Please check my (dodgy) proof! (Field theory)

Unparseable latex formula:

$$[K : \mathbb{Q}] = 2$$

Where

K

is a number field, show

K = \mathbb{Q}(\sqrt{d})

for some

d \in \mathbb{Z}

,

d

squarefree.
EDIT: I forgot to mention that theta exists by the simple extension theorem :smile:

I begin by proving:

Unparseable latex formula:

\forall q \in \mathbb{Q}:[br] [br]$\mathbb{Q}(q\theta) = \mathbb{Q}(\theta)$[br]$\mathbb{Q}(q + \theta) = \mathbb{Q}(\theta)$[br] [br]

Then I say that because of the above properties and the fact that

Unparseable latex formula:

$$[K : \mathbb{Q}] = 2$$

we can assume

\theta

to be not in

\mathbb{Q}

and:

Unparseable latex formula:

\forall q \in \mathbb{Q} - \{0\}[br]$$\theta \neq \phi + q$$ [br]

and

Unparseable latex formula:

\forall q \in \mathbb{Q} -\{1\}[br]$$ \theta \neq q \phi $$

So then since the degree of the field extension is 2, there's a minimal polynomial for theta over Q:

p_{\theta}(x) = x^2 + \alpha x - \beta

clearly beta isn't 0 (as it contradicts irreducibility over Q) and by my assumption for theta we find alpha must be 0 (by checking the roots of the polynomial.
So,

p_{\theta}(x) = x^2 - \beta

is irreducible over Q iff there are no rational roots implying the square root of beta to be irrational.

So:

\theta = \sqrt (\beta)

and by using again the assumption I imposed for theta it's simple to show that beta will have to be square free, and we can get an integer by just rationalising the denominator and ignoring the rational coefficient.

I feel like this is pretty dodgy but I'm not sure, any advice? :smile:

(edited 5 years ago)

Reply 1

DFranklin

18

Original post by Ryanzmw

Unparseable latex formula:

$$[K : \mathbb{Q}] = 2$$

Where

K

is a number field, show

K = \mathbb{Q}(\sqrt{d})

for some

d \in \mathbb{Z}

,

d

squarefree.
EDIT: I forgot to mention that theta exists by the simple extension theorem :smile:

I begin by proving:

Unparseable latex formula:

\forall q \in \mathbb{Q}:[br] [br]$\mathbb{Q}(q\theta) = \mathbb{Q}(\theta)$[br]$\mathbb{Q}(q + \theta) = \mathbb{Q}(\theta)$[br] [br]

Then I say that because of the above properties and the fact that

Unparseable latex formula:

$$[K : \mathbb{Q}] = 2$$

we can assume

\theta

to be not in

\mathbb{Q}

and:

Unparseable latex formula:

\forall q \in \mathbb{Q} - \{0\}[br]$$\theta \neq \phi + q$$ [br]

and

Unparseable latex formula:

\forall q \in \mathbb{Q} -\{1\}[br]$$ \theta \neq q \phi $$

So then since the degree of the field extension is 2, there's a minimal polynomial for theta over Q:

p_{\theta}(x) = x^2 + \alpha x - \beta

clearly beta isn't 0 (as it contradicts irreducibility over Q) and by my assumption for theta we find alpha must be 0 (by checking the roots of the polynomial.
So,

p_{\theta}(x) = x^2 - \beta

is irreducible over Q iff there are no rational roots implying the square root of beta to be irrational.

So:

\theta = \sqrt (\beta)

and by using again the assumption I imposed for theta it's simple to show that beta will have to be square free, and we can get an integer by just rationalising the denominator and ignoring the rational coefficient.

I feel like this is pretty dodgy but I'm not sure, any advice? :smile:

I'm not seeing how alpha=0. Note that if you have a primitive element theta, then theta + q is also primitive for any choice of q. So theta isn't as "fixed" as you might think.

