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Thread: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

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    Super Member Bernhard's Avatar
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    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Example 13 from Nicholson: Introduction to Abstract Algebra, Section 6.2, page 282 reads as follows: (see attachment)

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    Example 13: If $\displaystyle u = \sqrt[3]{2} $ show that $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2 $

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    The solution comes down to the following:

    Given $\displaystyle \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q} $

    so $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] $

    Now Nicholson shows that $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = 3 $ and $\displaystyle [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] = 3 $

    so $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 $

    Then Nicholson (I think) concludes that $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2 $

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    My problem is as follows:

    How (exactly) does it follow that:

    $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 \Longrightarrow \mathbb{Q}(u) = \mathbb{Q}(u)^2 $

    Can someone help?

    Peter
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    Last edited by Bernhard; Oct 15th 2013 at 05:46 PM.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Think minimal polynomials.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by SlipEternal View Post
    Think minimal polynomials.
    Thanks SlipEternal ... well, given that $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 $ the degree of minimal polynomial of the root u over $\displaystyle \mathbb{Q}(u)^2 $ is 1.

    Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in $\displaystyle \mathbb{Q}(u)^2 $???)

    How does this lead us to $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2 $?

    I think I need a bit more help?

    Peter
    Last edited by Bernhard; Oct 15th 2013 at 06:01 PM.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal ... well, given that $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 $ the degree of minimal polynomial of the root u over $\displaystyle \mathbb{Q}(u)^2 $ is 1.

    Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in $\displaystyle \mathbb{Q}(u)^2 $???)

    How does this lead us to $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2 $?

    I think I need a bit more help?

    Peter
    That implies that $\displaystyle u \in \mathbb{Q}(u^2)$ since $\displaystyle u$ is the coefficient of $\displaystyle x^0$ in that minimal polynomial.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    Quote Originally Posted by SlipEternal View Post
    That implies that $\displaystyle u \in \mathbb{Q}(u^2)$ since $\displaystyle u$ is the coefficient of $\displaystyle x^0$ in that minimal polynomial.
    Sorry SlipEternal ... still not completely following.

    Let me restate where I am (thanks to your help)

    We have that the degree of the minimal polynomial m(u) of $\displaystyle \mathbb{Q}(u) $ over $\displaystyle \mathbb{Q}(u^2) $ is 1.

    Thus, the minimal polynomial m(u) is a monic polynomial of degree 1 with coefficients in $\displaystyle \mathbb{Q}(u^2) $ such that m(u) = 0.

    So then $\displaystyle m(u) = a_0 + a_1 u = 0 $ where $\displaystyle a_0, a_1 \in \mathbb{Q}(u^2) $

    How do we (formally) show that $\displaystyle u \in \mathbb{Q}(u^2) $

    Peter
    Last edited by Bernhard; Oct 15th 2013 at 06:50 PM.
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    A monic polynomial means $\displaystyle a_1 = 1$.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

    oh ... of course!

    So $\displaystyle m(u) = a_0 + u = 0 $ where $\displaystyle a_0 \in \mathbb{Q}(u^2) $

    So $\displaystyle u = -a_0 $

    and $\displaystyle a_0 $ and of course $\displaystyle - a_0 \in \mathbb{Q}(u^2) $

    Thanks!

    Peter
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