# 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!

• What does "of algebraically equal status" mean? Feb 26 '16 at 8:30
• I think I better phrasing would be to just say that you are looking for a concrete "reason" that the Galois group cannot be all of $S_n$ in some example. Feb 26 '16 at 8:52
• There is no clear definition and it depends on as you feel it. I don't know how to express it. But I have some feeling that they are not equal algebraically. For example, consider irreducible polynomial $p(t)=t^4-2\in \mathbb{Q}[t]$. Let $\alpha=2^{\frac{1}{4}}$, $\beta=2^{\frac{1}{4}}i$, $\gamma=-2^{\frac{1}{4}}$. Then $\alpha$, $\beta$, $\gamma$ are all roots of $p(t)$, but their reciprocal relation over $\mathbb{Q}$ is a bit different. For example, $\alpha+\gamma=0$ whereas $(\frac{\alpha}{\beta})^2+1=0$. Feb 26 '16 at 8:53
• The distinctive status of each roots I think is the whole relationship of each roots of some irreducible polynomial with other elements in $\bar{\mathbb{Q}}$ over $\mathbb{Q}$. Feb 26 '16 at 8:59

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 $\alpha_1=\sqrt{2}$, $\alpha_2=-\sqrt{2}$, $\alpha_3=i\sqrt{2}$, and $\alpha_4=-i\sqrt{2}$. 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 $F$.
• +1 Nice observation. From a group-theoretic point of view, I think the idea is that the Galois group is the dihedral group $G$ of order $8$, which acts transitively, but imprimitively on the roots. If ones sees $G$ as acting on the vertices of a square, a block system is given by the diagonals $\{ \alpha_{1}, \alpha_{2}\}$ and $\{ \alpha_{3}, \alpha_{4}\}$. Feb 26 '16 at 9:09