The example I'm thinking of is adjoining a cube root of unity w (which is also a valid primitive element). We have w^2+w+1 = 0 and so w^2 certainly isn't in Q, so you can't use w directly to get the quadratic extension in the way you want.

The way I'd proceed is note that using the quadratic formula on the minimum polynomial gives you an explicit value you need to be able to take the square root of to solve the quadratic.

Reply 2

Ryanzmw

OP

16

Original post by DFranklin

I'm not seeing how alpha=0. Note that if you have a primitive element theta, then theta + q is also primitive for any choice of q. So theta isn't as "fixed" as you might think.

The example I'm thinking of is adjoining a cube root of unity w (which is also a valid primitive element). We have w^2+w+1 = 0 and so w^2 certainly isn't in Q, so you can't use w directly to get the quadratic extension in the way you want.

The way I'd proceed is note that using the quadratic formula on the minimum polynomial gives you an explicit value you need to be able to take the square root of to solve the quadratic.

Sorry, I mean to say that I impose the criteria that theta must have no purely ration sum term and no purely rational (non 1) coefficient just because I can express the same field extension in the same way and so I should be able to impose these restrictions of theta in the same way one can say if a is rational then a = p/q with gcd(p,q) = 1, with the gcd restriction being analogous to my restrictions. (so I guess I think I'm imposing my restrictions for theta without a loss of generality?)

I found alpha to be 0 since if you take my restrictions for theta and the fact that

p_{\theta}

is the minimal polynomial for

\theta

over

\mathbb{Q}

:

P_{\theta}(\theta) = \theta^2 + \alpha \theta - \beta = 0

\Rightarrow \theta = \dfrac{-\alpha}{2} \pm \dfrac{\sqrt{\alpha^2 + 4\beta}}{2}

but I assumed theta has no purely rational sum term so if

P_{\theta}

is the minimal polynomial for the specific theta I'd like we require

\alpha = 0

so:

P_{\theta}(x) = x^2 - \beta

which is the minimal polynomial so we need sqrt(beta) to be irrational (since degree 2,3 polynomial is irreducible iff no rational roots)

P_{\theta}(\theta) = 0 = \theta^2 - \beta \Rightarrow \theta = \sqrt{\beta}

(by assumption the only rational coefficient allowed is 1)

and if

\beta \in \mathbb{Q} \Rightarrow \beta = \dfrac{\beta_1}{\beta_2}

\gcd(\beta_1, \beta_2) = 1

\beta_1, \beta_2 \in \mathbb{Z}

so actually to impose the condition about theta having no non 1 rational coefficient:

\theta = \sqrt{\dfrac{\beta_1}{\beta_2}} = \dfrac{\sqrt{\beta_1 \beta_2 }}{\beta_2}

,

so in accordance with my restriction for theta :

\theta = \sqrt{\beta_1}

(which should be square free otherwise it'll have a rational coefficient again) and can't be reduced further :smile:

(edited 5 years ago)

Reply 3

DFranklin

18

Original post by Ryanzmw

Sorry, I mean to say that I impose the criteria that theta must have no purely ration sum term and no purely rational (non 1) coefficient just because I can express the same field extension in the same way and so I should be able to impose these restrictions of theta in the same way one can say if a is rational then a = p/q with gcd(p,q) = 1, with the gcd restriction being analogous to my restrictions. (so I guess I think I'm imposing my restrictions for theta without a loss of generality?)

You still haven't convinced me you can do this - to my mind assuming this is trivialising the question. Fundamentally, what do you *mean* by theta has "no rational part"? Because it seems to me you're essentially meaning "it's the square root of an element of Q", which is definitely begging the question.

You know you can solve for theta by solving a quadratic equation with rational coefficients. It follows that theta is in Q[sqrt(b^2-4ac)] which is what you need to show.

