If $K = \mathbb{Q}(\alpha)$, then must $\mathcal{O}_K$ contain $\alpha$, and conversely?

Question

Suppose $K = \mathbb{Q}(\alpha)$ with $[K:\mathbb{Q}] = n > 1.$ Then must its ring of integers $\mathcal{O}_K$ contain $\alpha$? I think it must, and here is my reasoning: suppose it did not. Now $K$ is the field of fractions of $\mathcal{O}_K$ and $\mathbb{Q} \cap \mathcal{O}_K = \mathbb{Z},$ so this means $\mathcal{O}_K = \mathbb{Z},$ hence $K = \mathbb{Q},$ a contradiction as the degree of the extension is greater than $1.$ Is this correct?

Also, I was also wondering if the converse holds, i.e. if $\mathcal{O}_K = \mathbb{Z}[\alpha],$ then is $K = \mathbb{Q}(\alpha)?$ I believe the same reasoning (that $K$ is the field of fractions of $\mathcal{O}_K$) gives us this result again?

Answer 1

If $K = \mathbb{Q}(\alpha)$ for some $\alpha \in K$, then you can multiply or divide $\alpha$ by elements $c \in \mathbb{Z}$ so that the result is or is not in $\mathcal O_K$, as you wish. You can see this by combining four facts:

$\bullet$ $0 \neq c \in \mathbb{Q}$, then $\mathbb{Q}(c \alpha) = \mathbb{Q}(\alpha)$.

$ \bullet$ If $\mathfrak p$ is a prime of $\mathcal O_K$, and $x \in K$, let $\nu_{\mathfrak p}(x)$ be the largest integer $m$ such that $x \in \mathfrak p^m$. Then $\nu_{\mathfrak p}(xy) = \nu_{\mathfrak p}(x) + \nu_{\mathfrak p}(y)$.

$\bullet$ If $x \in K$, then $x \in \mathcal O_K$ if and only if $\nu_{\mathfrak p}(x) \geq 0$ for all primes $\mathfrak p$.

$\bullet$ If $\mathfrak p$ is a prime of $\mathcal O_K$, then $\mathfrak p \cap \mathbb{Z} \neq 0$. In particular, $\mathfrak p \cap \mathbb{Z} = p\mathbb{Z}$ for some prime number $p$.

So, you can always write $K = \mathbb{Q}(\alpha)$ for some $\alpha \in \mathcal O_K$. For such $\alpha$, you always have $\mathcal O_K \supseteq \mathbb{Z}[\alpha]$. A more interesting question is whether it is always possible to choose $\alpha$ so that these rings are equal (e.g. for quadratic extensions this is possible, the Gaussian integers satisfy $\mathcal O_{\mathbb{Q}[i]} = \mathbb{Z}[i]$). The answer is surprisingly no!

To your second question (as I indicate in the previous paragraph, this is not a true converse: if $\alpha \in \mathcal O_K$, then you only have $\mathbb{Z}[\alpha] \subseteq \mathcal O_K$, not necessarily equal), the answer is yes.

In general, if $R$ is an integral domain with quotient field $K$, and $x$ is an element in a field $L$ containing $K$, then the quotient field $E$ of $R[x]$ is $K(x)$. Certainly $E$ must contain the inverses of elements of $R$ as well as the inverse of $x$, so $K(x) \subseteq E$. On the other hand, $R[x] \subseteq K[x]$, so we have $E = \textrm{quot } R[x] \subseteq \textrm{quot } K[x] = K(x)$.

Answer 2

The first question has been answered in the comments: no, the $\alpha$ in $\mathbb Q(\alpha)$ need not be an algebraic integer.

The second question does have an affirmative answer, since $\mathcal O_K \otimes \mathbb Q \simeq K$. For the same reason, this is true if we only require $\mathcal O_K$ to be an order in $K$, rather than the full ring of integers. (E.g., $\mathcal O=\mathbb Z[\sqrt{-3}]$ is not the full ring of integers of $\mathbb Q(\sqrt{-3})$.)