# Deducing [Field of Algebraic Numbers : Q] is infinite

• PsychonautQQ
In summary: Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.:wink:Well, isn't it a fundamental aspect of field theory that given an irreducible polynomial over Q, that there exists a splitting field in some extension? I think what I'm getting at is does the polynomial of degree n have to have NO roots in common with all the polynomials of lesser degree? Isn't it only important that it has at least one root that was not achievable with the roots before?That is correct. An irreducible polynomial of degree n must have at least one root
PsychonautQQ

## Homework Statement

Use Eisenstein's criterion to show that there exists irreducible polynomials over Q or arbitrarily large degree, and from this deduce that the field of algebraic numbers is an infinite extension of Q

none

## The Attempt at a Solution

Note that x^n+4x+2 is irreducible for all positive n>1 by Eisenstein Criterion with p=2.

I was wondering if each of these irreducible polynomials is the minimal polynomial for some element of the field of algebraic numbers? If so, then I think we could argue that for any given degree of a minimal element in the field of algebraic numbers, there is an element of higher degree, and therefore it is an infinite extension.

Am I on the right track here?

PsychonautQQ said:

## Homework Statement

Use Eisenstein's criterion to show that there exists irreducible polynomials over Q or arbitrarily large degree, and from this deduce that the field of algebraic numbers is an infinite extension of Q

none

## The Attempt at a Solution

Note that x^n+4x+2 is irreducible for all positive n>1 by Eisenstein Criterion with p=2.

I was wondering if each of these irreducible polynomials is the minimal polynomial for some element of the field of algebraic numbers? If so, then I think we could argue that for any given degree of a minimal element in the field of algebraic numbers, there is an element of higher degree, and therefore it is an infinite extension.

Am I on the right track here?
I'd say yes.

But you will have to consider the (im)possibilty, that new elements in an extension with ##x^n+4x+2## don't cover old ones of ##x^m+4x+2## with ##m < n## or are already given by all of the ##x^m+4x+2## together, which are already adjoint before. This means irreducibility over ##\mathbb{Q}## isn't enough. ##x^n+4x+2## has to be irreducible over all ##\mathbb{Q}[a_1, \dots ,a_k]## where the ##a_i## are all the roots of all ##x^m+4x+2## with ##m < n##. Eventually you will have to restrict ##n## to prime numbers. (Just a thought, I haven't done the math.)

PsychonautQQ
fresh_42 said:
I'd say yes.

But you will have to consider the (im)possibilty, that new elements in an extension with ##x^n+4x+2## don't cover old ones of ##x^m+4x+2## with ##m < n## or are already given by all of the ##x^m+4x+2## together, which are already adjoint before. This means irreducibility over ##\mathbb{Q}## isn't enough. ##x^n+4x+2## has to be irreducible over all ##\mathbb{Q}[a_1, \dots ,a_k]## where the ##a_i## are all the roots of all ##x^m+4x+2## with ##m < n##. Eventually you will have to restrict ##n## to prime numbers. (Just a thought, I haven't done the math.)
If I restrict n to prime numbers from the start won't the argument still work because their are infinite prime numbers? I guess what I'm asking is if I only use prime numbers will that somehow ensure that none of the m<n degree polynomial/s will have roots that can't be generated from the roots that came before it? Or at least that it will have one root that didn't come before it?

PsychonautQQ said:
If I restrict n to prime numbers from the start won't the argument still work because their are infinite prime numbers? I guess what I'm asking is if I only use prime numbers will that somehow ensure that none of the m<n degree polynomial/s will have roots that can't be generated from the roots that came before it? Or at least that it will have one root that didn't come before it?
I don't know whether primes are needed. One could number all of them ##p_1, \dots , p_n , \dots## and make an induction along their numbering. But I wouldn't start with it, rather I would keep it in mind as a possibility, if questions whether ##m## and ##n## are coprime should become important.

You will have to show it anyway that ##x^n + 4x +2## is irreducible over ##\mathbb{Q}(a_1,\dots ,a_k)## where the ##a_i## are all roots of previous ##x^m+4x+2##, i.e. ##m < n##.

As I look at it now, I think it might be easier to show, that ##x^m + 4x +2## and ##x^n + 4x +2## cannot have a common root if ##m \neq n##. Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.

PsychonautQQ
fresh_42 said:
I don't know whether primes are needed. One could number all of them ##p_1, \dots , p_n , \dots## and make an induction along their numbering. But I wouldn't start with it, rather I would keep it in mind as a possibility, if questions whether ##m## and ##n## are coprime should become important.

You will have to show it anyway that ##x^n + 4x +2## is irreducible over ##\mathbb{Q}(a_1,\dots ,a_k)## where the ##a_i## are all roots of previous ##x^m+4x+2##, i.e. ##m < n##.

As I look at it now, I think it might be easier to show, that ##x^m + 4x +2## and ##x^n + 4x +2## cannot have a common root if ##m \neq n##. Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.

Well, isn't it a fundamental aspect of field theory that given an irreducible polynomial over Q, that there exists a splitting field in some extension? I think what I'm getting at is does the polynomial of degree n have to have NO roots in common with all the polynomials of lesser degree? Isn't it only important that it has at least one root that was not achievable with the roots before?
fresh_42 said:
I don't know whether primes are needed. One could number all of them ##p_1, \dots , p_n , \dots## and make an induction along their numbering. But I wouldn't start with it, rather I would keep it in mind as a possibility, if questions whether ##m## and ##n## are coprime should become important.

