Is there an elementary way to see that a number ring $\mathcal{O}_K$ of a number field $K$ is integrally closed in $K$?

In other words, let $\alpha \in \mathbb{C}$ and $a_i$ are algebraic integers. How can one see easily and with basic tools that if $\alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_{1}\alpha +a_0 = 0$, then $\alpha$ is also an algebraic integer?

  • $\begingroup$ Integrality is transitive. $\endgroup$ – Angina Seng May 30 at 13:01
  • $\begingroup$ I can't use this unfortunately. $\endgroup$ – EinStone May 30 at 13:07
  • $\begingroup$ @EinStone why not? It's very elementary. You can reduce it to showing that finite extensions are transitive and that finite extensions are integral. The first fact is trivial and the second uses the "determinant trick," which is maybe a little tricky to discover for oneself but certainly elementary. $\endgroup$ – Badam Baplan May 30 at 14:10
  • $\begingroup$ I am asking from a basic number theory course perspective, there should be no commutative algebra / module theory be involved in the proof. I am looking for something like Theorem 3.3.4 in here (where I unfortunately don't understand the proof): normalesup.org/~page/Enseignement/Universite/2016-2017/Poly.pdf $\endgroup$ – EinStone May 30 at 15:20
  • $\begingroup$ Are you familiar with basic Galois theory? i.e. the way that field automorphisms interact with roots of polynomials. $\endgroup$ – Qwertiops May 30 at 16:54

You mentioned in the comments that in the proof that you linked, you don't understand why the coefficients of $Q(x)$ are rational and algebraic integers. I have come up with my own proof, but it relies in small part on unique prime factorisation of ideals, which is far from elementary. As such, I'll just try to explain why the coefficients of $Q$ are rational integers.

Each coefficient of $Q(x)$ is a polynomial expression (i.e. a sum of products) in the coefficients of the polynomials $P^\sigma$. These coefficients are of the form $\sigma(a)$ where $\sigma$ is an embedding of $K$ in $\mathbb{C}$ and $a$ is an algebraic integer. As such, $\sigma(a)$ is a root of the minimum polynomial of $a$, which is a monic integer polynomial, hence $\sigma(a)$ is an algebraic integer. Therefore, each coefficient of $Q(x)$ is a polynomial expression in some algebraic integers, and hence is certainly an algebraic integer (because algebraic integers form a ring).

Now for rationality. Let $N$ be the normal closure of $K$ over $\mathbb{Q}$, and let $G$ be the Galois group $\text{Gal}(N / K)$. Let $g \in G$. Then for each embedding $\sigma$, it is clear that $g\circ \sigma$ is also an embedding of $K$. Furthermore, if $g \circ \sigma = g \circ \tau$ for embeddings $\sigma, \tau$, then $\sigma = \tau$ since $g$ is injective (since $g(\sigma(x)) = g(\tau(x)) \forall x$ so $\sigma(x) = \tau(x) \forall x$). Thus, $g$ actually permutes the embeddings of $K$, which means that

$$ Q^g(x) = \Big(\prod_{\sigma}P^\sigma(x)\Big)^g = \prod_{\sigma}P^{g \circ \sigma}(x) = \prod_{\sigma '}P^{\sigma'}(x) = Q(x) $$ where we make the substitution $\sigma' = g \circ \sigma$ at the end.

Therefore, the coefficients of $Q$ are fixed by every element of $G$, and hence they are rational (by the Fundamental Theorem of Galois Theory).

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