For each polynomial $p \in K[t]$, there exists another polynomial $g$ such that $\{p(\epsilon_i): 1\leq i \leq n\}$ is its set of roots

I'd like to solve the following problem:

Let $K$ be a field, $f \in K[t]$ a polynomial of degree $n$ and $E$ a splitting field of $f$ over $K$ in which $f$ has $n$ distinct roots $\epsilon_1,\dots,\epsilon_n$. Show that for each polynomial $p\in K[t]$ there exists another one, $g \in K[t]$ of degree $n$, such that $\{p(\epsilon_i): 1\leq i \leq n\}$ is the set of roots of $g$.

I thought that the following statement coulp be helpful to prove it:

Let $L\vert K$ a Galois extension and consider for each $\sigma \in \mathrm{Gal}(L \vert K)$, the ring homomorphism $$\overline{\sigma}: L[t] \to L[t], \quad \sum_{i=1}^da_it^i \mapsto \sum_{i=1}^d\sigma(a_i)t^i$$ Then a polynomial $h\in K[t]$ belongs to $K[t]$ if and only if $\overline{\tau}(h) = h$ for every $\tau \in \mathrm{Gal}(L \vert K)$.

Now since $E$ is a splitting field of $f$ over $K$. The field extension $E \vert K$ is Galois. Now, let

$$g := \prod_{i=1}^n(t-p(\epsilon_i))$$

So to show that $g \in K[t]$, we need to take a $\tau \in \mathrm{Gal}(L \vert K)$ and check that $\overline{\tau}(g)=g$. So

\begin{align*} \overline{\tau}(g) & = \overline{\tau}\left( \prod_{i=1}^n(t-p(\epsilon_i))\right)= \prod_{i=1}^n\overline{\tau}((t-p(\epsilon_i))) = \prod_{i=1}^n(t-\overline{\tau}(p(\epsilon_i)))\\ & = \prod_{i=1}^n\left(t-\overline{\tau}\left(\sum_j a_j\epsilon_i^j\right)\right)=\prod_{i=1}^n\left(t-\sum_j a_j\tau(\epsilon_i)^j\right)= \prod_{i=1}^n(t-p(\tau(\epsilon_i))) \end{align*}

But I can't continue further from here, so any help would be appreciated. Thanks in advance!

• The title seems to be incomplete: "...there exists...[??]...such that..."? May 21 '16 at 17:13
• @Joanpemo Edited. "...another polynomial". Question too long for a title D: May 21 '16 at 17:15

$\tau$ permutes the roots of $f$, so $\prod_{i=1}^n (t - p(\tau(\varepsilon_i)))$ is exactly the same as $\prod_{i=1}^n (t - p(\varepsilon_i)).$