How can we show that all the roots of some irreducible polynomial are not of algebraically equal status? In studying Galois theory, I found that all roots of some irreducible polynomial are not of algebraically equal status, because the Galois group of some irreducible polynomial may not be full symmetric group $S_n$.

But I am searching a concrete example elucidating that all roots are not algebraically equal and should be distinct in an algebraic way.

One of my attempt is like this.
Let $p(t)$ be an irreducible separable polynomial over $F$ and $E$ its spliting field over $F$. Let $\alpha_1,\alpha_2,\cdots,\alpha_n$ are all roots of $p(t)$. Then $E=F(\alpha_1,\alpha_2,\cdots,\alpha_n)$ and we consider the tower of fields $F\le F(\alpha_1) \le F(\alpha_1,\alpha_2)\le \cdots \le F(\alpha_1,\alpha_2,\cdots,\alpha_n)$.
I guess that the $\deg (irr(\alpha_i,F(\alpha_j)))$ may differ depending on the choice of $\alpha_i$ and $\alpha_j$. If this is true, I can suggest this to support my claim that all roots are not algebraically equal. 
But I am not able to find an apt example supporting my guess.
Do you know some example verifying my guess? Or if you have any idea which helps convince that all roots are not algebraically equal, please share with me.
Thanks for reading my question and any comment will be appreciated!
 A: Sure, what you're asking for can happen.  For a simple example, take $F=\mathbb{Q}$ and $p(t)=t^4-2$.  The roots are $\alpha_1=\sqrt[4]{2}$, $\alpha_2=-\sqrt[4]{2}$, $\alpha_3=i\sqrt[4]{2}$, and $\alpha_4=-i\sqrt[4]{2}$.  Note then that $\alpha_2\in F(\alpha_1)$ (so its minimal polynomial over $F(\alpha_1)$ has degree $1$), while $\alpha_3\not\in F(\alpha_1)$ (and its minimal polynomial over $F(\alpha_1)$ has degree $2$).
However, I would object somewhat to your phrasing that this means the roots are "algebraically distinct".  It is always true that the Galois group acts transitively on the roots (as long as $p$ is irreducible): that is, for any $i$ and $j$, there is an automorphism of $E$ over $F$ that maps $\alpha_i$ to $\alpha_j$.  So you can't really distinguish $\alpha_i$ from $\alpha_j$ (at least, from the perspective of $F$).  All you can really say is that you can distinguish certain subsets of the roots from other subsets of the roots of the same size.  For instance, in the example above, the set $\{\alpha_1,\alpha_2\}$ is in a strong sense distinguishable from $\{\alpha_1,\alpha_3\}$ over $F$.
