working out a product of polynomials

Question

Suppose $u(x,y)$ is a polynomial in two variables $x$ and $y$ over a field $GF(q)$ containing a primitive $s$-th root of unity $\zeta$. If one defines $$u^*(x,y)=u(x,y)u(x,\zeta y)u(x,\zeta^2 y)\ldots u(x,\zeta^{s-1} y)$$ then clearly $u^*(x,\zeta y)=u^*(x,y)$ which suggests that $u^*(x,y)$ only has terms in which the power of $y$ is a multiple of $s$. How can one prove that? According to the text I am reading, if $u^*(x,y)$ has a term $$ u_{i_0j_0}x^{i_0}y^{j_0}.u_{i_1j_1}x^{i_1}(\zeta y)^{j_1}....u_{i_{s-1}j_{s-1}}x^{i_{s-1}}(\zeta^{s-1} y)^{j_{s-1}}$$ than it also should have the term $$ u_{i_0j_0}x^{i_0}(\zeta y)^{j_0}.u_{i_1j_1}x^{i_1}(\zeta^2 y)^{j_1}....u_{i_{s-1}j_{s-1}}x^{i_{s-1}}y^{j_{s-1}}$$ which is the former multiplied by $\zeta^{j_0+j_1+\ldots+j_{s-1}}$ and so on. However, if I try to work out a simple example, like, say $$ (1+x^2+xy+y^2)(1+x^2+x\zeta y+\zeta^2y^2)(1+x^2+x\zeta^2 y+\zeta^4y^2)$$ things don't work out the way they should.

Answer 1

I originally said: ''$u^*$ is fixed under the action of $\mathrm{Gal}(\mathbb{F}_q/\mathbb{F}_p(\zeta^s))$ and so lies in $\mathbb{F}_p(\zeta^s)$'', which explains the commentary below. What I meant so say was: ''$u^*$ is fixed under the action of $\mathrm{Gal}(\mathbb{F}_q(x,y)/\mathbb{F}_q(x,y^s))$, which is generated by $y\mapsto \zeta y$, and so lies in $\mathbb{F}_q(x,y^s)$''.

Answer 2

After taking a couple of antidepressants, I finally had the courage to think it through. If $u(x,y)$ has $m$ terms, then $u^*(x,y)$ will have $m^s$ terms (all possible combinations of terms in all factors). If $S= \{1,2,\ldots,m\}^s$ then, on $S$, define the relation $(i_0,i_1,\ldots,i_{s-1})\sim (j_0,j_1,\ldots,j_{s-1}) $ if and only if $(j_0,j_1,\ldots,j_{s-1}) $ can be obtained from $(i_0,i_1,\ldots,i_{s-1})$ by a finite number of consecutive cyclic shifts. Since $(i_0,i_1,\ldots,i_{s-1})$ will turn into itself after a finite number of consecutive cyclic shifts, this relation is reflexive and symmetric. Obviously, it's transitive as well, hence an equivalence relation. Consider the equivalence classes in $S$. First type is a singleton $\{(i,i,\ldots,i)\}$. This will give rise to terms of the form $$u_{i_0j_0}x^{i_0}y^{j_0}.u_{i_0j_0}x^{i_0}(\zeta y)^{j_0}.\ldots.u_{i_0j_0}x^{i_0}(\zeta^{s-1}y)^{j_0}=ax^{si_0}y^{sj_0}$$ The second type is $\{(i_0,i_1,\ldots,i_{s-1}),(i_{s-1},i_0,\ldots,i_{s-2}),\ldots,(i_1,\ldots,i_{s-1},i_0)\}$. This will give rise to the sum of terms $$u_{i_0j_0}x^{i_0}y^{j_0}.u_{i_1j_1}x^{i_1}(\zeta y)^{j_1}.\ldots.u_{i_{s-1}j_{s-1}}x^{i_{s-1}}(\zeta^{s-1}y)^{j_{s-1}}+u_{i_{s-1}j_{s-1}}x^{i_{s-1}}y^{j_{s-1}}.u_{i_0j_0}x^{i_0}(\zeta y)^{j_0}.\ldots.u_{i_{s-2}j_{s-2}}x^{i_{s-2}}(\zeta^{s-1}y)^{j_{s-2}}+\ldots+u_{i_{1}j_{1}}x^{i_{1}}y^{j_{1}}.\ldots.u_{i_{s-1}j_{s-1}}x^{i_{s-1}}(\zeta^{s-2}y)^{j_{s-1}}.u_{i_0j_0}x^{i_0}(\zeta^{s-1} y)^{j_0}$$ In the latter sum, the second term is the first multiplied by $\zeta^{j_0 + j_1 + \ldots + j_{s-1}}=\zeta^{j}$ if we put $j=j_0 + j_1 + \ldots + j_{s-1}$, the third term is the second multiplied by $\zeta^{j}$, thus the first one multiplied by $\zeta^{2j}$ and so on, hence we get $$bx^{i_0 + i_1 + \ldots + i_{s-1}}y^j(1+ \zeta^j + \zeta^{2j}+\ldots+\zeta^{(s-1)j})=bx^{i_0 + i_1 + \ldots + i_{s-1}}y^j(1+ \zeta^j + (\zeta^{j})^2+\ldots+(\zeta^{j})^{s-1})$$ Since $\zeta$ is a primitive $s$-th root of unity $1+ \zeta^j + (\zeta^{j})^2+\ldots+(\zeta^{j})^{s-1}=0$ if $j$ is not a multiple of $s$. We are done.