Deducing [Field of Algebraic Numbers : Q] is infinite

[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

Homework Equations

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?

 

[fresh_42]

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

Homework Equations

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]

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?

 

[fresh_42]

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]

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?

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.

 

[PsychonautQQ]

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

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 $r^m+4r+2=r^n+4r+2=0$.

 

[fresh_42]

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]

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]

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.