A question referring the proof of class number ![enter image description here]( In this proof, second paragraph, why do we just need to consider the primes for $2\le x\le \sqrt{|d|}$, how about the primes larger than...--prophetes.ai

A question referring the proof of class number ![enter image description here]( In this proof, second paragraph, why do we just need to consider the primes for $2\le x\le \sqrt{|d|}$, how about the primes larger than $\sqrt{|d|}$ ? And why the the primes for $2\le x\le \sqrt{|d|}$ inert implies the class number 1? (the reason he gives doesn't make sense to me, what does he mean that no primes inert? At least we don't know the case larger than $\sqrt{|d|}$, and even all the primes are inert, why is the class number 1?)

It follows from the observation he made in page 149

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Which in turn follows from Minkowski's theorem and Corolary I (in pag 135)

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This is because if $A$ is an ideal class and $\mathfrak{a}$ is a representative such that $N(\mathfrak{a})< \sqrt{\mid d \mid}$. Then for a prime ideal $\mathfrak{p}$ dividing $\mathfrak{a}$ with $ p \,\mathbb{Z}= \mathfrak{p} \cap \mathbb{Z} $, we have $p\leq N(\mathfrak{p})\mid N(\mathfrak{a})< \sqrt{\mid d \mid }$ so $\mathfrak{a}$ is a product of prime ideals $\mathfrak{p} $ each factor of some prime $p< \sqrt{\mid d \mid}$ and so $A$ is the product of the corresponding classes.

In this case all this primes $p$ are inertial (thus their only factor is principal) and therefore every class is the identity.