### Theorem 4.2

We have . In other words, the maps and are discriminant preserving inverse bijections between isomorphism classes of cubic fields and binary cubic forms belonging to U.

### Proof

Let K be a cubic number field, and the image by of K. We will first show that . As in Proposition 4.1, write with a fundamental discriminant.

Let p be a prime. If , then , so . Hence we now assume that .

By Proposition 4.1 (1), it follows that p is totally ramified. By Proposition 2.3, this means that splits as the cube of a linear form, i.e that . We consider three cases.

p>3. In this case, it follows from Proposition 4.1 (2) and (3) that , hence that by Proposition 3.3 (3).

p=2. Since , we have by Proposition 4.1 (2), hence by Proposition 4.1 (3) we deduce that hence as before.

p=3. This is the only difficult case. Since 3 cannot be an inessential discriminantal divisor, there exists such that . Now since 3 is totally ramified, write for a prime ideal of degree 1. Since is isomorphic to , we have , hence either , or belongs to . Replacing if necessary by or (which does not change the property , we may assume that .

We clearly have

Assume first that . Thus has non-negative valuation for every prime ideal except perhaps those above 3, but the only prime ideal above 3 is and the -adic valuation of is at least equal to 1, hence . By the above formula we have , and since , it follows that . Since is the only prime ideal above 3, this means that the -adic valuation of is at least equal to 3, hence that , in other words that . Now since , 3 splits in K as the characteristic polynomial of splits modulo 3. It follows that for a certain . Write

We thus have and . Since we have . But this is absurd since we would then have

Thus . Since we still have and we obtain

and since we have so . Furthermore by our choice of we have

hence

Let be the integral basis used to define the form . Thus for some u, v and w in , and hence by definition we have . We have the following lemma:

Lemma 4.3 Assume that F is a primitive cubic form and p a prime such that . Then if and only if there exists such that with .

Assuming this lemma for a moment, we see that since represents an integer congruent to modulo 9, we have , thus finishing the proof that .

Let us prove the lemma. By lifting the condition to , we can write

with , , in and G an integral cubic form. Then if and only if . Thus if we have with .

Conversely, assume that there exists such that with . Then , hence . Since F is primitive, hence . Again F being primitive, and cannot both be divisible by p, from which it follows that there exists such that and , and since we have . But then hence and so , so , thus proving the lemma.

To finish the proof of the Davenport-Heilbronn Theorem 4.2, we must now prove that if , there exists a cubic field K such that F is equivalent to . For this, we introduce a definition.

Definition 4.4 We will say that two cubic forms and are rationally equivalent if there exists such that for .

We first show the following lemma.

Lemma 4.5 Let F be any form in ( i.e primitive and irreducible). Then there exists a number field K such that F is rationally equivalent to .

Proof In some algebraic closure of , write . Since F is irreducible, is a cubic irrationality, and we will take . Write so that (we have seen above that this is always possible). Since K is a -vector space of dimension 3, there exist 4 integers k, l, m and n not all zero such that . Taking conjugates, we obtain the same equality with and . Using , we obtain

with . Furthermore the determinant kn-lm is nonzero since otherwise either or would be in . Thus is rationally equivalent to F.

Finally, we have the following lemma.

Lemma 4.6 Let and be two forms belonging to U. Then if and are rationally equivalent, they are equivalent.

Since belongs to U, it follows from Lemma 4.5 and this lemma that any form in U is equivalent to for a certain K, and this finishes the proof of Davenport-Heilbronn's Theorem 4.2.

Proof Assume that for some and . Since we want to show that and are equivalent, we can replace them by equivalent forms. In other words, without changing the equivalence classes of and we may replace M by any matrix of the form UMV with U and V in . The elementary divisor theorem ( i.e the Smith normal form) tells us that we can choose U and V so that

Replacing by , we may assume . To summarize, by replacing and by equivalent forms and modifying and M, we may assume that with .

If m=1, then and, since and are in U, they are primitive. Hence so and are equivalent.

Otherwise, there exists a prime p such that . Write and with and prime to p ( i.e of zero p-adic valuation). Since we have . Now, writing for i=1 or 2, , the equality is equivalent to

where the are rational numbers prime to p.

Assume l-k>0, hence . Then and . If on the other hand then hence and . Replacing by we obtain again and .

The formula for the discriminant implies . If p>2 this immediately implies . If p=2, the formula for the discriminant shows that and this is congruent to 0 or 1 modulo 4, so .

Since we must have , and with . Since , means that , hence . But then, since , , contradiction. This finishes the proof of Lemma 4.6, and hence of Davenport-Heilbronn's Theorem 4.2.