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How many equivalence classes of $\Bbb F_2[x,y]$ polynomials with $x$ degree $n_x$ and $y$ degree $n_y$ are there such that each $y^i$ coefficient (polynomial in $\Bbb Z[x]$) is distinct and $x^i$ coefficient (polynomial in $\Bbb Z[y]$) is distinct?

Two polynomials are equivalent if there is a member of $S_{n_x+1}\times S_{n_y+1}$ changes permutes coefficient of one polynomial to other.

Is there a way to give the standard basis for these polynomials (are these a vector space in any sense)?

Consider $0/1$ as integers how are these classes organized under $GL_{n_x}(\Bbb R)\times GL_{n_y}(\Bbb R)$ action (muliply coefficient matrix on left and right)?

Is there a $c>0$ such that if $n_y>n_x^c$ then there is no such polynomial for large enough $n_x$, $n_y$?

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  • $\begingroup$ Are the polynomials meant to be of degree $n$? Where does the problem come from? Is this a homework question? $\endgroup$
    – Anthony Quas
    Commented Jun 15, 2016 at 20:48
  • $\begingroup$ @AnthonyQuas does it look like a hw problem? I will correct degree part. $\endgroup$
    – user76479
    Commented Jun 15, 2016 at 20:51
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    $\begingroup$ There's really no reason to call these polynomials: this is a problem about $(1+n_x) \times (1+n_y)$ arrays of values $\pm 1$ with distinct rows and distinct columns, and you count equivalence classes under permutation of rows and columns. Each equivalence class has $n_x! n_y!$ members, so the number of equivalence classes is the number of such arrays divided by $n_x! n_y!$. $\endgroup$
    – Robert Israel
    Commented Jun 15, 2016 at 21:21
  • $\begingroup$ @RobertIsrael what about the second and last problem? $\endgroup$
    – user76479
    Commented Jun 15, 2016 at 21:22
  • $\begingroup$ ... or that should be $(n_x+1)!(n_y+1)!$ if you really meant to allow permutations of all the rows and all the columns, rather than just $n_x$ and $n_y$ of them. $\endgroup$
    – Robert Israel
    Commented Jun 15, 2016 at 21:23

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