For the proof of a new theorem about solutions of transcendental equations of elementary functions, I could need the statement below.
How can the following statement be proved?
Let
$P_1(z,y)\in\overline{\mathbb{Q}}[z,y]$ non-constant and without univariate factor over $\overline{\mathbb{Q}}$,
$P_2(x)\in\overline{\mathbb{Q}}[x]$ non-constant.
Then a non-constant $P(x,y)\in\overline{\mathbb{Q}}[x,y]$ without univariate factor over $\overline{\mathbb{Q}}$ exists so that $P_1(P_2(x),y)=P(x,y)$.
$P(x,y)$ is non-constant and uniquely determined. I here only want to see how to prove that $P(x,y)$ has no univariate factor over $\overline{\mathbb{Q}}$.
A mathematical expression is said to be univariate if the number of different variables it contains is one.
I checked thousands of randomly generated polynomials but couldn't find any evidence to the contrary in this way.
I'm familiar with transforming polynomials, but I don't know where to start here. I'm not so familiar with polynomials of several variables.
Suppose $P(x,y)$ has a univariate factor $f(x)$ over $\overline{\mathbb{Q}}$ so that $P_1(P_2(x),y)=f(x)Q(x,y)$, where $\deg f > 0$. Then there is a root of $f$ in $\overline{\Bbb Q}$, name it $x_0$. The set $\{z = P_2(x_0)\}$ is a subset of the zero set of $P_1(z,y)$ in $\overline{\Bbb Q}^2$. Then, by Nullstellensatz, $z - P_2(x_0)$ would divide $P_1(z,y)$, this is a contradiction.
The case $P_1(P_2(x),y)=g(y)Q(x,y)$ is obvious.