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So correct if I am wrong, but these sets are:

$\mathbb Q (\sqrt2 + \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 +d \sqrt6 : a,b,c,d \in \mathbb Q \}$

$\mathbb Q (\sqrt2 , \sqrt3 )=\{a+b\sqrt2 + c\sqrt3 : a,b,c \in \mathbb Q \}$

Was trying to see if the LHS is a subset of RHS and vise versa.

Let $A \in \mathbb Q (\sqrt2 , \sqrt3 )$ with $A=a+b\sqrt2 + c\sqrt3= a+b\sqrt2 + c\sqrt3 + d \sqrt6 - d \sqrt6 = a+b\sqrt2 + c\sqrt3 + d \sqrt6 - d \sqrt3 \sqrt2 $ but i can't really do anything after this. If you factor the root $3$ or $2$, it doesn't seem to help.

snowman
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    A field that contains $\sqrt{2}+\sqrt{3}$ contains its reciprocal. Compute the reciprocal. – André Nicolas Feb 18 '16 at 17:40
  • @AndréNicolas why isnt the reciprecol $(a+ b\sqrt2 + c\sqrt3 + d\sqrt6 )^{-1}$? dont u take the general form of each element in the set? – snowman Feb 18 '16 at 17:51
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    It suffices to show that $\sqrt{2}$ and $\sqrt{3}$ are in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$. (The other direction is trivial.). If $\sqrt{2}+\sqrt{3}$ is in the field, its reciprocal $\sqrt{3}-\sqrt{2}$ is in, therefore the sum $2\sqrt{3}$ is in, therefore $\sqrt{3}$ is in. Note this is the same idea as a couple of the answers, except I am using division instead of squaring and manipulating. – André Nicolas Feb 18 '16 at 17:59
  • @AndréNicolas I cant see how the other direction is trivial. Can you show me how that works please! – snowman Feb 18 '16 at 19:56
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    If $\sqrt{2}$ and $\sqrt{3}$ are in a certain field, their sum is in the field. – André Nicolas Feb 18 '16 at 20:13
  • @AndréNicolas I feel like this question has nothing to do with looking how the elements in the two sets actually look like in general terms. Let me get this straight: If we have $$\mathbb Q (\sqrt2, \sqrt3 )$$ then $\sqrt2 , \sqrt3 \in \mathbb Q (\sqrt2, \sqrt3 )$ which means that their sum which is $$\sqrt2 + \sqrt3 \in \mathbb Q (\sqrt2 + \sqrt3)$$ ??? Because of the operations within a field. – snowman Feb 18 '16 at 20:19
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    We are proving that each of the fields is a subfield of the other. The easier part is to show that $Q(\sqrt{2}+\sqrt{3})$ is a subfield of $Q(\sqrt{2},\sqrt{3}$. For clearly $\sqrt{2}$ and $\sqrt{3}$ are in the field on the right, so their sum is, so all of $Q(\sqrt{2}+\sqrt{3})$ is. – André Nicolas Feb 18 '16 at 20:35
  • @AndréNicolas So for the hard part, in words, can you explain what we are doing please. Are we taking an inverse element of $\mathbb Q (\sqrt2 + \sqrt3)$ and then manipulating it so that it is in $\mathbb Q (\sqrt2 , \sqrt3)$? – snowman Feb 18 '16 at 20:48
  • Also why are we allowed to take the inverse? – snowman Feb 18 '16 at 20:51
  • No, we are showing that both $\sqrt{2}$ and $\sqrt{3}$ are elements of $Q(\sqrt{2}+\sqrt{3})$, which implies all the elements of $Q(\sqrt{2},\sqrt{3})$ are in $Q(\sqrt{2}+\sqrt{3})$. – André Nicolas Feb 18 '16 at 20:52
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    Why are we allowed to take the inverse? $\sqrt{2}+\sqrt{3}$ is in the field $Q(\sqrt{2}+\sqrt{3})$, so its inverse is. – André Nicolas Feb 18 '16 at 20:53
  • @AndréNicolas Can I ask, what are the basis of each of the two sets? Are we meant to just show if the basis of each set exists within the other set then they are equal? – snowman Feb 18 '16 at 21:09
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    Let $H$ and $K$ be fields, with $H$ a subfield of $K$. Let $\alpha\in K$. By definition $H(\alpha)$ is the intersection of all subfields of $K$ that contain $H$ and $\alpha$. Similar for $H(\alpha,\beta)$. – André Nicolas Feb 18 '16 at 21:13
  • @AndréNicolas In exactly the same types of words you just used, can you define what $H(\alpha + \beta )$ is please. – snowman Feb 18 '16 at 21:20
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    The intersection of all subfields of $K$ that contain $H$ and $\alpha+\beta$. – André Nicolas Feb 18 '16 at 21:21
  • @AndréNicolas So in general, when showing two extended fields are equal, say $X[._x] = Y[._y]$, we show that the $._x$ is in the $Y[._y]$ and the $._y$ is in the $X[._x]$??? Sorry for the bad notation. – snowman Feb 18 '16 at 21:25
  • Probably! I cannot read what you wrote. But depending on the detailed situation, one may use some other strategy. By the way, the procedure you started, if carried out correctly, would have worked too. But it gets messy. – André Nicolas Feb 18 '16 at 21:39
  • @AndréNicolas I see. And thanks for letting me know that my way would have worked too. – snowman Feb 18 '16 at 21:45

3 Answers3

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Observe that $\sqrt2+\sqrt3\in\mathbb{Q}(\sqrt2,\sqrt3)$ so by minimality $\mathbb{Q}(\sqrt2+\sqrt3)\subseteq\mathbb{Q}(\sqrt2,\sqrt3)$. For the reverse inclusion consider $(\sqrt2+\sqrt3)^2$ for example and so on.

Sri-Amirthan Theivendran
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It will suffice to show $\sqrt{2}, \sqrt{3} \in \mathbb{Q(\sqrt{2}+\sqrt{3})}$ and that $\sqrt{2}+\sqrt{3} \in \mathbb{Q}(\sqrt{2}, \sqrt{3})$. The latter is obvious the former is a bit trickier.

siegehalver
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I myself had this problem a while ago, but this is because I didn't quite get the definitions. Instead of trying to explicitly describe these sets, remember the definitions.

First we take an algebraic closure of $\mathbb{Q}$, say $\mathbb{C}$. Then $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$, $\sqrt{2}$ and $\sqrt{3}$. Similarly, $\mathbb{Q}(\sqrt{2},+\sqrt{3})$ is the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and $\sqrt{2}+\sqrt{3}$. Well, since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is a field and contains $\sqrt{2}$ and $\sqrt{3}$, the sum $\sqrt{2}+\sqrt{3}$ is in there and so $\mathbb{Q}(\sqrt{2}+\sqrt{3})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$. To show reverse inclusion, it suffices to show that $\sqrt{2}$ and $\sqrt{3}$ are in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ (why?). Now, $\sqrt{2}+\sqrt{3}$ is in there, so its square $5+2\sqrt{6}$ is in there, so its square $-5$= $2\sqrt{6}$ is in there. Then $2\sqrt{6}(\sqrt{2}+\sqrt{3})=2(2\sqrt{3}+3\sqrt{2})$ is in there. Subtracting $\sqrt{2}+\sqrt{3}$ from this a few times and scaling you get what you want. Play around and have fun!

EDIT: André Nicolas gave a way simpler argument for showing the "hard part" in the comments. Anyways, it is always good to play around and discover things for yourself

Shoutre
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