# Question about the number of roots of irreducible polinomials.

Suppose $F|K$ is a field extension and let $\alpha \in F$ be an algebraic element over $K$ and let $\beta \in K(\alpha)$. Let $f=Irr(\alpha, K)$, $g=Irr(\beta, K)$, $h=Irr(\alpha, K(\beta))$. Let $L$ be a splitting field of $fg$ over $F$. Let $n_f, n_g, n_h$ be the numbers of distinct roots of each polynomial. Show that $n_f=n_g.n_h$.

Well, I tried to write the three polynomials but I can't see exactly what to do. It's clear that the proposition is true if all the polynomials are separable. What should I do?

• you used $g$ twice. – Rene Schipperus Jun 8 '14 at 23:46
• Just fixed, thank you. – Vinicius Rodrigues Jun 8 '14 at 23:47
• now you used h twice. – Rene Schipperus Jun 8 '14 at 23:47
• Oops, I think I fixed it now. – Vinicius Rodrigues Jun 8 '14 at 23:48
• Well if your assuming separability, then the number of roots in each case is the degree of the extension, and we know this is multiplicative. – Rene Schipperus Jun 8 '14 at 23:52

I think I made a proof, but I think there are more solutions very different from this one, therefore I will leave the question open for a while :)

It follows that:

• If $\tau:K\rightarrow \bar K$ is an homomorphism and $\bar K$ is an algebraic closure of $K$, $n_f$ is the number of extensions from $\tau$ to $K(\alpha)$.
• If $\tau:K\rightarrow \bar K$ is an homomorphism and $\bar K$ is an algebraic closure of $K$, $n_g$ is the number of extensions from $\tau$ to $K(\beta)$.
• If $\tau:K(\beta)\rightarrow \overline{ K(\beta)}$ is an homomorphism and $\overline{ K(\beta)}$ is an algebraic closure of $K(\beta)$, $n_h$ is the number of extensions of $\tau$ to $K(\alpha)$.

Let $\phi: K\rightarrow \bar K$ be an homomorphism where $\bar K$ is an algebraic closure of $K$. Let $\sigma_1, \cdots \sigma_{n_g}$ be all the distinct extensions of $\phi$ to $K(\beta)$. For each $i$, $\sigma_i:K(\beta)\rightarrow \bar K$ is an homomorphism and with each of these homomorphisms, $\bar K$ is an algebraic closure of $K(\beta)$, therefore, there are exactly $n_h$ extensions of $\sigma_i$, call them $\sigma_{i, 1}, \cdots, \sigma_{i, n_h}$. Let $X=\{\sigma_{i, j}: 1\leq i \leq n_g, 1\leq j \leq n_h\}$. If $(i, j) \neq (m, n)$, then $\sigma_{i, j} \neq \sigma_{m, n}$: If $i \neq m$ then $\sigma_i \neq \sigma_m$, and therefore their extensions are also different. If $i=m$,then by construction, $\sigma_{i, j} \neq \sigma_{i, n}=\sigma_{m, n}$. Therefore, $|X|=n_g.n_h$. It's also easy to see that each $\sigma_{i, j}$ extends $\phi$, since for each $k \in K$, $\sigma_{i, j}(k)=\sigma_i(k)=\phi(k)$. Therefore, $n_f\geq n_g.n_h$.

To prove the equality, it suffices to show that if $\sigma: K(\alpha)\rightarrow \bar K$ extends $\phi$ then $\sigma$ is one of the $\sigma_{i, j}$.

Pick such $\sigma$. It's clear that $\sigma|_{K(\beta)}:K(\beta)\rightarrow \bar K$ extends $\phi$, so there exists $i$ such that $\sigma|_{K(\beta)}=\sigma_i$. Since $\sigma$ extends $\sigma_i$, there exists $j$ such that $\sigma=\sigma_{i, j}$, which completes the proof.