Is there any possibility to get $\Gamma \subset \Lambda$?

Question

Let $f,g$ be $n$-degree and $m$-degree polynomials over $\mathbb C$ respectively, with $n>m$.

• Set $\Lambda=\{x \in \mathbb C~:~ f(x)=0 \}$ and $\Gamma=\{x \in \mathbb C~:~ g(x)=0 \}$.

• Assume $h \circ f=g \circ h$, where $h:~\Lambda \to \Gamma$ is a surjection.

Is there any possibility to get $\Gamma \subset \Lambda$ ?

Let $\gamma \in \Gamma$, we need to show $g(\gamma)=0$.

By the surjective property of $h$ for any $\lambda \in \Lambda$ there is some some $\alpha \in \Gamma$ so that $h(\alpha)=\lambda$ but this does not help me.

Any comments please.

Answer 1

Let $\lambda \in \Lambda$, which means $f(\lambda) = 0$. The equality

$$h \circ f = g \circ h$$ translates into

$$h(0) = h(f(\lambda)) = (h \circ f)(\lambda) = (g \circ h)(\lambda) = g(h(\lambda)) =0$$ as $h(\lambda) \in \Gamma$. We get that $0$ belongs to $\Lambda$ and $h(0)=0$.

Now clearly, we need not have $\Lambda \subset \Gamma$. And any $f,g$ such that $f(0)=g(0) =0$ where $g$ doesn't divide $f$ will provide a counterexample. Like $f(x) = x(x-1)(x-2)$ and $g(x) = x(x-3)$ with $h(0)=0$ and $h(1)=h(2)=3$.