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Most of the theorems in algebraic number theory seem to generalize to arbitrary base fields apart from $\mathbb{Q}$ apart from one. The characteristic of the residue field is equal to positive prime generating the ideal over $\mathbb{Q}$. This even stops making sense over other base fields.

Of course, the obvious generalization is that residue fields have the same characteristic as the residue field of the primes over which they lie. This is because they both lie over the same primes in $\mathbb{Q}$.

Edited:

I would like some other general examples where $\mathbb{Q}$ is much simpler than general number fields or has some property that does not hold in general.

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  • $\begingroup$ What symmetry are you referring to? What "arbitrary" base fields? $\endgroup$ – Brandon Carter Mar 12 '15 at 16:33
  • $\begingroup$ By symmetry I simply meant that almost all the results I know for algebraic number fields over Q generalize to number fields over any algebraic number field. $\endgroup$ – Asvin Mar 12 '15 at 16:56
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    $\begingroup$ Then where does the symmetry break? There is a uniform description that for a prime $\mathfrak{p}$ in a number field $K$, the residue field has characteristic equal to the positive generator of $\mathfrak{p} \cap \mathbf{Z}$. As there is no well-defined question here, I have voted to close. $\endgroup$ – Brandon Carter Mar 12 '15 at 17:14
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    $\begingroup$ Many questions over $\mathbb{Q}$ are more complicated over general number fields. For example, the ring of integers of a quadratic extension of $\mathbb{Q}$ is easily described explicitly, while quadratic extensions of number fields are probably more complicated, e.g. see mathoverflow.net/questions/81887/… $\endgroup$ – Victor Wang Mar 13 '15 at 18:16
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    $\begingroup$ Your question is pretty garbled. It'd be better to be more specific than to be intentionally vague in this case, e.g., asking for a property of $\mathbf Q$ that does not hold "in general" is hopeless because the list is so long! A general number field does not have class number one or a finite unit group or no everywhere unramified finite proper extension (no abelian condition assumed) or one real embedding or no nonreal complex embeddings etc. etc. $\endgroup$ – KCd Mar 14 '15 at 2:26
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Different nonarchimedean places of $\mathbf Q$ have different residue field characteristics. This is not true of any other number field.

Every finite abelian extension of $\mathbf Q$ is contained in a cyclotomic extension. This is false with $\mathbf Q$ replaced by any other number field. The analogy can be recovered using ray class fields instead, but such extensions of general number fields are not explicit like cyclotomic extensions.

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intersection all $\mathbb{Q}_p$= $\mathbb{Q}$ , where $p$ are all places.

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    $\begingroup$ Different $p$-adic fields do not really have a natural intersection. Can you explain what $\mathbf Q_3 \cap \mathbf Q_7$ means? $\endgroup$ – KCd Mar 14 '15 at 2:17
  • $\begingroup$ You can consider $\Pi_{p}\mathbb{Q}_{p}$ is a bigger field under this sense. $\endgroup$ – Jacob Mar 14 '15 at 22:49
  • $\begingroup$ A product of fields is not a field. Please state a precise result, rather than sentences with unclear hypotheses or conclusions. $\endgroup$ – KCd Mar 15 '15 at 0:45

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