You will have to show it anyway that ##x^n + 4x +2## is irreducible over ##\mathbb{Q}(a_1,\dots ,a_k)## where the ##a_i## are all roots of previous ##x^m+4x+2##, i.e. ##m < n##.

As I look at it now, I think it might be easier to show, that ##x^m + 4x +2## and ##x^n + 4x +2## cannot have a common root if ##m \neq n##. Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.

So I need to show that they have NO roots in common? Wouldn't be enough to show that there is at least one new root in the polynomial of degree n that can not be achieved with all the roots of polynomials in lesser degree's? I'll work on trying to find a contradiction if they share a root.

fresh_42 said:
I don't know whether primes are needed. One could number all of them ##p_1, \dots , p_n , \dots## and make an induction along their numbering. But I wouldn't start with it, rather I would keep it in mind as a possibility, if questions whether ##m## and ##n## are coprime should become important.

You will have to show it anyway that ##x^n + 4x +2## is irreducible over ##\mathbb{Q}(a_1,\dots ,a_k)## where the ##a_i## are all roots of previous ##x^m+4x+2##, i.e. ##m < n##.

As I look at it now, I think it might be easier to show, that ##x^m + 4x +2## and ##x^n + 4x +2## cannot have a common root if ##m \neq n##. Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.

Cont
fresh_42 said:
I don't know whether primes are needed. One could number all of them ##p_1, \dots , p_n , \dots## and make an induction along their numbering. But I wouldn't start with it, rather I would keep it in mind as a possibility, if questions whether ##m## and ##n## are coprime should become important.

You will have to show it anyway that ##x^n + 4x +2## is irreducible over ##\mathbb{Q}(a_1,\dots ,a_k)## where the ##a_i## are all roots of previous ##x^m+4x+2##, i.e. ##m < n##.

If x^n+4x+2 and x^m+4x+2 share a root r, then (x-r)(n(x)) = (x-r)(m(x)) in a field extension with r in it,and thus n(x) = m(x) in this splitting field which is a contradiction because deg(n(x)) = n - 1 and deg(m(x)) = m-1 where m<n. Correct?

As I look at it now, I think it might be easier to show, that ##x^m + 4x +2## and ##x^n + 4x +2## cannot have a common root if ##m \neq n##. Maybe you could find a contradiction if you assume there was a common root ##r## that satisfies ##r^m+4r+2=r^n+4r+2=0##.

PsychonautQQ said:
Well, isn't it a fundamental aspect of field theory that given an irreducible polynomial over Q, that there exists a splitting field in some extension? I think what I'm getting at is does the polynomial of degree n have to have NO roots in common with all the polynomials of lesser degree? Isn't it only important that it has at least one root that was not achievable with the roots before?
Yes. But why to bother about the difference? This only makes things unnecessarily more complicated. We only need one more root with every polynomial, that's right. Which or how many isn't important here.
So I need to show that they have NO roots in common? Wouldn't be enough to show that there is at least one new root in the polynomial of degree n that can not be achieved with all the roots of polynomials in lesser degree's? I'll work on trying to find a contradiction if they share a root.
Yes. And by the way: You will also have to show, that they don't split over ##\mathbb{Q} ## for otherwise you would only extend ##\mathbb{Q}## by elements of ##\mathbb{Q}##.

PsychonautQQ
fresh_42 said:
Yes. And by the way: You will also have to show, that they don't split over ##\mathbb{Q} ## for otherwise you would only extend ##\mathbb{Q}## by elements of ##\mathbb{Q}##.
Well doesn't showing that it's irreducible over Q due to Eisenstein mean that it obviously can't split over Q?

fresh_42
PsychonautQQ said:
Well doesn't showing that it's irreducible over Q due to Eisenstein mean that it obviously can't split over Q?
Yes, Eisenstein is a possibility.

PsychonautQQ

## 1. What is an algebraic number?

An algebraic number is a number that can be expressed as a root of a polynomial equation with integer coefficients. For example, the number √2 is algebraic because it is a root of the polynomial equation x^2 - 2 = 0.

## 2. How do we deduce that the field of algebraic numbers is infinite?

We can deduce that the field of algebraic numbers is infinite by using the Fundamental Theorem of Algebra, which states that every non-constant polynomial with complex coefficients has at least one complex root. Since there are infinitely many polynomials with integer coefficients, there must also be infinitely many algebraic numbers.

## 3. Is there a specific method for constructing infinite algebraic numbers?

There is no specific method for constructing infinite algebraic numbers, as there are infinitely many possible polynomial equations with integer coefficients that can produce algebraic numbers. However, some common methods include using the quadratic formula or applying the Rational Root Theorem.

## 4. Can any real number be expressed as an algebraic number?

No, not all real numbers can be expressed as algebraic numbers. For example, π and e are transcendental numbers, meaning they cannot be expressed as the roots of any polynomial equation with integer coefficients. However, all rational numbers (numbers that can be expressed as a ratio of two integers) are also algebraic numbers.

## 5. How does the infinitude of algebraic numbers relate to other fields of mathematics?

The infinitude of algebraic numbers has implications in various fields of mathematics, such as number theory, algebra, and analysis. It is also connected to the concept of transcendental numbers and the unsolvability of certain mathematical problems, such as the trisecting of an angle using only a compass and straightedge.

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