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Recently the following was explained to me by Melanie Wood. (What follows is a summary of some results from her prepint preprint here; see in particular the discussion at the top of page 3.)

If $f$ is a primitive irreducible binary $n$-ic form with coefficients in $\mathbb Z$, then we can consider the locus $S_f$ in $\mathbb P^1_{\mathbb Z}$ that $f$ cuts out. This will be finite over $\mathbb Z$, and hence is equal to Spec $R$, where $R = H^0(S_f, \mathcal O)$. The finite $\mathbb Z$-algebra $R$ will be an order in a number field of degree $n$. Not all orders in all number fields arise this way (if $n >3$), but certainly some do!

Now it turns out that restriction of $\mathcal O(n-2)$ to $S_f$ (which is an invertible sheaf on $S_f$, and hence gives an ideal class of $R$) is canonically isomorphic to the inverse different of $R$, and hence if $n$ is even then we can consider the restriction to $S_f$ of $\mathcal O((n-2)/2)$, and so obtain a square root of the inverse different of $R$.

As is explained in the above linked preprint, $R$ alone does not suffice to determine $f$; rather $f$ is determined by $R$ together with some extra ideal-theoretic strucure. So (as far as I know) this construction of the square root does not depend on $R$ alone, but on extra data. Nevertheless, it gives a very interesting answer to my question in those cases where it applies.

1

Recently the following was explained to me by Melanie Wood. (What follows is a summary of some results from her prepint here; see in particular the discussion at the top of page 3.)

If $f$ is a primitive irreducible binary $n$-ic form with coefficients in $\mathbb Z$, then we can consider the locus $S_f$ in $\mathbb P^1_{\mathbb Z}$ that $f$ cuts out. This will be finite over $\mathbb Z$, and hence is equal to Spec $R$, where $R = H^0(S_f, \mathcal O)$. The finite $\mathbb Z$-algebra $R$ will be an order in a number field of degree $n$. Not all orders in all number fields arise this way (if $n >3$), but certainly some do!

Now it turns out that restriction of $\mathcal O(n-2)$ to $S_f$ (which is an invertible sheaf on $S_f$, and hence gives an ideal class of $R$) is canonically isomorphic to the inverse different of $R$, and hence if $n$ is even then we can consider the restriction to $S_f$ of $\mathcal O((n-2)/2)$, and so obtain a square root of the inverse different of $R$.

As is explained in the above linked preprint, $R$ alone does not suffice to determine $f$; rather $f$ is determined by $R$ together with some extra ideal-theoretic strucure. So (as far as I know) this construction of the square root does not depend on $R$ alone, but on extra data. Nevertheless, it gives a very interesting answer to my question in those cases where it applies.