What are the roots of an irreducible $p\in F[X]$ in the field extension $F[X]/\langle p\rangle$?

Question

We know the following fact from field theory.

Let $F$ be a field and $p(X)$ an irreducible polynomial in $F[X]$. Then we can find a field extension $L$ of $F$ such that $p(X)$ has a root in $L$.

Proof: Clearly the ideal $\langle p(X)\rangle$ generated by $p(X)$ is maximal in $F[X]$. So $F[X]/\langle p(X)\rangle$ is a field. Let us denote it by $L$. So $L$ is a field extension of $F$ since we have $F \hookrightarrow F[X] \xrightarrow{\rm \pi} L$ in a natural way. Note that $F[X] \leq L[X]$ and $\pi(X)$ is a root of $p(X)$ in $L[X]$.

1. Is it true that $\pi(X)$ is the unique root of $p(X)$ in $L$ ?
2. What is the relation between $\pi(X)$ and other roots of $p(X)$ if $deg(p) > 1$ ?

Thanks!

Answer 1

What you constructed is a rupture field for the irreducible polynomial; it has dimension $n=\deg p$ as vector space over the ground field $F$. A minimal extension field where $p$ has $n$ roots (if we suppose it to be separable) is called a splitting field for$~p$; it's dimension is some divisor of $n!$, so it can be considerably larger than a rupture field. However when $n=2$ one has $n!=n$, so the rupture field is now the same as the splitting field, and one will have both roots in the rupture field.

For $n>2$ it depends very much on the polynomial whether the rupture field contains any roots of$~p$ other than the one that was constructed. Generically the answer is no (and the splitting field has dimension$~n!$) but specific polynomials are not generic. Any other roots of $p$ in $L$ can be found by applying automorphisms of $L/F$, but again there may be only the trivial automorphism. Computing Galois groups and their action is not easy.

Answer 2

1. is most definitely not true. Consider $\mathbb{Q}[X]/(X^2+1)$ for instance. $\pi(-X)$ is also a solution. In fact if the extension is 'galois' then it's most definitely false (if deg $p(X)>1 $ anyways. Actually it's part of definition.)

2. If the extension is Galois then all roots of p(x) must exist in $L$ (sorry I can't remember if that extension is necessarily galois; it's been a long time since I took a course in Galois theory) Then the Galois group act transitively on the sets of all roots of $p(X)$.