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Math Help - Extension Fields

  1. #1
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    Extension Fields

    Prove that
    \mathbb{Q}( \sqrt{2}, \sqrt{3}) is isomorphic to \mathbb{Q}( \sqrt{2} + \sqrt{3})

    ATTEMPT:

    Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that \sqrt{2}, \sqrt{3} and \sqrt{2} + \sqrt{3} are all zero that would be great.

    I really have no idea what to do can you guys give me hints?
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  2. #2
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    Quote Originally Posted by mulaosmanovicben View Post
    Prove that
    \mathbb{Q}( \sqrt{2}, \sqrt{3}) is isomorphic to \mathbb{Q}( \sqrt{2} + \sqrt{3})

    ATTEMPT:

    Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that \sqrt{2}, \sqrt{3} and \sqrt{2} + \sqrt{3} are all zero that would be great.

    I really have no idea what to do can you guys give me hints?


    I figured it out:

    We will show each set is a subset of the other:

    elements of \mathbb{Q}( \sqrt{2}, \sqrt{3}) have the form (a+b \sqrt{2}) + (c+d \sqrt{2}) \sqrt{3}

    let q be an element of \mathbb{Q}( \sqrt{2} + \sqrt{3}) then q has the form:

    q = a + b( \sqrt{2} + \sqrt{3}) = a + b \sqrt{2}+b \sqrt{3} which is of the form mentioned above with a=a, b=b, c=0, d=b. Therefore \mathbb{Q}( \sqrt{2} + \sqrt{3}) is a subset of \mathbb{Q}( \sqrt{2}, \sqrt{3})


    Now consider ( \sqrt{2} + \sqrt{3})^3 = 11 \sqrt{2} + 9 \sqrt{3} ( I did the math)

    subtract 9( \sqrt{2}+ \sqrt{3}) from both sides and you get:

    2 \sqrt{2} = ( \sqrt{2}+ \sqrt{3})^3 - 9( \sqrt{2}+ \sqrt{3}) so \sqrt{2} is an element of \mathbb{Q}( \sqrt{2} + \sqrt{3}) similar procedure can show that \sqrt{3} is also an element of that set and therefore \mathbb{Q}( \sqrt{2}, \sqrt{3}) is a subset of \mathbb{Q}( \sqrt{2} + \sqrt{3})

    So the sets are the same
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  3. #3
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    your conclusion is correct, but your reasoning is slightly off.

    q has the form a + b√2 + c√3 + d√6, so √2 is the vector (0,1,0,0) and √3 is the vector (0,0,1,0) relative to the basis {1,√2,√3,√6}.

    the implication the other way is fine.

    a more in-depth look at Q(√p,√q) for p,q distinct primes was posted here: http://www.mathhelpforum.com/math-he...ml?pagenumber=
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