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(Apologies if this question isn't quite research-level: a colleague came across it while preparing a non-examinable bonus lecture on class field theory for an undergraduate algebraic number theory course.)

Let $K$ be a number field. Is there always a finite extension $L / K$ such that $L$ has class number 1?

If $K$ has finite class field tower (i.e. the tower of fields $(K_n)_{n \ge 0}$, where $K_0 = K$ and $K_{n+1}$ is the Hilbert class field of $K_n$, eventually terminates) then that solves the problem. But it's a well-known theorem of Golod and Shafarevich that the class field tower of $K$ doesn't terminate if $K$ is an imaginary quadratic field with enough primes ramified.

The textbook my colleague has been using claims that it follows from Golod-Shafarevich that these fields $K$ cannot be embedded in any number field with class number 1, but this implication isn't clear to me. Golod-Shafarevich shows that $K$ has no finite, solvable, everywhere-unramified extension with class number 1, but that's a much weaker statement, isn't it?

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I just want to mention that this question came up in a totally different context! It was about proving the (weak) Mordell-Weil Theorem for elliptic curves over number fields. There is a proof that works only when $K$ has class number $1$. Of course, if we can embed $K$ in a field with class number $1$, then the theorem follows a fortiori in $K$ because of its truth in the larger field. –  David Corwin Jul 9 '12 at 5:59

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up vote 13 down vote accepted

See Proposition 1 on p.231 of Cassels and Frohlich for a proof of the claim in the textbook:

The point is that if such an $L$ exists then $K_1L$ is abelian and unramified over $L$ so it is contained in the Hilbert classfield of $L$ which is $L$ itself. By induction, this implies that $K_i \subset L$ for all $i$ so the class field tower of $K$ must be finite.

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Ah, of course: the class field tower of $L$ has to be at least as big as the class field tower of $K$. Thanks! –  David Loeffler Mar 26 '12 at 10:09

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