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    Can someone show me a full proof of this please explaining as much as possible. I have seen some proofs on stack exchange but I don't fully understand them.

    So we have to show that the left is a subset of the right and vise versa. But it seems as if it is not simply just taking a general element of the left and then manipulating so that it is an element of the right, and vise versa.

    Also, they say one of the directions is trivial but I don't see how so please explain this too!

    Please help!
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    (Original post by cooldudeman)
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    Can someone show me a full proof of this please explaining as much as possible. I have seen some proofs on stack exchange but I don't fully understand them.

    So we have to show that the left is a subset of the right and vise versa. But it seems as if it is not simply just taking a general element of the left and then manipulating so that it is an element of the right, and vise versa.

    Also, they say one of the directions is trivial but I don't see how so please explain this too!

    Please help!
    I'm going to refer you back to stackexchange - as this thread here has a pretty complete description of what is going on; especially look at the highest rated answer. I think that the first thing you need to get sorted out is your descriptions of the two extensions in terms of basis elements - if you do that, you will see that one way round ( \mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})) is straightforward.

    Let us know if you're still stuck after reading that thread.
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    (Original post by Gregorius)
    I'm going to refer you back to stackexchange - as this thread here has a pretty complete description of what is going on; especially look at the highest rated answer. I think that the first thing you need to get sorted out is your descriptions of the two extensions in terms of basis elements - if you do that, you will see that one way round ( \mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})) is straightforward.

    Let us know if you're still stuck after reading that thread.
    So the basis of Q(sqrt2 , sqrt3) is sqrt2 and sqrt3.

    Basis of Q(sqrt2 + sqrt3) is simply sqrt2 + sqrt3 ??
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    (Original post by cooldudeman)
    So the basis of Q(sqrt2 , sqrt3) is sqrt2 and sqrt3.
    There may be more to this but there is one more element for your basis as:

    

\mathbb{Q} (\sqrt{2}, \, \sqrt{3} ),

    is a field extension and you need closure under multiplication fir any elements in this field extension.
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    (Original post by cooldudeman)
    So the basis of Q(sqrt2 , sqrt3) is sqrt2 and sqrt3.

    Basis of Q(sqrt2 + sqrt3) is simply sqrt2 + sqrt3 ??
    No. Let's try and sort this out before we go any further. We'll think about how you might go about constructing a basis for a field extension over the rationals.

    So, lets look at \mathbb{Q}(\alpha). If you extend the rationals by this \alpha then, because the field extension is closed under multiplication, we must have \alpha, \alpha^2, \alpha^2, \ldots, \alpha^n, \ldots also in the extension. This appears to make it go on forever! But if we think about \mathbb{Q}(\sqrt{2}), for a moment, we see that it doesn't. This is because \alpha = \sqrt{2} satisfies a polynomial equation - namely  x^2 - 2 = 0. So whenever you get to \alpha^2 in the field extension you can replace it by 2. This immediately gives us a basis for \mathbb{Q}(\sqrt{2}), namely  \{a + b \sqrt{2} | a,b, \in \mathbb{Q} \}. Had we considered  \alpha = \pi then this would not have happened. \pi does not satisfy a polynomial equation in the same way.  \mathbb{Q}(\pi) is infinite dimensional over \mathbb{Q}.

    Now consider  \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2})(\sqrt{3}). Think through the definition of this double extension and you'll see that we need  \sqrt{2}, \sqrt{3} and \sqrt{2} \times \sqrt{3} = \sqrt{6} in the basis.

    Now consider  \mathbb{Q}(\sqrt{2} + \sqrt{3}) . If we set \alpha = \sqrt{2} + \sqrt{3}, we need to consider  \alpha, \alpha^2, \alpha^2, \ldots, \alpha^n, \ldots again. Does this series go on, or does it terminate? It terminates because \alpha is algebraic over \mathbb{Q}. It satisfies a polynomial equation. In fact it satisfies a polynomial of minimal degree called the minimal polynomial. Do you know what this polynomial is for  \sqrt{2} + \sqrt{3}? In fact it has degree four, so you know that a basis for  \mathbb{Q}(\sqrt{2} + \sqrt{3}) is given by 1, \alpha, \alpha^2, \alpha^3. Now you start doing the exercise - the point of which is to show that this basis can in fact be written in terms of the basis of  \mathbb{Q}(\sqrt{2}, \sqrt{3}) .

    That is what is being shown in the highest rated answer in the stackexchange thread.
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    (Original post by Gregorius)
    No. Let's try and sort this out before we go any further. We'll think about how you might go about constructing a basis for a field extension over the rationals.

    So, lets look at \mathbb{Q}(\alpha). If you extend the rationals by this \alpha then, because the field extension is closed under multiplication, we must have \alpha, \alpha^2, \alpha^2, \ldots, \alpha^n, \ldots also in the extension. This appears to make it go on forever! But if we think about \mathbb{Q}(\sqrt{2}), for a moment, we see that it doesn't. This is because \alpha = \sqrt{2} satisfies a polynomial equation - namely  x^2 - 2 = 0. So whenever you get to \alpha^2 in the field extension you can replace it by 2. This immediately gives us a basis for \mathbb{Q}(\sqrt{2}), namely  \{a + b \sqrt{2} | a,b, \in \mathbb{Q} \}. Had we considered  \alpha = \pi then this would not have happened. \pi does not satisfy a polynomial equation in the same way.  \mathbb{Q}(\pi) is infinite dimensional over \mathbb{Q}.

    Now consider  \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2})(\sqrt{3}). Think through the definition of this double extension and you'll see that we need  \sqrt{2}, \sqrt{3} and \sqrt{2} \times \sqrt{3} = \sqrt{6} in the basis.

    Now consider  \mathbb{Q}(\sqrt{2} + \sqrt{3}) . If we set \alpha = \sqrt{2} + \sqrt{3}, we need to consider  \alpha, \alpha^2, \alpha^2, \ldots, \alpha^n, \ldots again. Does this series go on, or does it terminate? It terminates because \alpha is algebraic over \mathbb{Q}. It satisfies a polynomial equation. In fact it satisfies a polynomial of minimal degree called the minimal polynomial. Do you know what this polynomial is for  \sqrt{2} + \sqrt{3}? In fact it has degree four, so you know that a basis for  \mathbb{Q}(\sqrt{2} + \sqrt{3}) is given by 1, \alpha, \alpha^2, \alpha^3. Now you start doing the exercise - the point of which is to show that this basis can in fact be written in terms of the basis of  \mathbb{Q}(\sqrt{2}, \sqrt{3}) .

    That is what is being shown in the highest rated answer in the stackexchange thread.
    Ok i THINK I understand now. Thank you.

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    (Original post by cooldudeman)
    Ok i THINK I understand now. Thank you.

    Posted from TSR Mobile
    There's a very nice little book "Fields and Galois Theory" by John Howie in the Springer Undergraduate series. It has lots of worked examples on this sort of stuff and your university library may have free access to the pdf version.
 
 
 
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