= Solution
For $K=\mathbb Q(\sqrt{-163})$, the <fundamental discriminant> is $-163$ and the <Minkowski bound for ideal classes> is
$$
\frac2\pi\sqrt{163}<9.
$$
Thus every ideal class contains an integral ideal of norm at most eight.
The <ring of integers of a quadratic field> is $\mathbb Z[\theta]$ for $\theta=(1+\sqrt{-163})/2$, whose minimal polynomial is $X^2-X+41$. Its discriminant $-163$ is a quadratic nonresidue modulo each of $3,5,$ and $7$; directly modulo two, $X^2+X+1$ has no root. Hence these primes are inert by the <splitting of rational primes in a quadratic field>. The only prime ideal over one of them with norm at most eight is $(2)$, which has norm four and is principal. Unique <prime ideal factorization> now makes every ideal of norm at most eight principal. The Minkowski bound therefore proves
$$
\boxed{\ \operatorname{Cl}(\mathbb Q(\sqrt{-163}))=1.\ }
$$
This is the <Ideal class group of Q of square root of minus one hundred sixty-three>.
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