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Math Help - Field Theory - Algebraic Extensions - D&F Section 13.2 - Example 2 - page 526

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    Super Member Bernhard's Avatar
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    Field Theory - Algebraic Extensions - D&F Section 13.2 - Example 2 - page 526

    I am reading Dummit and Foote on algebraic extensions. I am having some issues understanding Example 2 on page 526 - see attachment.

    Example 2 on page 526 reads as follows:

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

    (2) Consider the field  \mathbb{Q} ( \sqrt{2}, \sqrt{3} ) generated over  \mathbb{Q} by  \sqrt{2} and  \sqrt{3} .

    Since  \sqrt{3} is of degree 2 over  \mathbb{Q} the degree of the extension  \mathbb{Q} ( \sqrt{2}, \sqrt{3} ) is at most 2 and is precisely 2 if and only if  x^2 - 3 is irreducible over  \mathbb{Q} ( \sqrt{2} ) . ... ... etc etc

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

    My question is: why exactly does it follow that the degree of the extension  \mathbb{Q} ( \sqrt{2}, \sqrt{3}  ) is at most 2 and is precisely 2 if and only if  x^2 - 3  is irreducible over  \mathbb{Q} ( \sqrt{2} ) ?

    Although I may be being pedantic I also have a concern about why exactly  \sqrt{3} is of degree 2 over  \mathbb{Q} . I know it is intuitively the case or it seems the case that the minimal polynomial is in this case  x^2 - 3 but how do we demonstrate this for sure - or is it obvious? (I may be overthinking this??)

    Can someone help with the above issues/problems?

    Peter
    Last edited by Bernhard; September 27th 2013 at 08:42 PM.
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    Re: Field Theory - Algebraic Extensions - D&F Section 13.2 - Example 2 - page 526

    Read up on degrees of field extensions over other fields. The idea is, given an irreducible polynomial over a field, the degree of the field extension by a root of that polynomial is the degree of the minimal polynomial with that root. Just as you showed, x^2-3 has \sqrt{3} as a root. Now, can you prove that there is not a polynomial of smaller degree whose root is \sqrt{3}? Easily. Assume there is a first degree polynomial over \mathbb{Q} whose root is \sqrt{3}: ax+b, a,b\in \mathbb{Q}, a\neq 0. Now, plug in the root: a\sqrt{3}+b = 0 \Longrightarrow \sqrt{3} = -\dfrac{b}{a}\in \mathbb{Q}, which is false. So, there is no degree one polynomial with \sqrt{3} as a root.

    So, consider the field extension \mathbb{Q}(\sqrt{2},k) where k^2 \in \mathbb{Q}, but x^2-k^2 is reducible over \mathbb{Q}(\sqrt{2}). For example, if k = \sqrt{8}, then x^2-8 = (x+2\sqrt{2})(x-2\sqrt{2}). Now, consider the field extension \mathbb{Q}(\sqrt{2},\sqrt{8}) over \mathbb{Q}(\sqrt{2}). It clearly has order 1 since any element of the field extended by \sqrt{8} is still in the field extended by \sqrt{2}.

    So, chance are, you are probably overthinking it, because the book is just describing the expected behavior. It works just as you would expect it to. It turns out that these degrees are multiplicative. It is easy to check that \mathbb{Q}(\sqrt{2},\sqrt{3}) has degree 4 over \mathbb{Q}, and a minimal polynomial is (x^2-2)(x^2-3).
    Last edited by SlipEternal; September 28th 2013 at 08:56 AM.
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    Super Member Bernhard's Avatar
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    Re: Field Theory - Algebraic Extensions - D&F Section 13.2 - Example 2 - page 526

    Thanks SlipEternal, that post is most helpful!

    Peter
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