The addition of $p$-typical Witt vectors ($p$ a prime number) is given by universal polynomials $S\_n=S\_n(X\_0,\dots,X\_n;Y\_0,\dots,Y\_n)\in\mathbb{Z}[X\_0,X\_1,\dots;Y\_0,Y\_1,\dots]$ determined by the equalities 


$\Phi\_n(S\_0,\dots,S\_n)=\Phi\_n(X\_0,\dots,X\_n)+\Phi\_n(Y\_0,\dots,Y\_n)$ for all $n\ge 0$, 


where 


$\Phi\_n(T\_0,\dots,T\_n)=(T\_0)^{p^n}+p(T\_1)^{p^{n-1}}+\dots+p^nT\_n$. 


I guess that whoever sees the Witt vectors for the first time writes down explicitly $S\_0=X\_0+Y\_0$, $S\_1=X\_1+Y\_1+\frac{1}{p}((X\_0)^p+(Y\_0)^p-(X\_0+Y\_0)^p)$, maybe $S\_2$ if she/he is courageous, and then stops since the computation becomes extremely messy. I think that there is no reasonable explicit expression in general, but patterns seem to exist and my question is about making these patterns more precise. Before I ask, let me illustrate with $S\_2$. It is easy to see that there exists a unique sequence of polynomials $R\_n\in\mathbb{Z}[X,Y]$, $n\ge 0$, such that 


$X^{p^n}+Y^{p^n}=R\_0(X,Y)^{p^n}+pR\_1(X,Y)^{p^{n-1}}+\dots+p^nR\_n(X,Y)$. 


For example $R\_0=X+Y$ and $R\_1=\frac{1}{p}(X^p+Y^p-(X+Y)^p)$. Then: 


$S\_1=R\_0(X\_1,Y\_1)+R\_1(X\_0,Y\_0)$ 


$S\_2=R\_0(X\_2,Y\_2)+R\_1(X\_1,Y\_1)+R\_1(R\_0(X\_1,Y\_1),R\_1(X\_0,Y\_0))+R\_2(X\_0,Y\_0)$. 



> 
> Can someone make the shape of $S\_n$ more precise, e.g. in the form $S\_n=P\_0+\dots+P\_n$ presumably with $P\_0=R\_0(X\_n,Y\_n)$, $P\_n=R\_n(X\_0,Y\_0)$? The intermediary $P\_i$'s are more complicated but should be (uniquely) determined by a condition of the type  
> 
>  "$P\_i$ is an iterated composition of $R\_i$ involving only the variables $X\_0,\dots,X\_i$".  
> 
>  Maybe the polynomials $P\_i$ should be homogeneous w.r.t. some graduation. 
> 
> 
> 


Any hint or relevant reference will be appreciated. Thanks!



Polynomials for addition in the Witt vectors
I found a formula for $S\_n$ in terms of the previous $S\_i$'s and the polynomials $R\_i$, which I'm quite happy about. In fact, I need the multivariate version of the $R\_i$, which I will construct all at the same time. Let us consider the ring of formal power series in countably many variables $X\_1,X\_2,\dots$ with integer coefficients, that is $A=\mathbb{Z}[[X\_1,X\_2,\dots]]$. Note that there are several notions of power series in infinitely many variables; mine is that of Bourbaki, where the underlying module of $A$ is just the product of copies of $\mathbb{Z}$ indexed by (finitely supported!) multiindices. (Some people require that the homogeneous components of a power series be polynomials; this is not the case here.) Then one can see that there exists a unique sequence $(R\_n)$ of elements of $A$ such that for all $n\ge 0$ we have 


$X\_1^{p^n}+X\_2^{p^n}+\dots=R\_0^{p^n}+pR\_1^{p^{n-1}}+\dots+p^nR\_n$ 


where on the left is the sum of all $p^n$-th powers of the variables. This is a straightforward application of Bourbaki, Algèbre Commutative, Chapitre IX, $\S~1$, no 2, prop. 2, c) with the endomorphism $\sigma:A\to A$ defined by $\sigma(X\_i)=X\_i^p$ for all $i$. Now if we have finitely many (say $s$) variables, then we set $R\_n(X\_1,\dots,X\_s)=R\_n(X\_1,\dots,X\_s,0,0,\dots)$. Examples: 


$R\_0(X\_1,\dots,X\_s)=X\_1+\dots+X\_s$ and $R\_1(X\_1,\dots,X\_s)=\frac{X\_1^p+\dots+X\_s^p-(X\_1+\dots+X\_s)^p}{p}$. 


Assuming that the $R\_n$ are computable, I have an inductive recipe for $S\_n$ which is interesting because it shows that all the $p$-adic congruences implying integrality of the $S\_n$ are contained in the $R\_n$. The recipe goes like this. For each $i$, the polynomial $S\_i$ is a sum of $2i$ terms (it will be obvious below what these terms are) and assuming $S\_1,\dots,S\_{n-1}$ are known then 


$S\_n=R\_0Z\_n+R\_1Z\_{n-1}+\dots+R\_nZ\_0+R\_1S\_{n-1}+R\_2S\_{n-2}+\dots+R\_{n-1}S\_1$ 


where: $Z\_j$ is short for the pair of variables $(X\_j,Y\_j)$, $R\_iZ\_j$ is short for $R\_i(X\_j,Y\_j)$ (the bivariate $R\_i$) and $R\_iS\_j$ is the ($2j$-variate) polynomial $R\_i$ evaluated at the $2j$ terms of $S\_j$. I hope the following examples make it clear what this means, and how efficient it is: 


$S\_1 = R\_0 Z\_1 + R\_1 Z\_0$ 


$S\_2=R\_0 Z\_2 + R\_1 Z\_1 + R\_2 Z\_0 + R\_1 (R\_0 Z\_1 , R\_1 Z\_0 )$ 


$S\_3 = R\_0Z\_3+R\_1Z\_2+R\_2Z\_1+R\_3Z\_0 \\ \quad
+ R\_1(R\_0Z\_2,R\_1Z\_1,R\_2Z\_0,R\_1(R\_0Z\_1,R\_1Z\_0))+ R\_2(R\_0Z\_1,R\_1Z\_0)$ 


$S\_4 = R\_0Z\_4+R\_1Z\_3+R\_2Z\_2+R\_3Z\_1+R\_4Z\_0 \\ \quad
+ R\_1(R\_0Z\_3,R\_1Z\_2,R\_2Z\_1,R\_3Z\_0,R\_1(R\_0Z\_2,R\_1Z\_1,R\_2Z\_0,R\_1(R\_0Z\_1,R\_1Z\_0)),R\_2(R\_0Z\_1,R\_1Z\_0)) \\ \quad
+ R\_2(R\_0Z\_2,R\_1Z\_1,R\_2Z\_0,R\_1(R\_0Z\_1,R\_1Z\_0)) \\ \quad
+ R\_3(R\_0Z\_1,R\_1Z\_0)$ 


The proof that the recipe is correct is an exercise.