I found alpha to be 0 since if you take my restrictions for theta and the fact that

p_{\theta}

is the minimal polynomial for

\theta

over

\mathbb{Q}

:

P_{\theta}(\theta) = \theta^2 + \alpha \theta - \beta = 0

\Rightarrow \theta = \dfrac{-\alpha}{2} \pm \dfrac{\sqrt{\alpha^2 + 4\beta}}{2}

but I assumed theta has no purely rational sum term so if

P_{\theta}

is the minimal polynomial for the specific theta I'd like we require

\alpha = 0

so:

P_{\theta}(x) = x^2 - \beta

which is the minimal polynomial so we need beta to be irrational (since degree 2,3 polynomial is irreducible iff no rational roots)

P_{\theta}(\theta) = 0 = \theta^2 - \beta \Rightarrow \theta = \sqrt{\beta}

(by assumption the only rational coefficient allowed is 1)

and if

\beta \in \mathbb{Q} \Rightarrow \beta = \dfrac{\beta_1}{\beta_2}

\gcd(\beta_1, \beta_2) = 1

\beta_1, \beta_2 \in \mathbb{Z}

so actually to impose the condition about theta having no non 1 rational coefficient:

\theta = \sqrt{\dfrac{\beta_1}{\beta_2}} = \dfrac{\sqrt{\beta_1 \beta_2 }}{\beta_2}

,

so in accordance with my restriction for theta :

\theta = \sqrt{\beta_1}

(which should be square free otherwise it'll have a rational coefficient again) and can't be reduced further :smile:

Reply 4

Ryanzmw

OP

16

Original post by DFranklin

You still haven't convinced me you can do this - to my mind assuming this is trivialising the question. Fundamentally, what do you *mean* by theta has "no rational part"? Because it seems to me you're essentially meaning "it's the square root of an element of Q", which is definitely begging the question.

You know you can solve for theta by solving a quadratic equation with rational coefficients. It follows that theta is in Q[sqrt(b^2-4ac)] which is what you need to show.

I guess I mean that if you think of Q(theta) as a vector space over Q, the basis {1,theta} are sort of, orthogonal(?). In the sense that I want theta such that theta isn't phi *q + q' for phi in C/Q, q,q' in Q.

and in my mind my restriction of theta is basically the same as saying if I can find a basis element linearly independent from 1, I can impose that it be orthogonal (because the application of my claims is sort of like applying Gram-Schmidt and just removing the parts of theta that are in Q)

(edited 5 years ago)

Reply 5

DFranklin

18

I guess I mean that if you think of Q(theta) as a vector space over Q, the basis {1,theta} are sort of, orthogonal(?). In the sense that I want theta such that theta isn't phi *q + q' for phi in C/Q, q,q' in Q.

and in my mind my restriction of theta is basically the same as saying if I can find a basis element linearly independent from 1, I can impose that it be orthogonal (because the application of my claims is sort of like applying Gram-Schmidt and just removing the parts of theta that are in Q)I'm still totally unconvinced - if you want to apply Gram-Schmidt, what's your inner product?

To be blunt: your approach is not the way to do this. Use the quadratic formula. It will be shorter than what you've done and actually rigourous.

Reply 6

eugaurie

12

Original post by DFranklin

I'm still totally unconvinced - if you want to apply Gram-Schmidt, what's your inner product?

To be blunt: your approach is not the way to do this. Use the quadratic formula. It will be shorter than what you've done and actually rigourous.

How is this proof fundamentally different from: https://math.stackexchange.com/questions/164721/if-k-is-an-extension-field-of-mathbbq-such-that-k-mathbbq-2-prove

Reply 7

DFranklin

18

Original post by eugaurie

How is this proof fundamentally different from: https://math.stackexchange.com/questions/164721/if-k-is-an-extension-field-of-mathbbq-such-that-k-mathbbq-2-prove

What I've said to do is the same as the two upvoted answers. What the OP has tried to do is essentially the route that the question asker in the stackxechange link was trying to make work, and has people giving the same objections I did.

Please check my (dodgy) proof! (Field theory)

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