big Witt vectors for everyone (1/2)

Next time you visit your math-library, please have a look whether these books are still on the shelves : Michiel Hazewinkel's Formal groups and applications, William Fulton's and Serge Lange's Riemann-Roch algebra and Donald Knutson's lambda-rings and the representation theory of the symmetric group.

I wouldn't be surprised if one or more of these books are borrowed out, probably all of them to the same person. I'm afraid I'm that person in Antwerp...

Lately, there's been a renewed interest in $\lambda $-rings and the endo-functor W assigning to a commutative algebra its ring of big Witt vectors, following Borger's new proposal for a geometry over the absolute point.

However, as Hendrik Lenstra writes in his 2002 course-notes on the subject Construction of the ring of Witt vectors : "The literature on the functor W is in a somewhat unsatisfactory state: nobody seems to have any interest in Witt vectors beyond applying them for a purpose, and they are often treated in appendices to papers devoting to something else; also, the construction usually depends on a set of implicit or unintelligible formulae. Apparently, anybody who wishes to understand Witt vectors needs to construct them personally. That is what is now happening to myself."

Before doing a series on Borger's paper, we'd better run through Lenstra's elegant construction in a couple of posts. Let A be a commutative ring and consider the multiplicative group of all 'one-power series' over it $\Lambda(A)=1+t A[[t]] $. Our aim is to define a commutative ring structure on $\Lambda(A) $ taking as its ADDITION the MULTIPLICATION of power series.

That is, if $u(t),v(t) \in \Lambda(A) $, then we define our addition $u(t) \oplus v(t) = u(t) \times v(t) $. This may be slightly confusing as the ZERO-element in $\Lambda(A),\oplus $ will then turn be the constant power series 1...

We are now going to define a multiplication $\otimes $ on $\Lambda(A) $ which is distributively with respect to $\oplus $ and turns $\Lambda(A) $ into a commutative ring with ONE-element the series $~(1-t)^{-1}=1+t+t^2+t^3+\ldots $.

We will do this inductively, so consider $\Lambda_n(A) $ the (classes of) one-power series truncated at term n, that is, the kernel of the natural augmentation map between the multiplicative group-units $~A[t]/(t^{n+1})^* \rightarrow A^* $.
Again, taking multiplication in $A[t]/(t^{n+1}) $ as a new addition rule $\oplus $, we see that $~(\Lambda_n(A),\oplus) $ is an Abelian group, whence a $\mathbb{Z} $-module.

For all elements $a \in A $ we have a scaling operator $\phi_a $ (sending $t \rightarrow at $) which is an A-ring endomorphism of $A[t]/(t^{n+1}) $, in particular multiplicative wrt. $\times $. But then, $\phi_a $ is an additive endomorphism of $~(\Lambda_n(A),\oplus) $, so is an element of the endomorphism-RING $End_{\mathbb{Z}}(\Lambda_n(A)) $. Because composition (being the multiplication in this endomorphism ring) of scaling operators is clearly commutative ($\phi_a \circ \phi_b = \phi_{ab} $) we can define a commutative RING $E $ being the subring of $End_{\mathbb{Z}}(\Lambda_n(A)) $ generated by the operators $\phi_a $.

The action turns $~(\Lambda_n(A),\oplus) $ into an E-module and we define an E-module morphism $E \rightarrow \Lambda_n(A) $ by $\phi_a \mapsto \phi_a((1-t)^{-1}) = (1-at)^{-a} $.

All of this looks pretty harmless, but the upshot is that we have now equipped the image of this E-module morphism, say $L_n(A) $ (which is the additive subgroup of $~(\Lambda_n(A),\oplus) $ generated by the elements $~(1-at)^{-1} $) with a commutative multiplication $\otimes $ induced by the rule $~(1-at)^{-1} \otimes (1-bt)^{-1} = (1-abt)^{-1} $.

Explicitly, $L_n(A) $ is the set of one-truncated polynomials $u(t) $ with coefficients in $A $ such that one can find elements $a_1,\ldots,a_k \in A $ such that $u(t) \equiv (1-a_1t)^{-1} \times \ldots \times (1-a_k)^{-1}~mod~t^{n+1} $. We multiply $u(t) $ with another such truncated one-polynomial $v(t) $ (taking elements $b_1,b_2,\ldots,b_l \in A $) via

$u(t) \otimes v(t) = ((1-a_1t)^{-1} \oplus \ldots \oplus (1-a_k)^{-1}) \otimes ((1-b_1t)^{-1} \oplus \ldots \oplus (1-b_l)^{-1}) $

and using distributivity and the multiplication rule this gives the element $\prod_{i,j} (1-a_ib_jt)^{-1}~mod~t^{n+1} \in L_n(A) $.
Being a ring-qutient of $E $ we have that $~(L_n(A),\oplus,\otimes) $ is a commutative ring, and, from the construction it is clear that $L_n $ behaves functorially.

