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I'm hung up on a part of a proof for: If $P$ is a maximal ideal, then $A=PP^{-1} = (1)$. Some background: Definition of ideal I am working with is if $\alpha$ and $\beta$, both algebraic integers (roots in field $K$ of polynomials with rational integer coefficients), belong to an ideal $I$, then so do $\lambda \alpha + \mu \beta$, for all algebraic integers $\lambda$ and $\mu$ in $K$. Here $P$ is a maximal ideal (contained by no ideal other than $(1)$ and $P$ itself), and $P^{-1}$ is defined to be all numbers $\gamma$ (algebraic integers or not) in $K$ such that $\gamma \pi$ is an algebraic integer for every $\pi$ in $P$.

The first observation in the proof, which is not explicitly explained, is that $A$ is, in fact, an ideal. I'm having a hard time seeing this. Every $\gamma \in P^{-1}$ produces an integer when multiplying any $\pi \in P$. Suppose $\beta [\in A] = \gamma_1 \pi_1$ and $\zeta [\in A] = \gamma_2 \pi_2$. What needs to be shown is that for all integers $\lambda$ and $\mu \in K$, $\lambda \beta + \mu \zeta \in A$. $\pi_1 + \pi_2 \in P$, so $(\pi_1 + \pi_2)(\gamma_1 + \gamma_2)$ is an integer in $A$ (using that the cross terms $\gamma_2 \pi_1$ and $\gamma_1 \pi_2$ are integers by the definition of $P^{-1}$ as numbers that produce integers when multiplying ANY elements of $P$). But --- here is where I am stuck because I don't see how to get from this point to $\lambda \gamma_1 \pi_1 + \mu \gamma_2 \pi_2 \in A$ for all $\lambda, \mu \in K$ --- Even though all the terms in the expression $(\pi_1 + \pi_2)(\gamma_1 + \gamma_2)$ are integers, and the expression includes the part I am interested in ($\gamma_1 \pi_1 + \gamma_2 \pi_2$), since I don't "know" yet that $A$ is an ideal, I can't yet assume that any part of this larger expression belongs to $A$.

Am I making this way too hard? Am I going about this in the entirely wrong fashion? Should I be trying to use an integral basis for $P$ instead? Sorry if this question sounds confused but if it is the reason is: So am I! Thanks.

The proof is Lemma 8.23 From Harry Pollard's "The Theory of Algebraic Numbers".

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Since there were no responses to my question, I eventually figured out a (tentative) sketch of the proof, which is below (requires some background from the book I referenced). Previous Lemma 8.22 in the same book reads, If $P$ is a maximal ideal, $P^{-1}$ contains a number which is not an algebraic integer. In showing this, such non-algebraic integers (AI's) are of the form $\frac{\gamma}{\pi}$, where $P = (\pi)$ (since P is maximal it is principal), and $\gamma$ is an AI not in $P.$ Then any two elements of $A = PP^{-1}$, say $\pi' (\frac{\gamma'}{\pi})$, $\pi'' (\frac{\gamma''}{\pi})$, have common factor $\rho = \pi^k \cdot \frac{(\gamma', \gamma'')}{\pi}$, that is $\pi^k$ times the gcd of the $\gamma$'s, noting that $\pi$ and the $\gamma$'s are relatively prime, and $k \geq 1$. Further $\rho \in A$ because if not $\frac{(\gamma', \gamma'')}{\pi}$ multiplying some element of $P$ produces a non-AI but since $\gamma'$ and $\gamma''$ are AI's, this could not happen (using the aforementioned characterization of all non-AI's belonging to $P^{-1}$ as of form $\frac{\gamma}{\pi}$, and $P = (\pi)$). Since the gcd belongs to $A$, so will $\lambda \pi'(\frac{\gamma'}{\pi}) + \mu \pi''(\frac{\gamma''}{\pi})$ for AI's $\lambda$ and $\mu$, so the definition of $A$ as an ideal is met.