# Explanation about extensions between homomorphism

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.

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.