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Old 2017-09-18, 16:22   #4
Dr Sardonicus
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Feb 2017

132·23 Posts

If K is a number field, K/Q is a normal (Galois) extension, OK its ring of algebraic integers, p is a prime number, and pOK is the product of distinct prime ideals of norm p, then either all of these ideals are principal, or they are all non-principal. If they are non-principal, then there are no integers of norm p. In this case, let P be a prime ideal of norm p, and let c be its class in the class group. If there is a fractional ideal X of norm 1 whose class is c-1, then XP will be a principal ideal of norm p, but its principal generators will be non-integer fractions in K.

If Q is a prime ideal in the "numerator" of X, there has to be a corresponding Q' conjugate to Q in the "denominator" (this uses the assumption that K/Q is a normal extension).

Thus, any fractional ideal of norm 1 is a product of factors of the form Q/Q' where Q is a prime ideal in OK.

I don't know in general how to determine whether such a product can be in any given ideal class, but I do know one case where it is not possible. If K has class number h = 2, then in the fractional ideal Q/Q', either Q and Q' are both principal, or they are both in the sole non-principal class, so that Q/Q' is principal. So in this case, all fractional ideals with norm 1 are a product of principal ideals, and are therefore principal.

So if K/Q is normal with class number h = 2, and p a prime number with pOK the product of non-principal prime ideals of norm p, then there is no element of K having norm p.

The smallest cyclotomic example appears to be K = Q(\zeta_{39}) and p = 79.
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