All prime ideals of size less than the Minkowski bound being principal implies that the ring of integers is a PID

Question

So I am currently learning about ideal class groups and it is mentioned by the professor that to "show that the ring of integers $R$ is a PID, it suffices to show that all prime ideals of size less than or equal to the Minkoski bound is principal" I don't understand why this is the case and would appreciate if somebody can explain it to me.

From what I understand, the ideal class group $Cl(R)$ consists of all ideals in the ring of integers $R$ (where R can be shown as a Dedikind domain). Hence, all ideals in $Cl(R)$ have a finite unique factorization into prime ideals.

At the same time, the ideal class group $Cl(R)$ can be shown to be equal to the quotient group $J_K/P_K$ where $J_K$ is the group of fractional ideals of the ring of integers of $K$, and $P_K$ is its subgroup of principal ideals.

Any assistance on how I should proceed will be greatly appreciated. Thank you!

The class group $CL(K)$ is generated by prime ideals of norm at most $M_K$, the Minkowski bound. Therefore the claim follows. See for example here, page $4$. — Dietrich Burde, Mar 28 '23 at 18:24
@DietrichBurde I see..... But I don't see how every prime ideal of norm at most $M_k$ being principal implies that $CL(K)$ is trivial i.e. every ideal in $K$ being trivial. — Explorer1234, Mar 28 '23 at 23:50
No, trivial class group just means that they are all principal. Class number $1$ means PID for the ring of integers. — Dietrich Burde, Mar 29 '23 at 08:56
@DietrichBurde Oh yes. You are right. But I still dont see how every prime ideal of norm at most $M_k$ being principal implies that $CL(K)$ is trivial i.e. all the ideals are principal. — Explorer1234, Mar 29 '23 at 11:10
For an integral domain, if all prime ideals are principal, so is every ideal, see here. — Dietrich Burde, Mar 29 '23 at 12:15

All prime ideals of size less than the Minkowski bound being principal implies that the ring of integers is a PID

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