### Proposition 2.3

Let K be a cubic field, the associated cubic form, a root of such that (we have seen above that such a exists). Then:
• .
• is an integral basis of .
• A prime decomposes in K as F decomposes in . More precisely, if

is a decomposition of F into irreducible homogeneous factors in then we have

where the are distinct prime ideals of given as follows. Call any lift of in and set . If , then

If but , then

If , , then if or there exists such that (any if ), and then

Finally, if p=2 and , we can take

### Proof

(1) has been proved in the preceding section.

(2) is a root of . It follows from [3], Chapter 4 Exercise 15, and easily checked anyhow, that is an order in K, i.e it is an algebra and a -module of finite type, and in particular is a suborder of the maximal order . If denotes the 3 roots of , an easy computation shows that

hence .

(3) Assume first that . Set

Then f is a monic irreducible polynomial over with a root .

Now and

so (this also follows directly from (2)). Since p does not divide the index of , it follows from basic algebraic number theory that , where , where is an irreducible decomposition of f in .

But then

for some and so and . Finally, we note that for , we have

and the case where follows.

Assume now that . If we can find a matrix such that is such that , we can apply the preceding case since K is also generated by a root of G.

One easily checks that if but we can take , and if , but either or then there exists such that (any if ), and we then take . This immediately gives the formulas of the proposition.

Finally, if p=2 and , then 2 divides the coefficient of of any form equivalent to , and from the definition of this means that 2 divides the index of any , in other words that it is an inessential discriminantal divisor. We then know that 2 is totally split, and hence 2 still factors as modulo 2. To find the factors explicitly, we must split the étale algebra . All its elements are thus idempotents. If we set , and considered as elements of (they are in , see above), we check that in since c is odd and a and d even. It follows that , and are orthogonal idempotents of sum 1, thus giving the desired splitting of , hence of .