Suppose I have two polynomials $P(x,y)$ and $Q(x,y)$. When I compute the resultant of those two polynomials with respect to the variable $y$, I get a polynomial $f(x)$. Now I need to find the roots of $f(x)$, which I have. Let's say the roots are $\alpha_1$ and $\alpha_2$.
I'd like to do the same for the variable $y$, that is, find the roots of $\mathrm{res}_x(P,Q)$. However, when computing this last resultant, I get $0$. How do I interpret that result?
I've also tried to find the common roots of $P(\alpha_i, y)$ and $Q(\alpha_i, y)$ for $i=1,2$. For $\alpha_1$, I get $0$ no matter what the value for $y$ is, and for $\alpha_2$, the two polynomials have no common roots. I also don't know how to interpret this resulta.
So far, I've only seen that the resultant of two polynomials is zero if and only if the both share a common root, but that was the case when only one variable $x$ appeared.
If $\mathrm{res}_x (P(x, y), Q(x, y)) = 0$, then $P(x, y)$ and $Q(x, y)$ as elements of $R[y][x]$ have a common divisor of positive degree in $R[y][x]$, for example $P(x, y) = (x - C(y)) P_1(x, y)$ and $Q(x, y) = (x - C(y)) Q_1(x, y)$ for some $C(y) \in R[y]$. In particular, in this case it would follow that $P(x, \beta)$ and $Q(x, \beta)$ share a common root in $R$ for any $\beta \in R$, namely $C(\beta)$.
For example, consider $P(x, y) = x (y - x)$ and $Q(x, y) = x y (y - 1)$. Then $P(x, y)$ and $Q(x, y)$ as elements of $R[y][x]$ are both divisible by $x$ (which has degree $1$), so $\mathrm{res}_x (P(x, y), Q(x, y)) = 0$. But as elements of $R[x][y]$, they have no common divisor. For which $\alpha \in R$ do $P(\alpha, y)$ and $Q(\alpha, y)$ have a common divisor? Either for $\alpha = 0$, in which case they are both zero, or for $\alpha = 1$, in which case they are both divisible by $(y - 1)$. And indeed you can see that $\mathrm{res}_y (P(x, y), Q(x, y)) = x^4 (x - 1)$.