MHB Irreducible factors of polynomial

[mathmari]

Hey!

Let $K$ be a normal extension of $F$ and $f\in F[x]$ be irreducible over $F$.

1. Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$. Show that there exists $\sigma \in G(K/F)$ such that $g_2=\sigma (g_1)$.
2. If $f$ is reducible over $K$, show that all its irreducible factors in $K[x]$ have the same degree.

I don't really have an idea about that. Could you give me a hint? (Wondering)

 

[I like Serena]

Hey!

Let $K$ be a normal extension of $F$ and $f\in F[x]$ be irreducible over $F$.

1. Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$. Show that there exists $\sigma \in G(K/F)$ such that $g_2=\sigma (g_1)$.
2. If $f$ is reducible over $K$, show that all its irreducible factors in $K[x]$ have the same degree.

I don't really have an idea about that. Could you give me a hint? (Wondering)

Hey mathmari!

1. What is $G(K/F)$?
2. Did you mean: $f$ is reducible over $K[x]$?

 

[mathmari]

1. What is $G(K/F)$?

It is the Galois group.

2. Did you mean: $f$ is reducible over $K[x]$?

Oh yes, it must be a typo.

 

[I like Serena]

Doesn't (2) follow from (1)? (Wondering)

That is, suppose $f=g_1\cdot ... \cdot g_n$, where the $g_i$ are irreducible polynomials over $K[x]$.
Then each pair $g_i$ and $g_j$ must have an isomorphic mapping $\sigma$ between them, don't they?
And an isomorphic mapping implies that they are polynomials of the same degree, doesn't it?

 

[mathmari]

Doesn't (2) follow from (1)? (Wondering)

That is, suppose $f=g_1\cdot ... \cdot g_n$, where the $g_i$ are irreducible polynomials over $K[x]$.
Then each pair $g_i$ and $g_j$ must have an isomorphic mapping $\sigma$ between them, don't they?
And an isomorphic mapping implies that they are polynomials of the same degree, doesn't it?

Ahh yes! (Nerd) And for (1) do we use the fact that each element of the Galois group permutes the roots of $f$ in a unique way? (Wondering)

 

[I like Serena]

And for (1) do we use the fact that each element of the Galois group permutes the roots of $f$ in a unique way? (Wondering)

Maybe... (Thinking)

As an alternative approach I tried to find a counter example.
Suppose we pick $F=\mathbb Q$, $K=\mathbb Q(\sqrt[4] 2)$, and $f(x)=x^4-2$.
Then $f$ is irreducible over $\mathbb Q$.
And $f(x)=(x-\sqrt[4] 2)(x+\sqrt[4] 2)(x^2 + \sqrt 2)$ over $\mathbb Q(\sqrt[4] 2)$, isn't it?
But that seems to contradict (1), since the irreducible factors do not isomorphically map to each other. (Sweating)
Am I missing something?

 

[mathmari]

Maybe... (Thinking)

As an alternative approach I tried to find a counter example.
Suppose we pick $F=\mathbb Q$, $K=\mathbb Q(\sqrt[4] 2)$, and $f(x)=x^4-2$.
Then $f$ is irreducible over $\mathbb Q$.
And $f(x)=(x-\sqrt[4] 2)(x+\sqrt[4] 2)(x^2 + \sqrt 2)$ over $\mathbb Q(\sqrt[4] 2)$, isn't it?
But that seems to contradict (1), since the irreducible factors do not isomorphically map to each other. (Sweating)
Am I missing something?

$K$ is a normal extension. Doesn't this mean that the polynomial must be splitted in linear factors in $K[x]$ ? (Wondering)

 

[I like Serena]

$K$ is a normal extension. Doesn't this mean that the polynomial must be splitted in linear factors in $K[x]$ ? (Wondering)

Ah okay. However, according to wiki:

An algebraic field extension K/F is said to be normal if every polynomial that is irreducible over F either has no root in K or splits into linear factors in K.​

If $f$ splits into linear factors over $K$, then each factor has the same degree 1.
And according to the fact that 'each element of the Galois group permutes the roots of f in a unique way', it follows that we have an isomorphic mapping between each of the factors, doesn't it? (Wondering)

However, that leaves the possibility that $f$ has no roots in $K$...

For instance with $F=\mathbb Q, K=\mathbb Q(\sqrt 2), f(x)=x^4+4x^2+2$.
Then $f(x)=(x^2+2+\sqrt 2)(x^2+2-\sqrt 2)$.
How can we prove that in such cases we can also find an isomorphic mapping between the irreducible factors? (Sweating)

 

[Olinguito]

1. What is $G(K/F)$?

The Galois group $G(K/F)$ of the field extension $K/F$ is the group of all automorphisms of the field $K$ that fixes every element of the field $F$. Here, $K$ is an extension of the field $F$, i.e. $F$ is a subfield of $K$. You see, field theory is a bit different from group theory: in group theory, given a group $G$, we are interested in analysing the subgroups of $G$, whereas in field theory, given a field $F$, we are more interested in constructing larger fields containing $F$ as a subfield (or containing a subfield isomorphic to $F$); these larger fields are called extensions of $F$.

In this problem, $K/F$ is a normal extension, meaning that given an irreducible polynomial $f\in F[x]$, either $f$ has no roots in $K$ or else it does and can be linearly factorized in $L$, i.e. splits over $L$. For example: $\mathbb Q\left(\sqrt2\right)/\mathbb Q$ is a normal extension. The polynomial $x^2-2$ is irreducible over $\mathbb Q$ but has a root in $\mathbb Q\left(\sqrt2\right)$ and $x^2-2=\left(x+\sqrt2\right)\left(x-\sqrt2\right)$ in $\mathbb Q\left(\sqrt2\right)$. On the other hand, $\mathbb Q\left(\sqrt[3]2\right)/\mathbb Q$ is not a normal extension: the polynomial $x^3-2$, irreducible over $\mathbb Q$, has a root in $\mathbb Q\left(\sqrt[3]2\right)$ but does not split therein.