Lazard’s ring and Height

2 The computation

Here Lemma 1.1 is proven. To understand the map \(I/I^2 \to J/J^2\) in degree \(2n\), we will understand the functors they corepresent. Let \(M\) be an abelian group.

\[\Hom ((I/I^2)_{2n},M) = \Hom ((I/I^2)_{2n}^+,M_{2n}^+) = \Hom (L,M_{2n}^+) = FGL(M_{2n}^+)\]

Here \(M_{m}^+\) is the functor that takes an abelian group \(M\) to the ring which has a \(\ZZ \) in degree \(0\), \(M\) in degree \(m\), and trivial multiplication apart from the action of \(\ZZ \) on \(M\) and itself. Thus we need to understand formal group laws on \(M_{2n}^+\). These are given by \(f(x,y) = c_{i,j}x^iy^j\), \(c_{i,j} \in M\), but because of the grading, \(c_{i,j} = 0\) unless \(i+j = n+1\). Thus let us call \(c_{i,j}\) just \(c_i\) for short. Then \(c_0 = 0\) by the identity law, and commutativity tells us \(c_i = c_{n+1-i}\). Associativity amounts to the identity \(c_{i+j} {i+j\choose i} = c_{j+k} {j+k\choose j}\) whenever \(i+j+k=n+1\).

There are “obvious" solutions of these equations, namely those coming from the map \(I/I^2 \to J/J^2\). These are given by changing coordinates from the additive formal group via the power series \(g = t+mt^{n+1}\). These give the solutions \(c_i = {n+1 \choose i} m\) for \(1 \leq i \leq n\). Are these all the solutions? Well, let \(d_n\) be the gcd of \(n+1 \choose i\) for \(1\leq i\leq n\). Then \(c_i = \frac {{n+1 \choose i}}{d_n}m\) are solutions. So what is \(d_n\)? Recall that to compute binomial coefficients mod \(p\), we simply multiply the binomial coefficients of the coefficients in the \(p\)-adic expansions of the numerators and denominators. From this we get:

Lemma 2.1. \(p|{{i+j}\choose i}\) iff \(i+j\) can be added in base \(p\) without carrying.

If \(n+1\) is not a power of \(p\), we can write \(n+1= i+j\) where the nonzero digits of \(i,j\) in base \(p\) are distinct, so \(p\) doesn’t divide \(n+1\choose i\). If \(n+1=p^k\), then regardless of which \(i\) we choose, we will have to carry, so \(p\) divides the gcd of the \(n+1 \choose i\). In fact, the \(p\)-adic valuation of \(p^k \choose p^{k-1}\) is \(1\), so the gcd is \(p\). Summarizing:

Lemma 2.2. \(d_n\) is 1 if \(n+1\) isn’t a prime power, and is \(p\) if \(n+1 = p^k\).

Proposition 2.3. \(c_i = \frac {{n+1 \choose i}}{d_n}m\) gives an isomorphism \(\phi :M \cong \Hom ((I/I^2)_{2n},M)\).

Proof. To show it is an isomorphism, it suffices to check locally at a prime \(p\), so assume \(M\) be a \(Z_{(p)}\)-module. Each \(c_i\) gives a map \(c_i:\Hom ((I/I^2)_{2n},M) \to M\). By choosing \(i\) so that \(n+1\choose i\) has the smallest \(p\)-adic valuation, the composite \(c_i \circ \phi \) is multiplication by an unit, so \(\phi \) is injective. To show surjectivity, it suffices to show that this \(c_i\) is injective. If \(n+1 \neq p^k\), then we can take \(i = 1\), and otherwise take \(i = p^{k-1}\). From Lemma 2.1 and the associativity constraint, we obtain that \(c_j = 0\) implies \(c_{k+j} = 0\) whenever \(k+j\) can be added without carrying. Moreover, \(c_j = 0\) implies \(c_{n+1-j} = 0\). These together show that \(c_i= 0\) implies everything is \(0\). □

Thus \((I/I^2)_{2n}\) is \(\ZZ \), and the map to \((J/J^2)_{2n} = \ZZ \) is multiplication by \(d_n\).