1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Field Extensions - Dummit and Foote - Exercise 13, 13.2 ...

  1. May 21, 2017 #1
    1. The problem statement, all variables and given/known data

    I am reading Dummit and Foote, Chapter 13 - Field Theory.

    I am currently studying Section 13.2 : Algebraic Extensions

    I need some help with Exercise 13 of Section 13.2 ... ... indeed, I have not been able to make a meaningful start on the problem ... ...

    As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...


    Exercise 13 of Section 13.2 reads as follows:


    ?temp_hash=c965ff8e6de11c5cfb54df5b4a4fd426.png


    2. Relevant equations


    Definitions that may be relevant to solving this exercise include the following:

    ?temp_hash=c965ff8e6de11c5cfb54df5b4a4fd426.png


    A result which seems relevant is Lemma 16 plus the remarks that follow its proof ... Lemma 16 and the following remarks read as follows:


    ?temp_hash=c965ff8e6de11c5cfb54df5b4a4fd426.png



    3. The attempt at a solution


    As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...

    As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...

    One thought, though ... there must be some way to use ##\alpha_i^2 \in \mathbb{Q}## ... perhaps in establishing the dimension of ##F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_{k + 1} )## over ##F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_{k } )## ... and hence getting some knowledge of the dimension of ##F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_n )## over ##\mathbb{Q}## ... but even if we do gain such knowledge, how do we use it to show ##\sqrt [3]{2} \notin F## ... ...

    ... AND FURTHER ... anyway ... what is implied by ##\alpha_i^2 \in \mathbb{Q}## ... ... ???



    Help will be much appreciated ...


    Peter
     
    Last edited: May 21, 2017
  2. jcsd
  3. May 21, 2017 #2

    fresh_42

    Staff: Mentor

    Good idea. What can you say about ##\dim_\mathbb{Q} F\;##? And what about ##\dim_\mathbb{Q}\mathbb{Q}[\sqrt[3]{2}]\;##? Then what do you know about the fields in between, from ##\mathbb{Q}## to ##F\;:\;\mathbb{Q} \subseteq \mathbb{Q}[\alpha_1]\subseteq \mathbb{Q}[\alpha_1,\alpha_2]\subseteq \ldots \subseteq F\;##?
    It means that ##\alpha_i## is a root of ##x^2-r## for some rational ##r \in \mathbb{Q}##, ergo the minimal polynomial of ##\alpha_i## because ##\alpha_i\notin \mathbb{Q}## which rules out that the degree could be less than two.
     
  4. May 22, 2017 #3

    Thanks for the help, Andrew ... ...

    Just a simple clarification though ... ...

    You write, in the context of ##\alpha_i^2 \in \mathbb{Q}##, the following ... ...

    " ... ... It means that ##\alpha_i## is a root of ##x^2-r## for some rational ##r \in \mathbb{Q}##, ergo the minimal polynomial of ##\alpha_i## because ##\alpha_i\notin \mathbb{Q}## which rules out that the degree could be less than two ... ... "


    ... BUT ..

    ... how do we know that ##\alpha_i \notin \mathbb{Q}## ... indeed maybe ##\alpha_i## is in ##\mathbb{Q}## ... then ##\alpha_i## will satisfy a polynomial of degree 1 in ##\mathbb{Q}## ... namely ##x - \alpha_i## ... ...

    ... so doesn't ##\alpha_i^2 \in \mathbb{Q}## mean that

    ## [ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_i ) \ : \ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_{i - 1} ) ] ##

    is either 1 or 2 ... ...

    can you clarify/comment ?


    {Note : I have to confess that although I think it may well be true that ## [ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_i ) \ : \ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_{i - 1} ) ] ## is either 1 or 2 (i.e a positive integer less than or equal to 2) I do not know the precise argument for why it cannot be larger than 2 ... say 3 r 4 or 37 or something ... can you explain ...}


    Peter
     
    Last edited: May 22, 2017
  5. May 22, 2017 #4

    fresh_42

    Staff: Mentor

    Hi Michael!

    All cases in which ##\alpha_i \in \mathbb{Q}## can be omitted. Say ##\alpha_1 , \ldots \alpha_{k-1} \in \mathbb{Q}## and ##\alpha_k, \ldots , \alpha_n \notin \mathbb{Q}## then ##F=\mathbb{Q}[\alpha_1,\ldots , \alpha_n]=\mathbb{Q}[\alpha_k,\ldots , \alpha_n]## and we're back in the case where all ##\alpha_i \notin \mathbb{Q}##. (The case in which there isn't a proper extension at all is also trivial, since ##\sqrt[3]{2} \notin \mathbb{Q}## and nothing has to be shown.)

    So key to this exercise lies in the formula for finite field extensions:
    Let ##\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{L}## be finite field extensions, then ##[\mathbb{L}:\mathbb{F}] = [\mathbb{L} : \mathbb{K}]\,\cdot \,[\mathbb{K}:\mathbb{F}]\;##.

    In case you cannot use this equation, you might investigate how the roots of the minimal polynomial ##m_{\mathbb{Q}}(\sqrt[3]{2})(x)## can be or cannot be expressed by rationals and roots of the form ##x = \alpha_i = \sqrt{r_i}## with ##r_i \in \mathbb{Q}\;##.
     
  6. May 22, 2017 #5

    Thanks fresh_42 ... appreciate your help ...

    Just reflecting on what you have said ... but obviously most helpful ...

    Thanks again,

    Peter
     
  7. May 23, 2017 #6

    Thanks to Andrew and fresh_42 for all the help ...

    Now to finish the exercise based on the help you gave ...


    Essentially then, it follows that ...

    ##[ F \ : \ \mathbb{Q} ] = 2^k## where ##0 \le k \le n##


    Now ... ... suppose ##\sqrt [3]{2} \in F ## ... (and try for a contradiction)


    Then since ##\mathbb{Q} \subseteq \mathbb{Q} ( \sqrt [3]{2} ) \subseteq F##

    ... we have that ...

    ##[ F \ : \ \mathbb{Q} ] = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] [ \mathbb{Q} ( \sqrt [3]{2} ) \ : \ \mathbb{Q} ]##

    So that


    ##2^k = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] . 3##


    ... BUT ...


    ##3## does not divide ##2^k## ... ... Contradiction!


    So ##\sqrt [3]{2} \notin F##



    Can someone please confirm that the above is correct ... ...

    Peter
     
  8. May 23, 2017 #7

    fresh_42

    Staff: Mentor

    Yes, this is correct. At some place, an extension of degree three (or at least ##3m##) would be needed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Field Extensions - Dummit and Foote - Exercise 13, 13.2 ...
  1. Extension fields (Replies: 2)

  2. Field Extensions (Replies: 26)

  3. Extension Field (Replies: 2)

Loading...