# Proving properties of polynomial in K[x]

• I
• kmitza
In summary: Yes, you missed the fact that ##d(x)=0## if and only if the degree of the minimal polynomial is zero.
kmitza
TL;DR 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

kmitza said:
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
fresh_42 said:
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

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.

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.

fresh_42 said:
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?

Infrared said:
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.

fresh_42 said:
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.

Infrared said:
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.

## 1. What is a polynomial?

A polynomial is a mathematical expression consisting of variables and coefficients, combined using operations such as addition, subtraction, and multiplication. It can have one or more terms, and the degree of a polynomial is determined by the highest power of the variable.

## 2. What does it mean to prove properties of a polynomial?

Proving properties of a polynomial involves using mathematical techniques and principles to demonstrate that certain statements or equations are true for all possible values of the variables in the polynomial. This helps to establish the validity and accuracy of the polynomial in various contexts.

## 3. What are some common properties of polynomials?

Some common properties of polynomials include the degree, leading coefficient, and number of terms. Other properties include the roots or solutions of the polynomial, its symmetry, and its behavior as the variable approaches positive or negative infinity.

## 4. How can I prove properties of a polynomial in K[x]?

To prove properties of a polynomial in K[x], you can use techniques such as substitution, factoring, and algebraic manipulation. You can also use mathematical theorems and principles, such as the Remainder Theorem and the Fundamental Theorem of Algebra, to prove specific properties.

## 5. Why is it important to prove properties of polynomials?

Proving properties of polynomials is important because it helps to establish the validity and accuracy of the polynomial in different contexts. It also allows for a deeper understanding of the behavior and characteristics of polynomials, which can be useful in various mathematical applications and problem-solving situations.

• Linear and Abstract Algebra
Replies
12
Views
2K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
8
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
2K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K