Proving properties of polynomial in K[x]

[kmitza]

TL;DR

I have run into the following problem, I have managed to solve some of it but I don't have idea for the rest.

We have Galois extension $K \subset L$ and element $\alpha \in L$ and define polynomial $$f = \prod_{\sigma \in Gal(L/K)} (x - \sigma(\alpha))$$
Now we want to show that $f \in K[x]$ which is relatively easy to see because we can take $\phi(f)$ for any $\phi \in Gal(L/K)$ then $\phi \circ \sigma$ ranges over the galois group and all the roots stay fixed and we're done.

Further we want want to prove that $f$ is power of minimal polinomial of $\alpha$ and that it is equal to the minimal polynomial iff $L = K(\alpha)$.
Any hints or help will be appreciated

 

[fresh_42]

Summary:: I have run into the following problem, I have managed to solve some of it but I don't have idea for the rest.

We have Galois extension $K \subset L$ and element $\alpha \in L$ and define polynomial $$f = \prod_{\sigma \in Gal(L/K)} (x - \sigma(\alpha))$$
Now we want to show that $f \in K[x]$ which is relatively easy to see because we can take $\phi(f)$ for any $\phi \in Gal(L/K)$ then $\phi \circ \sigma$ ranges over the galois group and all the roots stay fixed and we're done.

Further we want want to prove that $f$ is power of minimal polinomial of $\alpha$ and that it is equal to the minimal polynomial iff $L = K(\alpha)$.
Any hints or help will be appreciated

The second part should be easy. Assume $L=K(\alpha ).$ Then compare the degrees of the field extension with the degree of $f(x)$. If $f(x)=m_\alpha (x)$ is the minimal polynomial of $\alpha ,$ then $f(x)\in K[x]$ is irreducible and
$$
\prod_{\sigma \in\operatorname{Gal}(L/K)}(x-\sigma (\alpha ))=\prod_{\sigma \in\operatorname{Gal}(K(\alpha )/K)}(x-\sigma (\alpha )).
$$
Now apply the fundamental theorem of Galois theory.

I'm having a bit of trouble with the first part. We know
$$
\underbrace{L\supseteq \underbrace{K(\alpha )\supseteq K}_{m_\alpha (x)}}_{f(x)} \text{ and } f(x)=m_\alpha(x)\cdot d(x) .
$$
I have difficulties to see $m_\alpha (x)\,|\,d(x),$ in other words $d(\alpha )\stackrel{?}{=}0.$

 

[kmitza]

The second part should be easy. Assume $L=K(\alpha ).$ Then compare the degrees of the field extension with the degree of $f(x)$. If $f(x)=m_\alpha (x)$ is the minimal polynomial of $\alpha ,$ then $f(x)\in K[x]$ is irreducible and
$$
\prod_{\sigma \in\operatorname{Gal}(L/K)}(x-\sigma (\alpha ))=\prod_{\sigma \in\operatorname{Gal}(K(\alpha )/K)}(x-\sigma (\alpha )).
$$
Now apply the fundamental theorem of Galois theory.

I'm having a bit of trouble with the first part. We know
$$
\underbrace{L\supseteq \underbrace{K(\alpha )\supseteq K}_{m_\alpha (x)}}_{f(x)} \text{ and } f(x)=m_\alpha(x)\cdot d(x) .
$$
I have difficulties to see $m_\alpha (x)\,|\,d(x),$ in other words $d(\alpha )\stackrel{?}{=}0.$

Yes you are right about the second one, thanks for the swift answer. I am working on doing the first one and I will reply with the solution if I find one, I can say that trying a few examples seems to indicate that the result holds

 

[Infrared]

Two factors $x-\sigma(\alpha)$ and $x-\tau(\alpha)$ are equal when $\sigma(\alpha)=\tau(\alpha),$ i.e. $\sigma$ and $\tau$ represent the same element of the quotient $\text{Gal}(L/K)/\text{Gal}(L/K(\alpha)).$ So every root occurs with the same multiplicity in the splitting field and the product is a power of the minimal polynomial.

It's been a while since I've done anything with fields, so I could be missing something.

@fresh_42 only dealt with half of the 'iff' but it sounds like you have that taken care of anyway.

 

[fresh_42]

The second statement is basically the correspondence of field dimensions with the degree of minimal polynomials in both directions. Equal fields lead to equal polynomials via degree comparison and the fact that one divides the other, and equal polynomials lead to equal fields, simply because there is no dimension left for an intermediate field.

The first statement says $f(x)=\prod_{\sigma \in\operatorname{Gal}(L/K)}(x-\sigma (a))=m_\alpha (x)^k.$

It is easy to see that $f(x)=m_\alpha (x)d(x)$ since $f(\alpha )=0,$ but why is $d(x)=m_\alpha (x)^{k-1}$ or (I think with a recursive argument) equivalently $d(\alpha )=0$? I have something like $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2},\sqrt{3})$ in mind.

 

[Infrared]

It is easy to see that $f(x)=m_\alpha (x)d(x)$ since $f(\alpha )=0,$ but why is $d(x)=m_\alpha (x)^{k-1}$ or (I think with a recursive argument) equivalently $d(\alpha )=0$?

I thought I proved this in the first paragraph of my last post. Did I miss something?

 

[fresh_42]

I thought I proved this in the first paragraph of my last post. Did I miss something?

Yes, but I didn't understand it. Where did $\tau$ come from? I mean, the first half of the first statement reads like a tautology to me.

 

[Infrared]

Yes, but I didn't understand it. Where did $\tau$ come from? I mean, the first half of the first statement reads like a tautology to me.

The minimal polynomial of $\alpha$ is the polynomial that has each $\sigma(\alpha)$ as a root with multiplicity $1.$ However, the roots have multiplicity since it is possible for two Galois elements $\sigma$ and $\tau$ to yield the same root $\sigma(\alpha)=\tau(\alpha).$ My post verifies that even though the roots have multiplicity, this multplicity is the same for each root, and so the whole polynomial is a power of the minimal polynomial.

 

[fresh_42]

The minimal polynomial of $\alpha$ is the polynomial that has each $\sigma(\alpha)$ as a root with multiplicity $1.$ However, the roots have multiplicity since it is possible for two Galois elements $\sigma$ and $\tau$ to yield the same root $\sigma(\alpha)=\tau(\alpha).$ My post verifies that even though the roots have multiplicity, this multplicity is the same for each root, and so the whole polynomial is a power of the minimal polynomial.

Thanks. My mistake was that I thought about some $\beta \in L\K$ and that it might appear in $f$. No idea, where I left the road.