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Thread: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

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    Super Member Bernhard's Avatar
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    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

    I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions.

    Example 13 on page 282 (see attachment) reads as follows:

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

    In the third line of the explanation - see page 282 of attachment - we read:

    "But $\displaystyle [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 $ because $\displaystyle u^2 \notin \mathbb{Q} $ ... ... "

    Can someone explain why it follows that $\displaystyle u^2 \notin \mathbb{Q} \Longrightarrow [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 $

    Peter
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

    $\displaystyle [\mathbb{Q}(k) : \mathbb{Q}] = 1$ implies that the minimal polynomial for $\displaystyle k$ over $\displaystyle \mathbb{Q}[x]$ has degree 1. Hence the minimal polynomial is $\displaystyle x-k$ which implies $\displaystyle k \in \mathbb{Q}$.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

    Thanks SlipEternal

    Just one further point of clarification:

    I can see why the minimal polynomial is x - k , but why exactly does that imply that $\displaystyle k \in \mathbb{Q} $?

    Peter
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

    Quote Originally Posted by Bernhard View Post
    Thanks SlipEternal

    Just one further point of clarification:

    I can see why the minimal polynomial is x - k , but why exactly does that imply that $\displaystyle k \in \mathbb{Q} $?

    Peter
    Since $\displaystyle (x-k) \in \mathbb{Q}[x]$, that means that the coefficients of each term are elements of $\displaystyle \mathbb{Q}$. Hence $\displaystyle -k \in \mathbb{Q}$, so $\displaystyle k \in \mathbb{Q}$.
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