Witt vectors addition confusing

Question

I raise this confusing because I try to understand the witt vectors for characteristic not equal to p.

Let us assume p=2. The Witt Polynomials is explicitly given by

$$ S_0=X_0+Y_0 $$

$$ S_1=X_1+Y_1-X_0Y_0 $$

Then consider truncated Witt vectors $W_2(\mathbb{Z}[X])$ for example. Let's add two vectors

$$ (X_0,0)+(X_0,0)=(2X_0, -X_0^2) $$

But I feel it should be $2(X_0,0)=(0,X_0^2)$ If $2=0$ this equation is perfectly true, but when we are working for $\mathbb{Z}$, what is wrong?

I understand Witt vectors when working in char p, but when it generalizes to $\mathbb{Z}$, I always got a trouble like this.

Answer 1

There is nothing wrong, the point of confusion as you noted lies in the characteristic of the ground ring $R = \mathbf{Z}[X]$. When $R$ has characteristic $0$, then the ghost map sending $W(R) \to R^\mathbf{N}$ is invertible whence an isomorphism. If you want or need more details on the ghost map, recall that $W_n(\underline{x}) = \sum_{i=0}^n p^i x_i^{p^{n-i}}$ is the $n$-th Witt polynomial, and the bizarre Witt sum and multiplication polynomials arise by insisting that the function $W(R) \to R^{\mathbf{N}}$ which sends $\underline{x}$ to $(W_0(\underline{x}), W_1(\underline{x}), \ldots)$ is a ring homomorphism, where we give $R^{\mathbf{N}}$ the usual component-wise operations. The key point is that when $p$ is a unit in $R$, then this map is also a bijection which can be seen by recursively unwinding the Witt polynomials. So the weird sum formula you have is the result of summing in an isomorphic copy of the usual component wise addition and then truncating.

In any event, you shouldn't expect 'usual' the formulation of "multiplication by $p$" in the ring of $p$-typical Witt vectors when the ground ring is {\it not} of characteristic $p$.

Note that the ghost map is completely useless if $p = 0$ in $R$. This is why to derive the sum and product formulas, one first works in a $\mathbf{Q}$-algebra, to get a universally valid formulation which then may be applied in any characteristic. If you want a great introduction to this, see Jacobson's "Basic Algebra II", Section 8.10, page 497 or Serre's classic "Corps Locaux"/"Local Fields".

Answer 2

A small addendum to the existing answer to address your specific question: extracting the $0$-th component of an element of $W_2(A)$ gives a ring homomorphism $R:W_2(A) \to A$, usually called the restriction map. In particular, $R$ is additive. Since $2X \neq 0$ in your $A$, you know that $R(2(X,0)) \neq 0$, so the $0$-th component of $2(X,0)$ is not $0$.

A comment on $W_2(-)$: there is a nice description of $W_2(A)$ (always $p$-typical) when $A$ is $p$-torsionfree that does not mention the Witt polynomials at all. Indeed, for such $A$, the ring $W_2(A)$ is the fibre product of the two maps $A \xrightarrow{can} A/p$ and $A \xrightarrow{\mathrm{Frob} \circ can} A/p$. Explicitly, this means that $W_2(A)$ is the subring $$\{(x,y) \in A \times A \mid x^p \equiv y \mod pA\} \subset A \times A,$$ which makes computations rather straightforward.