Explanation about extensions between homomorphism

Question

Let $f\in F[x] , E/F , \Omega/F$ are extensions of $F$ such that $E=F[\alpha_1,\alpha_2,\cdots,\alpha_n]$ $,\alpha_1,\alpha_2,\cdots,\alpha_n$ are roots of $f$ , $f$ is a splitting field over $\Omega$ as well.

Then, there are mostly $[E:F]$, $E\to \Omega$ homomorphism.

I consider $E$ is a splitting field implies there are exactly $[E:F]$ homomorphism (not entirly sure about it) I can't figure out why this statement is true.

For simplicity suppose $n=1$ and $f$ is a minimal polynomial and $\deg(f)=m$ and the root are distinct.

Hence $E=F[\alpha]$ and $[E:F]=m$.

Homomorphism $\Theta$ is defined by mapping $\alpha\in E$ to a root in $\Omega$.

Since there are $m$ different roots there are $m$ options hence there are $[E:F]$ different homomorphism.

In a different case $n=2$ there are $5 \choose 2$ different homomorphism , my Intuition is bad since there at most $[E:F]$ different homomorphism.

I will be grateful for a explanation and maybe an example,appreiciate.

Answer 1

You are asking about how to count the number of $F$-isomorphisms between two splitting fields $E$ and $\Omega$ of a polynomial $f(x)$ in $F[x]$. (Why use such strange notation as $E$ and $\Omega$? It's like calling two groups playing similar roles $G$ and $\Lambda$.)

You ask for an explanation, but have you at least looked at a proof about this issue in a textbook? Nearly every book on Galois theory will prove the theorem you're asking about (how to count the number of isomorphisms between two splitting fields). It proceeds by induction on $\deg f$. There is a nice answer when $f(x)$ is separable (distinct roots): $E$ and $\Omega$ have the same degree over $F$ and the number of $F$-isomorphisms between them is $[E:F]$. When $f(x)$ is inseparable, the answer can be less than $[E:F]$.

What I suspect you are not paying attention to is that as soon as you decide where one root of $f$ in $E$ is mapped to in $\Omega$, that affects where other roots can be sent.

Example. Let $F = \mathbf Q$, $f(x) = x^4 - 2$, and $E = \Omega = \mathbf Q(\sqrt[4]{2},i)$ (I take $E = \Omega$ to keep things simple), so $[E:F] = 8$. The roots of $f(x)$ are $\{\pm\sqrt[4]{2},\pm i\sqrt[4]{2}\}$. If we want $\sigma(\sqrt[4]{2}) = i\sqrt[4]{2}$, then necessarily $\sigma(-\sqrt[4]{2}) = -\sigma(\sqrt[4]{2}) = -i\sqrt[4]{2}$, so there are only two possible values for $\sigma(i\sqrt[4]{2})$: $\pm\sqrt[4]{2}$. And then $\sigma(-i\sqrt[4]{2}) = -\sigma(i\sqrt[4]{2})$ is completely determined. So $\sigma$ has at most $2$ values on $E$.

In the same way, for each of the $4$ values you pick for $\sigma(\sqrt[4]{2})$, there are at most $2$ choices for $\sigma$. This means the number of $F$-isomorphisms $E \to \Omega$ is at most $8$, which equals $[E:F]$. Actually proving all $8$ options for $\sigma$ really work is done by proving the theorem (by induction on $\deg f$) that two splitting fields $E/F$ and $\Omega/F$ of a separable polynomial in $F[x]$ have $[E:F]$ $F$-isomorphisms between them. See books on field theory or Galois theory.