Splitting Field of $g$ over $F$?

[mathmari]

Hey!

Let $F$ be a field, $f\in F[x]$ be non-constant and $K$ the splitting field of $f$ over $F$.
Let $g\in F[x]$ be a non-constant polynomial, that divides $f$.

I want to show that $g$ can be splitted into linear factors of $K[x]$. Is $K$ the splitting field of $g$ over $F$ ?
Since $g\in F[x]$ be a non-constant polynomial, that divides $f$, we have that $f(x)=g(x)h(x), h(x)\in F[x]$.

Since $K$ is the splitting field of $f$, $f$ can be decomposed into linear factors of $K[x]$.
So, $f(x)=\prod_i (x-a_i)^i$. Since $f(x)=g(x)h(x)$, we have that $\prod_i (x-a_i)^i=g(x)h(x)$.
Do we conclude from that that the product of some of these linear factors is equal to $g(x)$ ? (Wondering)

 

[Euge]

Hi mathmari,

Your expression for $f$ is not general. It's better to write $f(x) = c\prod (x-a_i)^{m_i}$ where $m_i$ are nonnegative integers (where at least one of the $m_i$ is positive) and $c$ is nonzero. You can argue that $g(x)$ can be factored as $\prod (x - a_i)^{n_i}$ where $0 \le n_i \le m_i$, as follows. Since $g$ is nonconstant, it has a prime factor. Further, since $g$ divides $f$, every prime factor of $g$ divides $f$. As the prime factors of $f$ are the $x - a_i$, we can write $g(x) = d\prod (x - a_i)^{n_i}$ where $d \neq 0$, $0 \le n_i \le m_i$ for all $i$, and one of the $n_i$ is positive.

 

[mathmari]

Your expression for $f$ is not general. It's better to write $f(x) = c\prod (x-a_i)^{m_i}$ where $m_i$ are nonnegative integers (where at least one of the $m_i$ is positive) and $c$ is nonzero. You can argue that $g(x)$ can be factored as $\prod (x - a_i)^{n_i}$ where $0 \le n_i \le m_i$, as follows. Since $g$ is nonconstant, it has a prime factor. Further, since $g$ divides $f$, every prime factor of $g$ divides $f$. As the prime factors of $f$ are the $x - a_i$, we can write $g(x) = d\prod (x - a_i)^{n_i}$ where $d \neq 0$, $0 \le n_i \le m_i$ for all $i$, and one of the $n_i$ is positive.

I got it! (Nerd)

Is $K$ the splitting field of $g$ over $F$ ?

Do we have that $K$ is the splitting field of $g$ over $F$ because of the folllowing?
$g$ is the product of some of the linear factors of $f$ and since $K$ is the splitting field of $f$ over $F$ it is also the splitting field of $g$ over $F$.

Is this correct? (Wondering)

 

[Euge]

Do we have that $K$ is the splitting field of $g$ over $F$ because of the folllowing?
$g$ is the product of some of the linear factors of $f$ and since $K$ is the splitting field of $f$ over $F$ it is also the splitting field of $g$ over $F$.

Is this correct? (Wondering)

What you have shown is that the splitting field of $g$ is contained in the splitting field of $f$. However, you haven't shown the reverse containment. That's ok though, because the reverse containment does not hold in general. For example, let $f(x) = (x^2 + 1)(x^2 + 2)$ and $g(x) = x^2 + 1$ in $\Bbb Q[x]$. The splitting field of $f$ is $\Bbb Q(i, \sqrt{2})$, but the splitting field of $g$ is $\Bbb Q(i)$.

 

[mathmari]

What you have shown is that the splitting field of $g$ is contained in the splitting field of $f$. However, you haven't shown the reverse containment. That's ok though, because the reverse containment does not hold in general. For example, let $f(x) = (x^2 + 1)(x^2 + 2)$ and $g(x) = x^2 + 1$ in $\Bbb Q[x]$. The splitting field of $f$ is $\Bbb Q(i, \sqrt{2})$, but the splitting field of $g$ is $\Bbb Q(i)$.

