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

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

pbe a prime. If , then , so . Hence we now assume that .By Proposition 4.1 (1), it follows that

pis totally ramified. By Proposition 2.3, this means that splits as the cube of a linear form,i.ethat . 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 since3is 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

We thus have and . Since we have . But this is absurd since we would then have1, 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 ,3splits inKas the characteristic polynomial of splits modulo 3. It follows that for a certain . WriteThus . Since we still have and we obtain

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

henceLet be the integral basis used to define the form . Thus for some

u,vandwin , and hence by definition we have . We have the following lemma:Lemma 4.3 Assume that

Fis a primitive cubic form andpa 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 andGan integral cubic form. Then if and only if . Thus if we have with .Conversely, assume that there exists such that with . Then , hence . Since

Fis primitive, hence . AgainFbeing primitive, and cannot both be divisible byp, 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

Ksuch thatFis 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

Fbe any form in (i.eprimitive and irreducible). Then there exists a number fieldKsuch thatFis rationally equivalent to .Proof In some algebraic closure of , write . Since

with . Furthermore the determinantFis irreducible, is a cubic irrationality, and we will take . Write so that (we have seen above that this is always possible). SinceKis a -vector space of dimension 3, there exist 4 integersk,l,mandnnot all zero such that . Taking conjugates, we obtain the same equality with and . Using , we obtainkn-lmis nonzero since otherwise either or would be in . Thus is rationally equivalent toF.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 inUis equivalent to for a certainK, 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

Replacing by , we may assume . To summarize, by replacing and by equivalent forms and modifying andMby any matrix of the formUMVwithUandVin . The elementary divisor theorem (i.ethe Smith normal form) tells us that we can chooseUandVso thatM, we may assume that with .If

m=1, then and, since and are inU, they are primitive. Hence so and are equivalent.Otherwise, there exists a prime

where the are rational numbers prime topsuch that . Write and with and prime top(i.eof zerop-adic valuation). Since we have . Now, writing fori=1or2, , the equality is equivalent top.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>2this immediately implies . Ifp=2, the formula for the discriminant shows that and this is congruent to0or1modulo 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.