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Let me point out that I would like to interpret "canonical" in a less strict sense. In order to explain what I mean, consider the following question:

Does there exist a canonical square root of $-1$ modulo primes $p = 4n+1$?

If canonical means that the result should be independent of any automorphism of the residue class group modulo $p$, then the answer is no: of $i$ is one answer, then the automorphism sending every residue class to its inverse will send $i$ to the other root $-i$.

Yet I would accept $i \equiv (\frac{p-1}2)! \bmod p$ as an answer to the question.

Edit 2 (28.07.10) I'm still thinking about what canonical should mean in this context. The ideal class of the square root is defined up to classes of order $2$; doesn't this mean that a "canonical" choice of this ideal class should be an element in the group $Cl(K)/Cl(K)[2]$?

Edit. Trying to generalize frictionless jellyfish's example I ended up with the following results (which so far I have only partially proved).

Let $p$ and $q$ be two prime numbers with $p \equiv q \equiv 1 \bmod 4$. There is a unique cyclic quartic extension $K/\mathbb Q$ with conductor $pq$ and discriminant $p^3q^3$. Let $\mathfrak p$ and $\mathfrak q$ denote the prime ideals in $K$ above $p$ and $q$. Then $diff(K/\mathbb Q) = {\mathfrak p}^3 {\mathfrak q}^3$. Moreover, ${\mathfrak p}^2 {\mathfrak q}^2 = (\sqrt{pq}\,)$ is principal, so the ideal class of the different is either trivial or has order $2$.

Theorem If $(p/q) = -1$, then $Cl_2(K) \simeq [2]$ if $p \equiv q \equiv 5 \bmod 8,$ and $Cl_2(K) \simeq [4]$ otherwise. In both cases, the $2$-class field of $K$ is abelian over $\mathbb Q$, hence is equal to its genus class field.

The ideal classes of each of the prime ideals above $\mathfrak p$ and $\mathfrak q$ generates the $2$-class group.

Taking $p=5$ and $q = 17$ gives an example of a "non-canonical" square root. I have not yet found a criterion that would tell me when the different is principal and when its class has order $2$. Both cases do occur.

The case where $p$ and $q$ are quadratic residues of each other is more involved, more interesting (and more conjectural):

Theorem Assume that $(p/q) = +1$, and that $p \equiv q \equiv 5 \bmod 8$. Then $Cl_2(K) \simeq [2,2]$ if $(p/q)_4 (q/p)_4 = -1$, $Cl_2(K) \simeq [4]$ if $(p/q)_4 = (q/p)_4 = -1$, and $Cl_2(K) \simeq [2^n], \ n \ge 3$ if $(p/q)_4 = (q/p)_4 = +1$.

If $(p/q)_4 = (q/p)_4 = -1$, then the ideal classes $[\mathfrak p]$ and $[\mathfrak q]$ are squares, but not fourth powers; in particular, the different is principal.

If $(p/q)_4 = (q/p)_4 = +1$, there are two cases:

1. $h_2(K) = h_2(k)= 2^n$; then $[\mathfrak p]$ and $[\mathfrak q]$ both have order $2$, and the different is principal.
2. $h_2(K) = 2h_2(k)= 2^n$; then either $\mathfrak p$ or $\mathfrak q$ is principal, whereas the other ideal generates a class of order $2$. In particular, the ideal class of the different has order $2$.
2 added 1993 characters in body

BTW, I very much like

Edit. Trying to generalize frictionless jellyfish's example , I ended up with the following results (which so far I have only partially proved).

Let $p$ and am currently working out $q$ be two prime numbers with $p \equiv q \equiv 1 \bmod 4$. There is a unique cyclic quartic extension $K/\mathbb Q$ with conductor $pq$ and discriminant $p^3q^3$. Let $\mathfrak p$ and $\mathfrak q$ denote the general conditions that make it workprime ideals in $K$ above $p$ and $q$. Then $diff(K/\mathbb Q) = {\mathfrak p}^3 {\mathfrak q}^3$. Moreover, ${\mathfrak p}^2 {\mathfrak q}^2 = (\sqrt{pq}\,)$ is principal, so the ideal class of the different is either trivial or has order $2$.

Theorem If $(p/q) = -1$, then $Cl_2(K) \simeq [2]$ if $p \equiv q \equiv 5 \bmod 8,$ and $Cl_2(K) \simeq [4]$ otherwise. In both cases, the $2$-class field of $K$ is abelian over $\mathbb Q$, hence is equal to its genus class field.

The ideal classes of each of the prime ideals above $\mathfrak p$ and $\mathfrak q$ generates the $2$-class group.

Taking $p=5$ and $q = 17$ gives an example of a "non-canonical" square root. I find anything have not yet found a criterion that would tell me when the different is principal and when its class has order $2$. Both cases do occur.

The case where $p$ and $q$ are quadratic residues of each other is more involved, more interesting (and more conjectural):

Theorem Assume that $(p/q) = +1$, and that $p \equiv q \equiv 5 \bmod 8$. Then$Cl_2(K) \simeq [2,2]$ if $(p/q)_4 (q/p)_4 = -1$, $Cl_2(K) \simeq [4]$ if $(p/q)_4 = (q/p)_4 = -1$, and $Cl_2(K) \simeq [2^n], \ n \ge 3$ if $(p/q)_4 = (q/p)_4 = +1$.

If $(p/q)_4 = (q/p)_4 = -1$, then the ideal classes $[\mathfrak p]$ and $[\mathfrak q]$ are squares, I'll put it herebut not fourth powers; in particular, the different is principal.

If $(p/q)_4 = (q/p)_4 = +1$, there are two cases:

• $h_2(K) = h_2(k)= 2^n$; then $[\mathfrak p]$ and $[\mathfrak q]$ both have order $2$, and the different is principal.
• $h_2(K) = 2h_2(k)= 2^n$; then either $\mathfrak p$ or $\mathfrak q$ is principal, whereas the other ideal generates a class of order $2$. In particular, the ideal class of the different has order $2$.
• 1

Let me point out that I would like to interpret "canonical" in a less strict sense. In order to explain what I mean, consider the following question:

Does there exist a canonical square root of $-1$ modulo primes $p = 4n+1$?

If canonical means that the result should be independent of any automorphism of the residue class group modulo $p$, then the answer is no: of $i$ is one answer, then the automorphism sending every residue class to its inverse will send $i$ to the other root $-i$.

Yet I would accept $i \equiv (\frac{p-1}2)! \bmod p$ as an answer to the question.

BTW, I very much like frictionless jellyfish's example, and am currently working out the general conditions that make it work. If I find anything interesting, I'll put it here.