I need help in showing that the converse of the theorem "if $F$ is an $(m-k)$-flat in $EG(m,2)$, the the corresponding Boolean polynomial $P_F$ has degree $k$" does not hold.
Or that is, "that there are Boolean polynomials $p(x_1,...,x_m)$ of degree $m-k$ whose corresponding subsets $F_p$ of $EG(m,2)$ are not flats."
Hint given in the book: Consider quadratic polynomials in three variables. Recall that in $EG(3,2)$, the hyperplanes correspond to the linear polynomials, so that if a quadratic polynomial does correspond to a flat, it must be a line (1-flat).
Since $EG(m,2) = \Bbb{F}_2^m$, then $m$ is the length of the codewords, or vectors, in a vector space $S$. And by definition, if $S$ is a subspace of $\Bbb{F}_2^m$ with $dim(S)=k$, then any coset $b+S$ is called a $k$-flat.
I wanted to show this proof by using an example.
Consider $EG(3,2) = \Bbb{F}_2^3$.
By a theorem, in $EG(3,2)$, the $(m-1)=(3-1)=2$-flats, or hyperplanes, correspond to the linear Boolean polynomials (degree $m-k=3-2=1$).
Thus, if there exists a quadratic (degree $2$) Boolean polynomial that corresponds to a flat, by definition, it must correspond to a $1$-flat ($m-k=2 =>3-2=k=1$).
However...
This was my sad attempt at the proof, by using this method (suggested in my textbook) I get stuck because I am unsure on how this relates to proving the statement. Should I show that the quadratic polynomial cannot correspond to a $1$-flat? And thus since it cannot correspond to a $1$-flat, any other subset it corresponds to will not be a flat?
Any help is appreciated, thank you in advanced.
I think you are way overthinking this. Stop thinking in abstract generalities and just write down an example! Try out some really simple quadratic polynomials in three variables, and see what their vanishing sets are. With a little trial and error you can find one whose vanishing set is not a flat.
In fact, I would consider the given hint kind of silly, since there are even easier counterexamples. You could instead try a polynomial of degree $0$, in any number of variables. Or, you could try a quadratic polynomial in only two variables.