Formal group law over $\mathbb{F}_p$ Let $p$ be a prime. For each $n > 0$ there is a unique 1-dimensional commutative formal group law $F$ over $\mathbf{Z}$, $F(X, Y) = X + Y + \dots \in \mathbf{Z}[[X, Y]]$, whose logarithm function is given by $$l(x) = \sum_{k \ge 0} \frac{x^{p^{nk}}}{p^k}.$$
Let $\bar{F} \in \mathbf{F}_p[[X, Y]]$ be the formal group over $\mathbf{F}_p$ given by reduction of $F$ modulo $p$. (Cf. Prop. 9.25 in
http://neil-strickland.staff.shef.ac.uk/courses/formalgroups/fg.pdf.) 
Is $\bar{F}$ an element of $\mathbf{F}_p[X] [[Y]]$?
Thank you for your answers.
 A: This is really a comment, but a bit too long.
The coefficient of $y$ in $F(x,y)$ is $1/l'(x)$.  If $n=1$ then $l'(x)=\sum_ix^{p^i-1}$ and $1/l'(x)$ is not polynomial mod $p$, so the answer is negative.
However, to my surprise, it seems like the answer might be positive for $n>1$.  (Note that $l'(x)=1\pmod{p}$ in that case.) For example, when $n=p=2$ one can calculate that the coefficient of $y^8$ in $\overline{F}(x,y)$ is $x^{14}+x^{20}+x^{26}+x^{56}\pmod{x^{1024}}$.  As there are no terms $x^i$ with $56<i<1024$, one might reasonably guess that there are no terms $x^i$ with $i>56$, in which case the coefficient of $y^8$ would be polynomial.  However, I do not know how to prove this.
A: Now multinomial-free, I believe that Ghassan Sarkis and I have a proof of the following
Theorem. Let $h\ge2$, and let $L(x)=x + x^{p^h}/p + x^{p^{2h}}/p^2+\cdots$ be the logarithm of the formal group $F(x,y)\in\Bbb Z_p[[x,y]]$. Then $F(x,y)\in\Bbb Z_p\{\{x\}\}[[y]]$, where $\Bbb Z_p\{\{x\}\}$ is the ring of convergent power series: those whose coefficients go to zero.
A word about this ring: it’s the completion of the polynomials with respect to the “Gauss norm”, i.e. the uniform norm on the closed unit disk; or, if you like, the $p$-adic completion of the ring of polynomials.
Since you get $\Bbb F_p[x]$ when you tensor the ring of convergent series with $\Bbb F_p$, Neil Strickland’s guess turns out to be correct, in a very strong way.
Now for an outline of the proof, which depends entirely on $L'(x)$ being a convergent series, but the proof I found depends also on the particular form of the logarithm.
(Perhaps I should say that the cognoscenti may look at all this and say, C’mon, it’s all clear ’cause the invariant differential is a convergent series, and it all drops out automatically from general facts. But I’m no cognoscente in anything, so I have to go through at least some of the motions. I add that Ghassan wonders whether the present result may be in Hazewinkel already, though in some indecipherable formulation.)
Treat $F(x,y)$ as an element of $\Bbb Z_p[[x]][[y]]$, so write it as
$$
F(x,y)=x +\sum_{m\ge 1}f_m(x)y^m\,.
$$
The aim is to show that each $f_m$ is in $\Bbb Z_p\{\{x\}\}$, not just in $\Bbb Z_p[[x]]$. The argument is by induction, starting with $f_1$, which we already know to be $1/L'(x)$, so convergent. We write out the fundamental property of the logarithm:
$$
L\bigl(F(x,y)\bigr)=L(x)+L(y)\,,
$$
and arrange the pieces differently:
$$
0=\sum_{N\ge0}\Bigl[F(x,y)^{p^{Nh}} - y^{p^{Nh}}\Bigr]\Big/p^N-L(x)\,.
$$
In the above, we want to look at the total coefficient-function of $y^s$, knowing inductively that all $f_m(x)$ for $m<s$ are in $\Bbb Z_p\{\{x\}\}$. In this, we’re not interested in the participation of any monomial with $y$-degree greater than $s$, so we may truncate, and again rearrange:
$$
-(x+\sum_{m=1}^sf_m(x)y^m)\equiv \sum_{N\ge1}\Bigl[(x+\sum_{m=1}^s f_my^m)^{p^{Nh}} - y^{p^{Nh}}\Bigr] - L(x)\pmod{y^{s+1}}\,.
$$
Now, when you look at the occurrence of $y^s$ for each piece with $N\ge1$, there’s only one of them, and lo and behold, the coefficient is $p^{N(h-1)}x^{p^{Nh-1}}$, one of the monomials in $L'(x)$. Collect them all on the other side, and get
$$
-f_s(x)L'(x) = \text{$y^s$-coefficient in}\sum_{N\ge1}\Bigl[x+\sum_{m=1}^{s-1} f_my^m\Bigr]^{p^{Nh}}\Big/p^N\,,
$$
though in case $s=p^{nh}$, one must add on the left $1/p^n$, an inconsequential change. But here, my friends, our tale is almost done.
The last display exhibits $f_s$ as a $\Bbb Q_p$-series in the series $f_1,\dots, f_{s-1}$. But look at the tail-end of the outer sum: because the degrees in $y$ are bounded, the binomial coefficients for the $p^{Nh}$-powers far overwhelm the denominators, and the total coefficients go to zero. So we know that the tail-end is convergent, just as a series of elements of $\Bbb Z_p\{\{x\}\}$. And the part before the tail-end? That is a polynomial with $\Bbb Q_p$-doefficients in the $s-1$ series in $\Bbb Z_p\{\{x\}\}$. Let’s call it $g(x)$ for the moment. We now have $-f_sL'=g$, and thus $f_s=-f_1g$, an element of $\Bbb Q_p\otimes_{\Bbb Z_p}\Bbb Z_p\{\{x\}\}$ that’s also in $\Bbb Z_p[[x]]$, since of course we know that $F$ has its coefficients in $\Bbb Z_p$. Thus $f_s(x)\in\Bbb Z_p\{\{x\}\}$, as desired.
What this result says is that there is an action of the formal group $F$ on the closed disk. Again, maybe the cognoscenti have known this all along, but I certainly didn’t. You certainly don’t expect such a thing for a random formal group, even (as here) of height greater than $1$.
