Explicit tensor product of isocrystals Let $L$ be the completion of the maximal unramified extension of $\mathbb Q_p$. Let $\sigma$ be a topological generator for the Galois group of $L/\mathbb Q_p$. Say that $(D, \phi)$ is an isocrystal over $L$ if $D$ is a finite dimensional vector space over $L$ and $\phi$ is a $\sigma$-linear automorphism of $D$.
Let $L\langle\phi\rangle$ be the twisted polynomial ring over $L$ such that $\phi c = \sigma(c) \phi$, for $c \in L$. Then we can define explicit isocrystals by setting $$D(r/s) = L \langle \phi \rangle / (\phi^r - p^s),$$ where $r/s \in \mathbb Q$ and $gcd(r, s) = 1$. As an $L$-vector space this is of dimension $r$. It is well known that every isocrystal over $L$ is a direct sum of isocrystals of this form. Furthermore, there is a tensor product given by
$$D(r/s) \otimes D(r'/s') = (D(r/s + r'/s'))^{\oplus gcd(r, r')},$$
see for example equation 15 here.
I've been struggling to make explicit the isomorphism giving rise to this tensor product. Let us assume for simplicity that $gcd(r, r') = 1$. Then we need to find an isomorphism
$$L\langle \phi \rangle /(\phi^r - p^s) \otimes_L L\langle \phi \rangle /(\phi^{r'} - p^{s'}) \to L\langle \phi \rangle /(\phi^{rs'+r's} - p^{ss'}).$$
How is this map defined?
 A: First of all, I will denote your $D(r/s)$ by $D(s/r)$, because it has slope $s/r$. We will show that $D(s/r)\otimes D(s'/r')\simeq D(s''/r'')^d$ where $d={\rm gcd}(r,r')$, $r'' = {\rm lcm}(r,r') = rr'/d$, and $s'' = (s'r+sr')/d$, so that $$ \frac s r + \frac{s'}{r'} = \frac{s''}{r''}. $$
$D(s/r)$ has basis $e_0, \ldots, e_{r-1}$ in which $\phi$ acts by $e_i \mapsto e_{i+1}$ if $i+1<r$ and $e_{r-1}\mapsto p^s e_0$. So the tensor product $D(s/r)\otimes D(s'/r')$ has basis $e_i\otimes e_{i'}$ ($0\leq i<r$, $0\leq i' < r'$) with $\phi$ acting by
$$ 
\phi(e_i\otimes e_{i'}) = \begin{cases}
e_{i+1}\otimes e_{i'+1} & i+1<r, \quad i'+1<r' \\
p^s e_0\otimes e_{i'+1} & i+1=r, \quad i'+1<r' \\
p^{s'} e_{i+1}\otimes e_0 & i+1=r, \quad i'+1<r' \\
p^{s+s'} e_0\otimes e_0 & i+1=r, \quad i'+1=r'
\end{cases}
$$
Keep applying this to $e_0\otimes e_0$ and see what happens. After $r''$ steps we will get $p^{a+a'} e_0\otimes e_0$ where $a$ is $s$ times the number the segment connecting $(0,0)$ and $(r'',r'')$ will intersect the vertical lines $r\mathbf{Z}\times \mathbf{R}$, and similarly for $a'$. A moment's thought gives $a=s\cdot r''/r = sr'/d$ and $a'=s'r/d$, so $s''=a+a'=(sr'+s'r)/d$. We can therefore change the basis by replacing $e_i\otimes e_{i'}$ with $p^{a_{ii'}}e_i\otimes e_{i'}$ for a suitably chosen exponent $p^{a_{ii'}}$ and then group them into $g$ groups corresponding to $D(s''/r'')$.
To give an explicit isomorphism, for $0\leq k<d$ (labelling the copy of $D(s''/r'')$ and $0\leq i''<r''$, set $e^k_{i''} = \phi^{i''}(e_0\otimes e_k)$. Using the notation $k\% r$ for the unique element of $k+r\mathbf{Z}$ in $\{0, \ldots, r-1\}$, we have $e^k_{i''} = p^{a_{i''}} e_{(k+i'')\% r}\otimes e_{i''\% r'}$ for some easy to compute exponent $a_{i''}$. Since $\phi(e^k_{i''}) = e^k_{i''+1}$ for $i''<r''-1$, to identify the span of $e^k_{i''}$ for fixed $k$ with $D(s''/r'')$ we only need to verify that $\phi^{r''}(e^k_0)=p^{s''}e^k_0$. This is done as in the previous paragraph.
