mathbb{Q}$. Since it is a cubic polynomial, if it were reducible it would have a rational root which is impossible by the rational root theorem. Let $\eta=\frac{\theta^{2}+\theta}{2}$, then by calculating the determinant and traces, note that $\eta^{3}-3\eta^{2}-10\eta-8=0$. The elements$1,\theta,\eta$ are independent over $\mathbb{Q}$ since $\theta$ does not satisfy a degree $2$ equation, and so $\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$ is a full rank subring of $\mathcal{O}_{K}$. I claim that $\mathcal{O}_{K}=\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$. To prove this, we calculate the discriminant.

Let $$ B=\left[\begin{array}{ccc} \text{Tr}_{K/\mathbb{Q}}\left(1\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta\right)\\ \text{Tr}_{K/\mathbb{Q}}\left(\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta^{2}\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta\eta\right)\\ \text{Tr}_{K/\mathbb{Q}}\left(\eta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta^{2}\right) \end{array}\right]. $$ Writing $\theta$ and $\eta$ in the basis $1,\theta,\theta^{2}$, we have that $$ M_{\eta}=\left[\begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2}\\ 4 & 1 & 1\\ 8 & 6 & 2 \end{array}\right],\ M_{\theta}=\left[\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 8 & 2 & 1 \end{array}\right], $$ and from this we find that $$ M_{\theta}=\left[\begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 8 & 2 & 1 \end{array}\right],\ M_{\theta}^{2}=\left[\begin{array}{ccc} 0 & 0 & 1\\ 8 & 2 & 1\\ 8 & 10 & 3 \end{array}\right],\ M_{\theta}M_{\eta}=\left[\begin{array}{ccc} 4 & 1 & 1\\ 8 & 6 & 2\\ 16 & 12 & 8 \end{array}\right],\ M_{\eta}^{2}=\left[\begin{array}{ccc} 6 & \frac{7}{2} & \frac{3}{2}\\ 12 & 9 & 5\\ 40 & 22 & 14 \end{array}\right]. $$ It follows that $\text{Tr}_{K/\mathbb{Q}}\left(1\right)=3$, $\text{Tr}_{K/\mathbb{Q}}\left(\theta\right)=1$, $\text{Tr}_{K/\mathbb{Q}}\left(\theta^{2}\right)=5$, $\text{Tr}_{K/\mathbb{Q}}\left(\eta\right)=3$, $\text{Tr}_{K/\mathbb{Q}}\left(\eta^{2}\right)=29$, $\text{Tr}_{K/\mathbb{Q}}\left(\theta\eta\right)=18$, and so $$ B=\left[\begin{array}{ccc} 3 & 1 & 3\\ 1 & 5 & 18\\ 3 & 18 & 29 \end{array}\right]. $$ Thus $$ \text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}\right)=\det B=-503. $$ As $503$ is a prime number, this implies that $\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$ equals the ring of integers, since otherwise we must have $\text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}\right)=m^{2}\text{disc}_{K/\mathbb{Q}}\left(\mathcal{O}_{K}\right)$ for some integer $m>1$.

Now, let $\delta=a+b\theta+c\eta$ be a general element of $\mathcal{O}_{K}$. Our goal is to show that $2|\text{disc}\left(\mathbb{Z}[\delta]\right)$. The matrix for $M_{\delta}$ in the basis $1,\theta,\eta$, equals

$$ M_{\delta}=\left[\begin{array}{ccc} a & b & c\\ 4c & a-b & 2b+2c\\ 4b+6c & 2c & a+2b+3c \end{array}\right]. $$ Reducing modulo $2$, we have that $$ M_{\delta}\equiv\left[\begin{array}{ccc} a & b & c\\ 0 & a-b & 0\\ 0 & 0 & a+c \end{array}\right]\text{ mod }2. $$ Since this upper triangular, it follows that $$ \text{Tr}\left(M_{\delta}^{k}\right)\equiv a^{k}+(a-b)^{k}+(a+c)^{k}\text{ mod }2. $$

$$ \equiv a-b+c\text{ mod 2}. $$ Now, if $a-b+c\equiv0\text{ mod }2$, then the last column in the matrix $$ A=\left[\begin{array}{ccc} \text{Tr}\left(1\right) & \text{Tr}\left(\delta\right) & \text{Tr}\left(\delta^{2}\right)\\ \text{Tr}\left(\delta\right) & \text{Tr}\left(\delta^{2}\right) & \text{Tr}\left(\delta^{3}\right)\\ \text{Tr}\left(\delta^{2}\right) & \text{Tr}\left(\delta^{3}\right) & \text{Tr}\left(\delta^{4}\right) \end{array}\right] $$ has each entry divisible by $2$, and hence $\text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}[\delta]\right)=\det A$ is divisible by $2$. If $a-b+c\equiv1\text{ mod }2$, then every element in $A$ is an odd number. As this determinant is a sum over permutations in $S_{3}$, we see that we are summing exactly $3!=6$ odd numbers, and so the determinant is even. In either case it follows that $2|\det A$ as well, and since $\text{disc}_{K/\mathbb{Q}}\left(\mathcal{O}_{K}\right)=-503$, we have shown that $\mathcal{O}_{K}\neq\mathbb{Z}[\delta]$ for any $\delta\in\mathcal{O}_{K}$ .

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