The roots after constructing an extension field by Kronecker's theorem

Question

Kronecker's theorem said that if $p(x)$ is irreducible over $F$, then $F[x]/\langle p(x)\rangle$ contains a root of $p(x)$. However, how many roots of $p(x)$ can it contain? For example, if $\deg p(x)=2$, will $F[x]/\langle p(x)\rangle$ contains both two roots? Why or why not? How can I get the further information about the roots of $p(x)$ after constructing an extension field? I'm stucking in such early steps in Galois theory. Need some help.

Answer 1

$\def\Gal{\operatorname{Gal}}\def\Stab{\operatorname{Stab}}\def\Aut{\operatorname{Aut}}\def\Fix\operatorname{Fix}$Here is the Galois theoretic answer.

If we consider the Galois group $G := \Gal(p)$ to be a permutation group on the roots $\lbrace\alpha_1,\ldots,\alpha_d\rbrace$ of $p$, then $p$ being irreducible is equivalent to $G$ being transitive. Further, the subgroup $H := \Stab_G(\alpha_1)$ has fixed field $E := F(\alpha_1) \cong F[x]/\langle p(x) \rangle$. Suppose $\alpha_i \in E$ is another root lying in $E$, then $E = F(\alpha_i)$, and this is the fixed field of $\Stab_G(\alpha_i)$. We deduce that $H = \Stab_G(\alpha_i)$, or to put it another way, $\alpha_i$ is a fixed point of $H$. We deduce:

The number of roots of $p$ lying in $E$ is the number of fixed points of $H$.

Also note that if $\alpha_i \in E$ is a root, then we have an automorphism $\sigma \in \Aut(E/F)$ defined by $\sigma(\alpha_i) = \alpha_1$. So the number of roots of $p$ in $E$ equals the size of the automorphsim group:

$$|\lbrace \alpha_i \in E\rbrace| = |\Fix(\Stab_G(\alpha_1))| = |\Aut(E/F)|$$

and in fact $$\Aut(E/F) \cong N_G(H)/H.$$

For $p$ quadratic, the only transitive permutation group of degree 2 is $S_2 = C_2 = \lbrace \mathrm{id}, (1\,2) \rbrace$. Then $H = \Stab_{S_2}(1)$ is the trivial subgroup $\lbrace \mathrm{id} \rbrace$, and $\Fix(H) = \lbrace 1, 2 \rbrace$ has two elements, and therefore $E$ contains both roots of $p$.

This latter fact you can see more directly: if $\alpha_1$ is a root of $$p(x) = x^2 - ax + b = (x-\alpha_1)(x-\alpha_2)$$ then $\alpha_2 = a-\alpha_1 = b/\alpha_1$ shows that $\alpha_2 \in F(\alpha_1) = E$.

For higher degree polynomials, different things can happen depending on the Galois group. For instance, in degree 3, the Galois group is either $S_3$ or $C_3$. In the first case, $E$ contains only $\alpha_1$; in the second case, $E$ contains all three roots of $p$.