Lemma: Let $I$ be a proper ideal in the ring $R$ of all algebraic integers, and $\alpha_1,\dots,\alpha_n$ algebraic over $\mathbb{Q}$. Then there exists $\beta\in K=\mathbb{Q}(\alpha_1,\dots,\alpha_n)$ such that $\beta\alpha_i\in R$ and not all $\beta\alpha_i$ belong to $I$.

Proof: (1) Let $R_K$ be the ring of integers in $\mathbb{Q}(\alpha_1,\dots,\alpha_n)$; it is a Dedekind domain.

(2) Let $A$ be the $R_K$-module generated by $\alpha_1,\dots,\alpha_n$.

(3) Since $R_K$ is Dedekind, there is a finitely generated $R_K$-module $B$ inside of $R_K$ such that $AB=R_K$.

(4) Since $AB\nsubseteq I$, there is $\beta\in B$ such that $\beta A\nsubseteq I$.

I don't understand (3) and (4), can one explain a little bit.

I was also doing in following way: $\alpha_1,\dots,\alpha_n$ are algebraic numbers; we can certainly find an integer $m$ in $\mathbb{Z}$ such that $m\alpha_i$ are all algebraic integers. After this I couldn't proceed to complete proof.


1 Answer 1


(3) is false as stated, it should be changed to "there is a finitely generated $ R_K $-module $ B $ inside $ K $". For a counterexample, take $ K = \mathbf Q(i) $ and $ A = (2i) $, for instance.

The modified version of (3) follows from the invertibility of ideals in Dedekind domains. Any finitely generated $ R_K $ submodule of $ K $ is a fractional ideal, and it has an inverse. Take $ B $ to be the inverse of the fractional ideal $ A $.

(4) follows because $ I $ is a proper ideal of $ R $, thus $ 1 \notin I $, for example. Since $ AB = R_K $, we cannot have that $ AB \subset I $ (because it contains $ 1 $.)


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