Aha ok... So, we can only say that it is a subset of $K$ and not if it is $K$ or not, right?

 

[Euge]

That's correct.

 

[mathmari]

Your expression for $f$ is not general. It's better to write $f(x) = c\prod (x-a_i)^{m_i}$ where $m_i$ are nonnegative integers (where at least one of the $m_i$ is positive) and $c$ is nonzero. You can argue that $g(x)$ can be factored as $\prod (x - a_i)^{n_i}$ where $0 \le n_i \le m_i$, as follows. Since $g$ is nonconstant, it has a prime factor. Further, since $g$ divides $f$, every prime factor of $g$ divides $f$. As the prime factors of $f$ are the $x - a_i$, we can write $g(x) = d\prod (x - a_i)^{n_i}$ where $d \neq 0$, $0 \le n_i \le m_i$ for all $i$, and one of the $n_i$ is positive.

I read your answer again and I have some questions.
By "prime factors" you mean "irreducible" ? (Wondering)
Also why can we write $g(x) = d\prod (x - a_i)^{n_i}$ ? How can we justify it? Isn't this an other formulation of the statement that we want to show? (Wondering)

 

[Euge]

The reason is that $F[x]$ is a UFD. Also, in a UFD, prime elements are the same as irreducible elements.

 

[mathmari]

It's better to write $f(x) = c\prod (x-a_i)^{m_i}$ where $m_i$ are nonnegative integers (where at least one of the $m_i$ is positive) and $c$ is nonzero.

$a_i$ are elements of $K$, or not? (Wondering)

The reason is that $F[x]$ is a UFD. Also, in a UFD, prime elements are the same as irreducible elements.

We have that $F$ is a field, therefore it is a UFD, and so $F[x]$ is a UFD, right? (Wondering)
In a UFD every non-zero non-unit element can be written as a product of prime (irreducible) elements.
We have that $g$ a non-zero non-unit (= non-constant) element, so it can be written as a product of prime elements.
Since $g$ divides $f$, every prime factor of $g$ divides $f$.
The prime factors of $f$ are the $x - a_i$. So, each prime factor of $g$ divides one of the prime factors of $f$, so $g$ must be of the form $d\prod (x - a_i)^{n_i}$.

Have I understood it correctly? (Wondering)

 

[Euge]

Yes, the $a_i$ belong to $K$. When I wrote $F[x]$ I meant $K[x]$, since we're doing factorizations over $K$. In any case, the key point is that a polynomial ring over a field is a UFD. You have understood the argument well. [emoji2]

 

[mathmari]

When I wrote $F[x]$ I meant $K[x]$, since we're doing factorizations over $K$.

So, it must be:

We have that $K$ is a field, therefore it is a UFD, and so $K[x]$ is a UFD, right? (Wondering)
In a UFD every non-zero non-unit element can be written as a product of prime (irreducible) elements.
We have that $g$ a non-zero non-unit (= non-constant) element, so it can be written as a product of prime elements.
Since $g$ divides $f$, every prime factor of $g$ divides $f$.
The prime factors of $f$ are the $x - a_i$. So, each prime factor of $g$ divides one of the prime factors of $f$, so $g$ must be of the form $d\prod (x - a_i)^{n_i}$.

right? (Wondering)

 

[Euge]

Yes, that looks good. [emoji106]

 

[mathmari]

We have that $c,d\in F$ and the factors of $f$ are irreducible over $K[x]$, or not?

 

[Euge]

Yes, except generally $c,d\in K$.

 

[mathmari]

Yes, except generally $c,d\in K$.

Why are they elements of $K$ not of $F$ ? Because the factorization of the polynomials is in $K$ and not in $F$ ? (Wondering)

 

[Euge]

Right. If it's more comfortable, you may assume without loss of generality that $f$ is monic. Then $c = d = 1$ and you don't have to worry about the constants.

 

[mathmari]

Right. If it's more comfortable, you may assume without loss of generality that $f$ is monic. Then $c = d = 1$ and you don't have to worry about the constants.

Ah ok... Thank you very much! (Smile)