Prove that for every non-zero polynomial $p$ (i.e there is a non-zero coefficient) in $3$ variables with real coefficients there are at most degree $p$ planes contained in $Z_p$. (where $Z_p$ is the set of zeros of $p$)
Attempt Here I stated that, let $p: \mathbb R^3 \rightarrow \mathbb R$ be a polynomial in $3$ variables such that: $$p(x,y,z) = \sum_{i,j,k} C_{i,j,k} x^i y^j z^k $$ And the zero set of $p$ is defined as: $$Z_p = \{ (x,y,z) \in \mathbb R^3 : p(x,y,z)=0 \}$$ I am stuck here. Please any help or hint on how to prove the problem is appreciated. Thanks !
Any plane is given by a single equation $h(x,y,z) = ax + by + cz + d = 0$. Let us fix one such plane, which we assume is contained in $Z_p$ as you call it. Now w.l.o.g we assume $a \neq 0$ (basically the same proof works with b or c ..). Note that we can now write $p = hq + r(y,z)$. Now claim is $r(y,z) = 0$ or equivalently $p = hq$. To see this observe that for any $(\alpha,\beta)$, we have $(-\frac{b\alpha+c\beta + d}{a}, \alpha,\beta)$ is a point that belongs to the plane $h = 0$ and hence also in $Z_p$,as we have assumed this plane is inside $Z_p$. Thus plugging in $(-\frac{b\alpha+c\beta + d}{a}, \alpha,\beta)$ in $p - hq + r$, we get $r(\alpha,\beta) = 0$. Hence $r$ has to be the zero polynomial. So now we know $h$ divides $p$. Now note that we can factorize multivariable polynomials and linear polynomials are irreducible. So it follows that if $k$ planes are contained in $Z_p$ with corresponding linear polynomials $h_i$, then $h_1h_2 \cdots h_k$ divides $p$ and by considering the degrees, we must have $k \leq deg(p)$