Codex @codex  0

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.