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Let $F$ be a field, $f(x)$ be an irreducible polynomial in $F[x]$ and $E/F$ be a normal extension. Show that if $g(x)$, $h(x)$ are irreducible factors of $f(x)$ in $E[x]$ then there exists an automorphism $\sigma$ of $E$ over $F$ such that $\sigma(g)=h$. Does this result hold if we do not assume normal extension?

What I've tried so far:

Let $\overline{F}$ be the algebraic closure of $F$. Then, by definition, $f(x)\in F[x]$ splits completely over $\overline{F}$. So

$$f(x)=(x-\alpha_1)\cdots(x-\alpha_n)(x-\beta_1)\cdots(x-\beta_m) (x-\gamma_1)\cdots(x-\gamma_k)$$

Since $g(x)$ and $h(x)$ are irreducible factors of $f(x)\in E[x]$ then we can write, without loss of generality, $n\leq m$ and

$$g(x)=(x-\alpha_1)\cdots(x-\alpha_n)\qquad h(x)=(x-\beta_1)\cdots(x-\beta_m)$$

I want to define a map $\sigma:E\rightarrow E$ which maps $\alpha_i$ to $\beta_i$ (with this I can conclude $\sigma(g)=h$, right?). But the problems are:

1) I don't know $n=m$.

2) I don't know $\alpha_i,\beta_i\in E$.

3) Even if $\alpha_i,\beta_i\in E$, I'd only have a map on a subset of $E$. I don't know if I can extend this map to the whole $E$.

EDIT: The question above can be found on Serge Lang's Algebra, Revised Third Edition, Volme 1, Chapter V, exercise 26, and is stated as:

Let $k$ be a field, $f(X)$ an irreducible polynomial in $k[X]$, and let $K$ be a finite normal extension of $k$. If $g$, $h$ are monic irreducible factors of $f(X)$ in $K[X]$, show that there exists an automorphism $\sigma$ of $K$ over $k$ such that $\sigma = h^\sigma$. Give an example when this conclusion is not valid if $K$ is not normal over $k$.

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    certainly not. For a nonnormal extension, the $g$ and $h$ need not have the same degree. Let $\lambda$ be a cube root of a prime $p$, and look at $\Bbb Q(\lambda)$. Then $X^3-p$ can be your $\Bbb Q$-irreducible polynomial, which factors as $(X-\lambda)(X^2+\lambda X+\lambda^2)$ over the nonnormal field $\Bbb Q(\lambda)$. – 2017-02-09
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    Thank you very much! I forgot to think on the degrees. Maybe this can help me to prove the actual proposition – 2017-02-09
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    I haven't figured out how to prove the proposition, but I notice that $(g), (h)$ are prime ideals of the ring $E[x]$ which lie over the prime ideal $(f)$ of $F[x]$. Maybe there is some argument involving Dedekind domains that can be made. – 2017-02-09
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    Factor $f$ in an algebraic closure, and think about what automorphisms of $\bar{F}/F$ do to roots of $f$. – 2017-02-09
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    @D_S, I don’t believe that the proposition on normal extensions has any arithmetic content: should be provable with nothing but Galois Theory. – 2017-02-09
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    You are right, the result just seemed very similar to the fact that if $B$ is the integral closure of a Dedekind domain $A$ in a finite Galois extension $L$ of the quotient field $K$ of $A$, with quotients fields $L/K$, then the group $\textrm{Gal}(L/K)$ acts transitively on the primes of $B$ – 2017-02-09
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    Thanks for the comments guys! I put a tentative solution of what I was thinking so far (using a more "brute force" approach). – 2017-02-10

2 Answers 2

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What you propose is already close to a full solution: There's an automorphism $\sigma\in Gal(K/F)$, $K$ a splitting field of $f$ over $E$ that sends $\sigma(\alpha_1)=\beta_1$; this works because both elements have $f$ as their minimal polynomial over $F$. Since $E$ is normal, it is invariant under $\sigma$, and thus we can restrict. As you suspected, it then follows that $\sigma(g)=h$ because these are two irreducible polynomials that share a root in an extension field, so they are both equal to the minimal polynomial of that root.