For rings $A $ such that $L_n(A)=\Lambda_n(A) $ we are done, but in general $L_n(A) $ may be strictly smaller. The idea is to use functoriality and do the relevant calculations in a larger ring $A \subset B $ where we can multiply the two truncated one-polynomials and observe that the resulting truncated polynomial still has all its coefficients in $A $.

Here's how we would do this over $\mathbb{Z} $ : take two irreducible one-polynomials u(t) and v(t) of degrees r resp. s smaller or equal to n. Then over the complex numbers we have
$u(t)=(1-\alpha_1t) \ldots (1-\alpha_rt) $ and $v(t)=(1-\beta_1) \ldots (1-\beta_st) $. Then, over the field $K=\mathbb{Q}(\alpha_1,\ldots,\alpha_r,\beta_1,\ldots,\beta_s) $ we have that $u(t),v(t) \in L_n(K) $ and hence we can compute their product $u(t) \otimes v(t) $ as before to be $\prod_{i,j}(1-\alpha_i\beta_jt)^{-1}~mod~t^{n+1} $. But then, all coefficients of this truncated K-polynomial are invariant under all permutations of the roots $\alpha_i $ and the roots $\beta_j $ and so is invariant under all elements of the Galois group. But then, these coefficients are algebraic numbers in $\mathbb{Q} $ whence integers. That is, $u(t) \otimes v(t) \in \Lambda_n(\mathbb{Z}) $. It should already be clear from this that the rings $\Lambda_n(\mathbb{Z}) $ contain a lot of arithmetic information!

For a general commutative ring $A $ we will copy this argument by considering a free overring $A^{(\infty)} $ (with 1 as one of the base elements) by formally adjoining roots. At level 1, consider $M_0 $ to be the set of all non-constant one-polynomials over $A $ and consider the ring

$A^{(1)} = \bigotimes_{f \in M_0} A[X]/(f) = A[X_f, f \in M_0]/(f(X_f) , f \in M_0) $

The idea being that every one-polynomial $f \in M_0 $ now has one root, namely $\alpha_f = \overline{X_f} $ in $A^{(1)} $. Further, $A^{(1)} $ is a free A-module with basis elements all $\alpha_f^i $ with $0 \leq i < deg(f) $.

Good! We now have at least one root, but we can continue this process. At level 2, $M_1 $ will be the set of all non-constant one-polynomials over $A^{(1)} $ and we use them to construct the free overring $A^{(2)} $ (which now has the property that every $f \in M_0 $ has at least two roots in $A^{(2)} $). And, again, we repeat this process and obtain in succession the rings $A^{(3)},A^{(4)},\ldots $. Finally, we define $A^{(\infty)} = \underset{\rightarrow}{lim}~A^{(i)} $ having the property that every one-polynomial over A splits entirely in linear factors over $A^{(\infty)} $.

But then, for all $u(t),v(t) \in \Lambda_n(A) $ we can compute $u(t) \otimes v(t) \in \Lambda_n(A^{(\infty)}) $. Remains to show that the resulting truncated one-polynomial has all its entries in A. The ring $A^{(\infty)} \otimes_A A^{(\infty)} $ contains two copies of $A^{(\infty)} $ namely $A^{(\infty)} \otimes 1 $ and $1 \otimes A^{(\infty)} $ and the intersection of these two rings in exactly $A $ (here we use the freeness property and the additional fact that 1 is one of the base elements). But then, by functoriality of $L_n $, the element
$u(t) \otimes v(t) \in L_n(A^{(\infty)} \otimes_A A^{(\infty)}) $ lies in the intersection $\Lambda_n(A^{(\infty)} \otimes 1) \cap \Lambda_n(1 \otimes A^{(\infty)})=\Lambda_n(A) $. Done!

Hence, we have endo-functors $\Lambda_n $ in the category of all commutative rings, for every number n. Reviewing the construction of $L_n $ one observes that there are natural transformations $L_{n+1} \rightarrow L_n $ and therefore also natural transformations $\Lambda_{n+1} \rightarrow \Lambda_n $. Taking the inverse limits $\Lambda(A) = \underset{\leftarrow}{lim} \Lambda_n(A) $ we therefore have the 'one-power series' endo-functor
$\Lambda~:~\mathbf{comm} \rightarrow \mathbf{comm} $
which is 'almost' the functor W of big Witt vectors. Next time we'll take you through the identification using 'ghost variables' and how the functor $\Lambda $ can be used to define the category of $\lambda $-rings.