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Let $K = \Bbb Q(\sqrt d)$ be a quadratic field. Let $O_j = \Bbb Z + f_j O_K$ be two orders in $O_K$.

Assume that $O_1 \cong O_2$ as rings. Does it follow that $O_1 = O_2$ (i.e. their conductors are the same)?

We know that $f_j = [O_K : O_j]$, but isomorphic subgroups don't need to have the same index in general (cf Isomorphic subgroups, finite index, infinite index).

Alphonse
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This is true. In general, if $R$ is an integral domain with fraction field $K$, the index of $R$ in its integral closure $R^0$ in $K$ is an isomorphism invariant of $R$.

hunter
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  • This is a cool fact. Would you have a hint about the proof? – Alphonse May 22 '21 at 18:48
  • I was thinking that an isomorphism $R \to R'$ extends (via $\otimes_{\Bbb Z} \Bbb Q$) to a $\Bbb Q$-algebra (?) morphism $K \to K$, but why would it leave $R$ invariant? Or how to use the fact $\mathrm{char}(R) = 0$ ? – Alphonse May 22 '21 at 18:52
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    @Alphonse the field of fractions is just the set of symbols $r/s$ and an isomorphims $\phi: R \to R'$ extends to an isomorphism from $K$ to $K'$ taking $r/s$ to $\phi(r)/\phi(s)$. This isomorphism must take the integral closure of $R$ in $K$ to the integral closure of $R'$ in $K'$ (the latter is just the set of roots in $K'$ of monic polynomials with coefficients in $R'$; pull the polynomials back along $\phi$.) – hunter May 22 '21 at 19:04
  • (also, sorry, should have said "integral domain" and not "ring of characteristic zero," have now edited answer!) – hunter May 22 '21 at 19:08