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Math Help - Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

  1. #1
    Super Member Bernhard's Avatar
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    Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

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

    Example 15 on page 282 (see attachment) reads as follows:

    ---------------------------------------------------------------------------------------------------------------------------------

    Example 15.

    Let  E = \mathbb{Q} ( \sqrt{2} , \sqrt{5} )  .

    Find  [E \ : \ \mathbb{Q} ] , exhibit a   \mathbb{Q}  -basis of E, and show that   E =  \mathbb{Q} ( \sqrt{2}  +  \sqrt{5} )  . Then find the minimum polynomial of   \sqrt{2}  +  \sqrt{5}  over   \mathbb{Q}  .

    -----------------------------------------------------------------------------------------------------------------------------------


    In the solution we read:

    Solution: We write  L = \mathbb{Q} ( \sqrt{2} )  for convenience so that  E = L(\sqrt{5})  ... ... etc

    ... ... ... We claim that  X^2 - 5   is the minimal polynomial of  \sqrt{5}   over L. Because  \sqrt{5}   and   - \sqrt{5}   are the only roots of  X^2 - 5   in  \mathbb{R} , we merely need to show that  \sqrt{5} \notin L   . ... ... etc




    My problem is the following:

    How does showing  \sqrt{5} \notin L   imply that  X^2 - 5   is the minimal polynomial of  \sqrt{5}   over L?


    Can someone please help with this issue?

    Peter
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    Last edited by Bernhard; October 18th 2013 at 03:39 PM.
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  2. #2
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    Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 15

    If \sqrt{5} \in L then X^2-5 = (X+\sqrt{5})(X-\sqrt{5}), which would imply that the minimal polynomial would be X-\sqrt{5}. So, \sqrt{5} \notin L implies X-\sqrt{5}\notin L[X], which implies the minimal polynomial must be at least degree 2. Since X^2-5 \in L[X] is a degree 2 polynomial with \sqrt{5} a root, it is the minimal polynomial.
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