Lazard’s ring and Height

3 Height

We saw that there is essentially one formal group for a \(\QQ \)-algebra, but what about in characteristic \(p\)? Tensoring our map \(L \to C\) with \(\ZZ /p\ZZ \), we see that \((I/I^2)_{2n} \to (J/J^2)_{2n}\) is an isomorphism unless \(n+1 = p^k\), when it is the zero map. The lack of injectivity and surjectivity suggests that not only are distinct formal group laws (surjectivity), but they can also have nontrivial strict automorphisms (injectivity), and that this failure is focused on the degrees of prime power.

The multiplicative formal group law \(x+y+xy\) and the additive formal group law \(x+y\) are isomorphic over \(\QQ \) via \(e^x-1\). This doesn’t have integral coefficients, so they are not isomorphic over \(\ZZ \) by triviality of automorphism groups over \(\QQ \). However this doesn’t tell us about if they are isomorphic over \(\ZZ /p\ZZ \).

To construct an invariant of formal group laws mod \(p\), we can first construct an invariant of maps \(F\) between formal group laws. First we should observe that a formal group law \(f\) has a unique translation invariant \(1\)-form \(\omega _f\) of the form \((1+O(t))dt\), spanning the invariant forms as an \(R\)-module. For example, for the additive formal group law it is \(dt\), and for the multiplicative one it is \(\frac {dt}{1+t}\).

Lemma 3.1. Given a formal group law \(f\) over \(R\), there is a unique translation invariant \(1\)-form \(\omega _f\) of the form \((1+O(t))dt\).

Proof. For a form \(g(t)dt = \sum g_it^idt\) to be invariant, we ask that \(f^*(g(t)dt) := g(f(x,y))(\frac {df}{dx}dx+\frac {df}{dy}dy) = g(x)dx+g(y)dy\). Setting \(x = dy = 0, dx = 1\), we get \(g(y)\frac {df}{dx}(0,y)= g(1)\) This gives a formula for \(g(y)\) in terms of \(g(1)\), showing the unique solution \(g\), and that the unique solution generates the invariant forms as an \(R\)-module. To see that the formula works, it suffices to show it on the universal formal group law on \(L\). But the formal group law on \(L \otimes \QQ \) has an invariant form since it is isomorphic to the additive formal group law, and the form must be given by the formula by uniqueness. One could equally well compute that the formula gives an invariant form. □

One can construct explicitly the isomorphism between a formal group law and the additive formal group law by integrating the invariant differential form. Namely if \(\sum _i g_i t^i dt\) is the invariant form, then \(\sum _i \frac {g_i}{i+1} t^{i+1}\), called the logarithm, is the strict isomorphism to the additive formal group law. To see this, if \(\omega _f\) is the invariant form, we can integrate the equation \(\omega _f(f(x,y))=\omega _f(x)+\omega _f(y)\) with respect to \(x\) to get \(\log _f(f(x,y))=\log _f(x)+c(y)\) for some constant of integration \(c\), which by symmetry has to be \(\log _f(y)\). Thus \(\log _f\) is a homomorphism to the additive formal group law, and has an inverse as it is an invertible power series. For example, \(MU\)’s logarithm is \(\sum _n \frac {[\CC \PP ^n]}{n+1}t^n\) which lets you compute the logarithm for any other complex oriented theory since \(MU\) is universal.

Now given a morphism \(F\) between formal group laws \(f,f'\), we have \(F^*(\omega _{f'})\) is an invariant form on \(f\), so is \(\lambda \omega _f\), where \(\lambda \) is the linear term of \(F\). Assume we are in characteristic \(p\) and \(\lambda = 0\). Then since \(0 = F^*(\omega _{f'}) = (1+O(t))dF\), we must have \(dF = 0\), which means \(F(t) = F_1(t^p)\) for some power series \(F_1\). But if \(f^p\) is the formal group law obtained from \(f\) by applying Frobenius to the coefficient ring, then \(F_1\) is a morphism from \(f^p\) to \(f'\), so we can repeat this argument to obtain:

Proposition 3.2. Given a nonzero morphism \(F\) between two formal group laws \(f,f'\) in characteristic \(p\), there is a unique \(k\) such that \(F(t) = F_k(t^{p^k})\), where \(F_k = \lambda t + O(t^2)\) with \(\lambda \neq 0\).

We call this \(k\) the height of the morphism \(F\), denoted \(ht_F\). We declare the \(0\) morphism to have infinite height. One easily sees from definition that if \(G,F\) are composable, then \(ht_G + ht_F = ht_{G\circ F}\). If \(\lambda \) is invertible, then the kernel of the map \(F\) on \(R[[x]]\) is \(R[[x]]/(F)\) which is a group scheme of rank \(p^{ht_F}\).

Since our formal group laws \(f\) are commutative, the \(n\)-series of \(f\), \([n]\), defined by \([1] = t, [n+1] = f([n](t),t)\) is an endomorphism that looks like \(nt+O(t^2)\). Thus in characteristic \(p\), \([p]\) is an endomorphism of height \(\geq 1\); this is called the height of the formal group law \(f\), and is easily seen to be an invariant.

For example, the additive group law is infinite height, and the multiplicative group law is height \(1\) since \([p] = (1+t)^p-1=t^p\).

Every elliptic curve \(E\) induces a formal group \(\hat {E}\) in the formal neighborhood of the identity via its addition law. From this point of view, the height of a map of formal group is an analog of the inseparable degree of an isogeny. More precisely,

Proposition 3.3. If \(E_1 \to E_2\) is an isogeny of elliptic curves, then the p-adic valuation of the inseparable degree of the map is the height of \(\hat {E}_1 \to \hat {E}_2\).

Since the degree of \([p]\) on an elliptic curve is \(p^2\), we see that there are two possible heights, \(1\) or \(2\). These two cases are called ordinary and supersingular. We see that in the super singular case, since the extension is purely inseparable, there are no nontrivial geometric \(p\)-torsion.