# Can one prove complex multiplication without assuming CFT?

The Kronecker-Weber Theorem, stating that any abelian extension of $\mathbb Q$ is contained in a cyclotomic extension, is a fairly easy consequence of Artin reciprocity in class field theory (one just identifies the ray class groups and shows that each corresponds to a cyclotomic extension). However, one can produce a more direct and elementary proof of this fact that avoids appealing to the full generality of class field theory (see, for example, the exercises in the fourth chapter of Number Fields by Daniel Marcus). In other words, one can prove class field theory for $\mathbb Q$ using much simpler methods than for the general case.

The theory of complex multiplication is similar to the theory of cyclotomic fields (and hence the Kronecker-Weber Theorem) in that it shows that any abelian extension of a quadratic imaginary field is contained in an extension generated by the torsion points of an elliptic curve with complex multiplication by our field. To prove this, one normally assumes class field theory and then shows that the field generated by the $m$-torsion (or, more specifically, the Weber function of the $m$-torsion) is the ray class field of conductor $m$.

My question is: Can one prove that any abelian extension of an imaginary quadratic field $K$ is contained in a field generated by the torsion of an elliptic curve with complex multiplication by $K$ without resorting to the general theory of class field theory? I.e. where one directly proves class field theory for $K$ by referring to the elliptic curve. Is there a proof in the style of the exercises in Marcus's book?

Note: Obviously there is no formal formulation of what I'm asking. One way or another, you can prove complex multiplication. But the question is whether you can give a proof of complex multiplication in a certain style.

## 1 Answer

(modified)

Historically, Complex Multiplication precedes Class Field Theory and many of the main theorems of CM for elliptic curves were proved directly. See Algebren (3 volumes) by Weber or Cox's book for an exposition.

Please also read Birch's article on the beginnings of Heegner points where he points this out explicitly (page three, paragraph beginning "Complex multiplication ...).

But not all. so the answer is no (unlike what I had mistakenly presumed at first and the comments below alerted me).

The actual history is quite complicated; see Schappacher.

• Historically, as far as I know, the first complete proof of complex multiplication for imaginary quadratic fields was given by Takagi in 1920, as a corollary of his class-field-theory. – user4245 Jun 22 '12 at 5:00
• @SGP: Are you sure they proved that every abelian extension of a quadratic imaginary field is contained in such an extension? – David Corwin Jun 22 '12 at 19:17
• @unknown(google) and Davidac897: Thanks! answer modified and reference of Schappacher added. – SGP Jun 23 '12 at 23:37
• Takagi's thesis was on the abelian extensions of Q(i); this is also covered in Silverman-Tate, if I recall it correctly. I would assume that for general complex quadratic base fields, the problem of class number > 1 (equivalent to the existence of the Hilbert class field) becomes a major obstacle. – Franz Lemmermeyer Jun 24 '12 at 9:48
• @Franz Lemmermeyer: Silverman-Tate mentions the result, but doesn't prove it. I wonder if they were thinking of the whole of class field theory, or if they were thinking of a simpler proof. – David Corwin Jun 29 '12 at 2:55