Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    17
    ReputationRep:
    Is it not correct to say this?

    

Q(\alpha,w) =(a+b\alpha+c\alpha^2+dw+

jw^2+ew\alpha+fw\alpha^2+

gw^{2}\alpha+hw^{2}{\alpha^2}:a,  b,c,d,e,f,g,h,j \in Q)



\alpha = \sqrt[3]{2}

w = e^{\frac{2\pi i}{3}}

    EDIT: The lines on latex are being a bit weird, but all of the w's should be omegas
    Offline

    2
    ReputationRep:
    Context?
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Mark85)
    Context?
    I am adjoining the rational field to those numbers
    Offline

    2
    ReputationRep:
    (Original post by 2710)
    I am adjoining the rational field to those numbers
    Aha, I should have looked at it more closely before asking. The title is misleading - this has nothing to with adjoints. I was thinking that this was going to be some kind of adjoint function and from that point of view it wouldn't make any sense.

    Also, you are adjoining those numbers to the rationals - not the other way round.

    Anyway, to answer the question, it is technically correct to say that:

    \mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+j \omega^2+e\omega\alpha+f\omega \alpha^2+g\omega^{2}\alpha+h \omega^{2}{\alpha^2}\mid a,b,c,d,e,f,g,h,j \in \mathbb{Q}\}

    but it is also misleading and even disingenuous to put it like that since the extension is only of degree six, not nine.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Mark85)
    Aha, I should have looked at it more closely before asking. The title is misleading - this has nothing to with adjoints. I was thinking that this was going to be some kind of adjoint function and from that point of view it wouldn't make any sense.

    Also, you are adjoining those numbers to the rationals - not the other way round.

    Anyway, to answer the question, it is technically correct to say that:

    \mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+j \omega^2+e\omega\alpha+f\omega \alpha^2+g\omega^{2}\alpha+h \omega^{2}{\alpha^2}\mid a,b,c,d,e,f,g,h,j \in \mathbb{Q}\}

    but it is also misleading and even disingenuous to put it like that since the extension is only of degree six, not nine.
    Damn my latex is bad compared to yours >__>

    I only put it like that because that is the way I think. I like to think concretely, and by putting it like that, I can see what that actually is. But ok, so it is technically correct, and thats all I need I guess :P

    Can I ask, so there will be 6 elements in the basis? Can I ask which 6?

    {1, a, a^2, w, w^2...?} (i'm using a as alpha) (Now that I think about it, wouldnt there be 9 elements in the basis?)
    Offline

    2
    ReputationRep:
    (Original post by 2710)
    I only put it like that because that is the way I think. I like to think concretely, and by putting it like that, I can see what that actually is. But ok, so it is technically correct, and thats all I need I guess :P
    Well, presumably this example arises in the context of working out the Galois group or something and in that case, or indeed any case, it is much better to work with a \mathbb{Q}-basis, it really will simplify and clarify anything you want to do with it.

    I mean, it is technically correct to say that

    \mathbb{R}^2= \{a(1,0)+b(3,5)+c(12,333)+d(\pi,  13.67)+e(\frac{\mathrm{sin}(34)}  {27.8888},3)+f(0,1)\mid a,b,c,d,e,f \in \mathbb{R}\}

    but you aren't really specifying \mathbb{R}^2 any more clearly than the obvious alternative.

    Or for an example closer to the spirit of the example, imagine wiriting:

    \mathbb{C}=\{a + bi + ci^2 + di^3 \mid a,b,c,d \in \mathbb{R}\}
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Mark85)
    Well, presumably this example arises in the context of working out the Galois group or something and in that case, or indeed any case, it is much better to work with a \mathbb{Q}-basis, it really will simplify and clarify anything you want to do with it.

    I mean, it is technically correct to say that

    \mathbb{R}^2= \{a(1,0)+b(3,5)+c(12,333)+d(\pi,  13.67)+e(\frac{\mathrm{sin}(34)}  {27.8888},3)+f(0,1)\mid a,b,c,d,e,f \in \mathbb{R}\}

    but you aren't really specifying \mathbb{R}^2 any more clearly than the obvious alternative.

    Or for an example closer to the spirit of the example, imagine wiriting:

    \mathbb{C}=\{a + bi + ci^2 + di^3 \mid a,b,c,d \in \mathbb{R}\}
    Hmm. ok, i get that in your exmaples, you have redundant terms. So in my expression, which ones are the redundant terms? And how would you write it?

    Like in your C example, ci^2 = -c, which I guess can be 'absorbed' by the a, making it redundant. Which terms of mine can be absorbed?
    Offline

    2
    ReputationRep:
    (Original post by 2710)
    Hmm. ok, i get that in your exmaples, you have redundant terms. So in my expression, which ones are the redundant terms? And how would you write it?

    Like in your C example, ci^2 = -c, which I guess can be 'absorbed' by the a, making it redundant. Which terms of mine can be absorbed?
    \omega is a primitive cube root of unity so in particular:

    1+\omega +\omega^2 = 0

    i.e. \omega^2 = -1-\omega
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Mark85)
    \omega is a primitive cube root of unity so in particular:

    1+\omega +\omega^2 = 0

    i.e. \omega^2 = -1-\omega
    Oh i see! Thank you, I understand now, I didnt realise that w^2 can be expressed like that.
    • Thread Starter
    Offline

    17
    ReputationRep:
    So it would be more appropiate to write this:

    

\mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+e \omega\alpha+f\omega\alpha^2\mid a,b,c,d,e,f \in \mathbb{Q}\}
    Offline

    2
    ReputationRep:
    (Original post by 2710)
    So it would be more appropiate to write this:

    

\mathbb{Q}(\alpha,\omega) = \{a+b\alpha+c\alpha^2+d\omega+e \omega\alpha+\omega\alpha^2+f\mi  d a,b,c,d,e,f \in \mathbb{Q}\}
    Exactly.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.