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    How do we know that there is such an automorphism $\sigma$ in $Gal(K/F)$? – 2017-02-10
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    @Jason: It follows because they are conjugates. If you want a more explicit argument, then the basic ingredient is the (easy) fact that if $a,b$ have the same minimal polynomial, then there is an $F$-isomorphism $\varphi: F(a)\to F(b)$ that sends $a\mapsto b$. This map can then be extended to the splitting field by using (a general version of) this fact repeatedly. – 2017-02-12
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    I understand that there is an $F$-isomorphism $\phi : F(\alpha_1) \to F(\beta_1)$. I guess where I'm stuck is how to extend this to the splitting field. Because to apply the general result that you cite, we need to know that $K$ is the splitting field for $f$ over $F(\alpha_1)$ and $F(\beta_1)$, do we not? And it doesn't seem like this is true (if $K$ is defined to be the splitting field of $f$ over $E$). Or maybe I'm not applying the result in the right way? – 2017-02-13
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    Or maybe we can take $K$ to be the splitting field of $f$ over $F$ rather than $E$, and then use the fact that both $K$ and $E$ will be contained in the algebraic closure of $F$? Hopefully this is making sense. :P – 2017-02-13
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    @Jason: $K$ *is* the splitting field over $F(\alpha_1), F(\beta_1)$. This follows directly from the definition: a splitting field over a ground field is also a splitting field over each intermediate field (the polynomial splits, and the extension is still generated by the roots). – 2017-02-13
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    But if $K$ is the splitting field for $f$ over $E$, then how do we know that $F(\alpha_1), F(\beta_1)$ are intermediate between $E$ and $K$? Because the assumption is that $F \subseteq E$. So I'm not sure how to show that $E \subseteq F(\alpha_1)$. – 2017-02-13
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    @Jason: Yes, I was sloppy about this point. If $E/F$ is finite (as in the quoted exercise from Lang), then we're in good shape because then $E$ will be the splitting field of some $g$, and then $K$ will be a splitting field over both $E$ (of $f$) and $F$ (of $fg$). – 2017-02-13
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    Excellent, thanks! – 2017-02-14
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    @Jason: In fact, I now think this argument works in general, but we will have to work with a collection of polynomials. – 2017-02-14
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Let $E = F(c_1, \ldots, c_k)$ and $f(X) = (X - a_1)\cdots (X - a_n)(X - b_1)\cdots (X - b_m)$ where $$h(X) = (X - a_1)\cdots (X - a_n) \text{ and } g(X) = (X - b_1)\cdots (X - b_m) (\text{ also assume } m > n).$$ Let $L$ be the splitting field of $f(X)$. Let $\sigma $ be an automorphism such that it takes $a_i \mapsto b_i$ for $i \leq n$ and $b_j \mapsto b_j$ for $n < j < m$. Since $E$ is a normal extension, it is a well know fact that $\sigma(E) = E$. Let $\sigma(h(X)) = g'(X)$. Then it follows that $g'(X)$ divides $g(X)$ and hence $g(X)$ is not irreducible- a contradiction unless $m = n$.

Remark: Existence of $\sigma:$ Over $F$, we have $F(a_i) \simeq F(b_j)$ as both $a_i$ and $b_i$ are roots of $f(X)$ with $a_i \mapsto b_i$. Therefore proceding similarly we have $F(a_1, \ldots, a_n) \simeq F(b_1, \ldots, b_n)$ with $a_i \mapsto b_i$ for $i \leq n$ and hence $F(a_1, \ldots a_n, b_1, \ldots, b_m, c_1, \ldots, c_k) \simeq F(a_1, \ldots a_n, b_1, \ldots, b_m, c_1, \ldots, c_k)$ where $a_i \mapsto b_i$ for $i \leq n$, $b_j \mapsto b_j$ for $n < j < m$ and $c_t \mapsto c_t$.

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    I would appreciate if someone could confirm my answer. I too faced a similar problem in my book and I was supposed to solve it without use galois group (or related fact as mentioned in the 1st answer). – 2018-09-13