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Definition: Given a normal field extension $E/k$, two elements $x,y\in E$ are conjugates if there is a $k$-automorphism $\sigma$ on $E$ such that $\sigma(x)=y$.

Two subextensions $F_1/k, F_2/k$ ($k\subset F_i\subset E$ ) are conjugates if there is a $k$-automorphism $\sigma$ on $E$ such that $\sigma(F_1)=F_2$.

Using this definition, I want to prove that:

  1. A subextension of the normal extension $E/k$ is normal if and only if it is equal to all its conjugates.
  2. Let $E/k$ be a finite and normal field extension, $a\in E$. Let $C=\{a_1,\dots,a_n\}$ be the set of all the conjugates of $a$. For every $\sigma\in Sym_n$ there is a $k$-automorphism on $E$, say $\varphi$, such that $\varphi(a_i)=a_{\sigma(i)}$ for each $i=1,\dots,n$.
  3. If $E/k$ is normal and $f\in k[x]$ is irreducible over $k$ with a zero $\alpha\in E$, then all the zeros of $f$ are conjugates of $\alpha$.

I know that an extension is normal iff is the minimal splitting field of some family of polynomial over the ground field. I have read that 1 and 3 are true but I don't know how to prove them. But 2 is one of my conjectures. I don't know if it is true.

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I am sorry I will not give here proofs of 1 and 3 (these can be found in many textbooks). Concerning your conjecture 2, it is false. It amounts to saying that all Galois groups are isomorphic to symmetric groups, and there are lots of examples where this fails, take for instance the cyclotomic extension K of the rationals generated by primitive p-th roots of unity, for p an odd prime. Then if your element a is one such root of unity, its conjugates are powers of a and all rational automorphisms on K are determined by its value on a, thus the group of such automorphisms (i.e., the Galois group of K/Q) is a cyclic group, not the full symmetric group. Perhaps you are interested in knowing that at least the case that you propose occurs over the rationals for any n, i.e., it is known that for any n there exists a suitable normal extension K of Q such that it has degree n, it is normal, and its Galois group is the full symmetric group in n letters. You can find in the literature many proofs of this, using for instance Hilbert's irreducibility theorem. The standard conjecture, by the way, is that exactly the opposite of what you propose in 2 should happen for normal extensions of the rationals, i.e., not only there are plenty of examples where Galois groups of normal extensions of Q fail to be the full symmetric group, but also every finite group in the universe should occur as a Galois group of some normal extension of Q, so for example all complicated finite groups that appear in the classification of finite simple groups, they should all show up among Galois groups of normal extensions of Q. This problem is wide open, even it there has been a lot of progress in it through the years. So, you have a lot to think about when you think about Galois groups!

P.S.: just to add a bit more of information that you can find useful, the structure of the Galois groups (i.e., groups of k-automorphisms of your normal extension E) depends a lot of what kind of fields you are considering. If you take for example the cases where your fields are finite fields (having both k and E finitely many elements) or when they are local p-adic fields (finite extensions of the field of p-adic numbers) it is known that in these situations ALL Galois groups are solvable groups. On the other hand, symmetric groups in n letters for n at least 5 are non-solvable groups. This proves that in the cases of finite or local p-adic fields, if n is at least 5, your conjecture 2 is completely wrong in the sense that it is impossible for the group of k-automorphisms to be the full symmetric group in n letters.

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Thanks @Luisinho for answering my question 2. I am going to write the proofs of 1 and of a generalization of 3 for the people who is interested. Please someone tell me if there is something wrong with these proofs.

Let's denote $Aut_k(E)$ the set of all the $k$-automorphisms of $E$.

Proposition 1: Let $E/k$ be a Normal extension, $f\in k[x]$ be irreducible over $k$ and $\alpha\in E$ be a root of $f$. Then $\beta\in E$ is a root of $f$ if and only if $\beta$ is a conjugate of $\alpha$. In other words, the set of roots of $f$ is precisely the set of conjugates of $\alpha$.

This proposition generalizes and answer my question 3.

Proof: Let $\sigma$ be in $Aut_k(E)$. If $f(\alpha)=0$ then $f(\sigma(\alpha))=\sigma(f(\alpha))=0$. Therefore every conjugate of $\alpha$ is a root of $f$ in $E$. Let $\beta\in E$ be a root of $f$. Because $f$ is irreducible over $k$, there is an k-isomorphism $\varphi :k(\alpha) \to k(\beta )$ such that $\varphi(\alpha)=\beta$. This extends to a $k$-homomophism $\varphi :k(\alpha)\to E$. Since $E/k$ is normal, $\varphi$ is extended to a $k$-automorphism of $E$. Therefore $\beta$ is a conjugate of $\alpha$. $\square$

Proposition 2: A subextension of the normal extension $E/k$ is normal if and only if it is equal to all its conjugates.

Proof: Suppose $k\subset F\subset E$ with $F/k$ a normal extension. Let $\sigma\in Aut_k(E)$.

First, we show $F^\sigma\subset F$ (where $F^\sigma$ is the notation for the image of $F$ under $\sigma$). Let $\alpha\in F$ and $f\in k[x]$ be the minimal polynomial of $\alpha$ over $k$. Then $\sigma(\alpha)$ is a root of $f$ and since splits over $F$, we have that $\sigma(\alpha)\in F$.

Second, we show that $F\subset F^\sigma$. Let $\alpha\in F$. Since all the roots of $f$ are in $F$ and $\sigma$ permutes the roots of $f$, then there is a root $\beta\in F$ of $f$ such that $\sigma(\beta)=\alpha\in F^\sigma$.

Thus we have shown that $F$ equals all its conjugates.

Now let's suppose that $F=F^\sigma$ for each $\sigma\in Aut_k(E)$. Since $E/k$ is algebraic ($E/k$ is normal) then $F/k$ is algebraic. Let $f\in k[x]$ be irreducible over $k$ with root $\alpha\in F$. We have that $f$ splits over $E$ and all the roots of $f$ are the conjugates of $\alpha$ (proposition 1). By our hypothesis we know that the conjugates of $\alpha$ belong to $F$. Hence $f$ splits over $F$. Therefore $F/k$ is normal.$\square$

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