Step in Proof of Existence of an Integral Basis

Question

This is Proposition 2.10 of Neukirch's Algebraic Number Theory. Let $A$ be an integral domain with field of fractions $K$, let $L/K$ be a finite extension, and let $B$ be the integral closure of $A$ in $L$. There is a step where he seems to use the fact that for any element $x \in L$, there is a nonzero $a \in A$ such that $ax \in B$. I see how this would be true it we were allowed to choose $a \in B$ to clear the denominator of $x$, but I'm not sure why we can choose an element $a$ in the smaller ring $A$.

Answer 1

Say your element is some fraction $x=\frac{\alpha}{\beta}\in L$ with $\beta \in B$. Multiply and divide $x$ by all the Galois conjugates of $\beta$, which are all in $B$ (because they all have the same minimal polynomial). Then you can rewrite $x$ as $$ x =\frac{\gamma}{\operatorname{N}_{L/K}(\beta)}$$ where $\gamma \in B$ and $\operatorname{N}_{L/K}(\beta)\in A$. Hence you may take $a=\operatorname{N}_{L/K}(\beta)$.

Remark: I used that $L/K$ is a separable extension, so that I can talk about Galois conjugates. I checked the reference you mention and this was also assumed by the author.

Answer 2

$x$ from $L$ is a root of a polynomial with coefficients in $K$. Since $K$ is the field of fractions of $A$ we can find by multiplying with a common denominator an equation for $x$ with coefficients in $A$ $$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = 0$$ Multiply this by $a_n^{n-1}$ and get $$(a_n x)^n + a_{n-1}(a_n x)^{n-1} + a_{n-2} a_n (a_n x)^{n-2}+ \cdots + a_0 a_n^{n-1}=0$$ that is, an integral equation for $a_n x$.