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Let $K$ be an infinite field positive characteristic and $F(X,Y)\in K[[X,Y]]$. Assume that $F(Z_1+U_1,Z_2+U_2)=F(Z_1,Z_2)+F(U_1,U_2)$ where $Z_1,Z_2,U_1,U_2$ are four indeterminates. Can one assert that $F(X,Y)=G(X)+H(Y)$ with $G,H\in K[[X]]$?

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Yes. It suffices to consider the case that $F$ is a homogeneous polynomial. Write $F(X,Y) = aX^d + Y^eP(X,Y)$, where $d = \deg(F)$ and $P$ is homogeneous of degree $d - e$, $e \geq 1$.

Claim: if $a \neq 0$, then $d$ is a power of the characteristic $p$ of $K$.

Proof: Otherwise the coefficient of $Z_1^{d-1}U_1$ in $F(Z_1+U_1, Z_2+U_2)$ is nonzero, whereas this coefficient is zero in $F(Z_1,Z_2) + F(U_1, U_2)$.

Now write $F = aX^d + bY^d + Q(X,Y)$, where both $X$ and $Y$ divide $Q$. We will show that $Q = 0$. Indeed, by the claim, $Q$ also satisfies: $Q(Z_1+U_1, Z_2+U_2) = Q(Z_1,Z_2) + Q(U_1, U_2)$. Plugging in $Z_2 = -U_2$ in the preceding identity yields that $Q(Z_1,-U_2) + Q(U_1, -U_2) = Q(Z_1+U_1, 0) = 0$. If $Q$ is a nonzero polynomial, this gives a contradiction.

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  • $\begingroup$ Thanks for the answer. But I do not see why it sufficient to consider the homogenous case. Can you explain? $\endgroup$
    – joaopa
    Commented Jan 9, 2020 at 6:15
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    $\begingroup$ if $F(Z_1 + U_1, Z_2 + U_2) = F(Z_1, Z_2) + F(U_1, U_2)$, then for each $d$, the homogeneous components of degree $d$ of LHS and RHS must be equal. $\endgroup$
    – pinaki
    Commented Jan 9, 2020 at 6:30

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