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Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$ ?

$$\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbb{Q}\}$$

$$\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbb{Q} \rbrace $$

So if an element is in $\mathbb Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?

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    $\begingroup$ $\mathbf{Q}(\sqrt{2}+\sqrt{3}) \not= \{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ because $\sqrt{2}+\sqrt{3}$ does not have degree 2 over $\mathbf{Q}$. $\endgroup$
    – lhf
    Dec 22, 2011 at 14:28
  • $\begingroup$ Hi lhf, then what is it? ? ? ? $\endgroup$
    – Tashi
    Dec 22, 2011 at 14:30
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    $\begingroup$ $\alpha=\sqrt{2}+\sqrt{3}$ has degree 4 over $\mathbf{Q}$ and so a basis is $1,\alpha,\alpha^2,\alpha^3$. $\endgroup$
    – lhf
    Dec 22, 2011 at 14:35
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    $\begingroup$ @Tashi: This may just be an issue of notation, since your argument in the question seems to suggest that you thougt that $\{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ also includes $\sqrt2\sqrt3$. (It doesn't.) $\endgroup$
    – joriki
    Dec 22, 2011 at 14:39

6 Answers 6

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$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3}) $.

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To recap the notation: $\mathbb{Q}[x]$ denotes the ring of polynomials with rational coefficients. The square bracket notation $\mathbb{Q}[\sqrt{2}]$ means $\{p(\sqrt{2}) : p \in \mathbb{Q}[x]\}$. It's easy to show that $\mathbb{Q}[\sqrt{2}] = \{a+b\sqrt{2}:a,b,\in \mathbb{Q}\}.$

A really nice fact is that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}[\sqrt{2}][\sqrt{3}],$ where \begin{array}{ccc} \mathbb{Q}[\sqrt{2}][\sqrt{3}] &=& \{a+b\sqrt{3} : a,b \in \mathbb{Q}[\sqrt{2}] \} \\ \\ &=& \{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6} : p,q,r,s \in \mathbb{Q}\}. \end{array} These all use square brackets because they are considered as rings. The round brackets give us the set of rational expressions, which are fields, e.g.

$$\mathbb{Q}(\sqrt{2},\sqrt{3}) = \left\{ \frac{\alpha}{\beta} : \alpha,\beta \in \mathbb{Q}[\sqrt{2},\sqrt{3}]\right\}$$

It turns out that, as sets, $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$.

In turns of the representation of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ we have seen that, as a set, we have $\{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}:p,q,r,s \in \mathbb{Q}\}$. There are many representations for this fiels, e.g. $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(1,\sqrt{2},\sqrt{3})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3})$, etc. We can show that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is also a representation of the same field too.

Think of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ as a $\mathbb{Q}$-vector space with $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ as a basis. Let $\gamma := \sqrt{2}+\sqrt{3}.$ We have $\gamma^2 = 5+2\sqrt{6},$ $\gamma^3 = 11\sqrt{2}+9\sqrt{3}$ and $\gamma^4 = 49 + 20\sqrt{6}$. Putting this together:

$$\left[\begin{array}{cccc} 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0 \\ 49 & 0 & 0 & 20 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

The 4-by-4 matrix on the left is non-singular, and so we can invert:

$$\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \frac{1}{2}\!\left[\begin{array}{cccc} 0 & 20 & 0 & -2 \\ -9 & 0 & 1 & 0 \\ 11 & 0 & -1 & 0 \\ 0 & -49 & 0 & 5 \end{array}\right]\left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

This tells us that $1$, $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ can all be expressed as rational polynomials in $\gamma = \sqrt{2}+\sqrt{3}$.

\begin{array}{ccc} 10\gamma^2-\gamma^4 &=& 1 \\ \tfrac{1}{2}(\gamma^3-9\gamma) &=& \sqrt{2} \\ \tfrac{1}{2}(11\gamma - \gamma^3) &=& \sqrt{3} \\ \tfrac{1}{2}(5\gamma^4-49\gamma^2) &=& \sqrt{6} \end{array}

It follows that $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6}) \cong \mathbb{Q}(\gamma) = \mathbb{Q}(\sqrt{2}+\sqrt{3}).$

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    $\begingroup$ I like this version, since it's clear how to extend to the situation where more roots are adjoined. $\endgroup$ May 28, 2013 at 19:23
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    $\begingroup$ I'm a bit late to the party but this is a beautiful answer - If I could +2 I would, however, +1 anyway. Best, Bacon. $\endgroup$
    – user284001
    Nov 3, 2015 at 9:26
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    $\begingroup$ the best explanation for a general set up $\endgroup$ Apr 27, 2016 at 14:27
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    $\begingroup$ How does it turn out that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$? How does one prove this? $\endgroup$ Sep 19, 2018 at 8:16
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    $\begingroup$ @HansStricker They are the same as sets, i.e. they have the same elements. $\mathbb Q[\sqrt 2,\sqrt 3]$ contains $a+b\sqrt 2+c\sqrt 3 + d\sqrt 6$, while $\mathbb Q(\sqrt 2,\sqrt 3)$ contains all fractions $$\frac{a+b\sqrt 2+c\sqrt 3 + d\sqrt 6}{e + f\sqrt 2 + g\sqrt 3 + h\sqrt 6}$$ By "rationalising the denominator" we can write every element of $\mathbb Q(\sqrt 2,\sqrt 3)$ in the form $r+s\sqrt 2 + t\sqrt 3 + u\sqrt 6$, meaning $\mathbb Q(\sqrt 2,\sqrt 3) \subseteq \mathbb Q[\sqrt 2,\sqrt 3]$. Clearly, $\mathbb Q[\sqrt 2,\sqrt 3] \subseteq \mathbb Q(\sqrt 2,\sqrt 3)$. $\endgroup$ Sep 19, 2018 at 15:54
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Hint $\ $ If a field $\rm F $ has two $\rm F$-linear independent combinations of $\rm\ \sqrt{a},\ \sqrt{b}\ $ then you can solve for $\rm\ \sqrt{a},\ \sqrt{b}\ $ in $\rm F$. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between $\rm F$ and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see here or here.

In this case it's simpler to notice $\rm\ E = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since

$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{a-b}{\sqrt{a}+\sqrt{b}}\ \in\ E = \mathbb Q(\sqrt{a}+\sqrt{b}) $$

To be explicit, notice that $\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in E\ $ so solving the linear system for the roots yields $\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (v-u)/2,\ $ both of which are clearly $\rm\in E,\:$ since $\rm\:u,\:v\in E\:$ and $\rm\:2\ne 0\:$ in $\rm\:E,\:$ so $\rm\:1/2\in E.\:$ This works over any field where $\rm\:2\ne 0,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.

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If it's allowed to use the Galois theory, it can be proved as following. Since the subgroup of the Galois group of the field extension $\mathbb{Q} (\sqrt2,\sqrt 3)$ over $\mathbb{Q}$ which the subfield $\mathbb{Q}(\sqrt 2+\sqrt 3)$ is trivial, therefor we know the result by the Galois theory. I admit it is not too trivial since one has to verify something as said above.

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Also note, $\sqrt{2}=\frac{(\sqrt{2}+\sqrt{3})^3-9(\sqrt{2}+\sqrt{3})}{2}$ and $-\sqrt{3}=\frac{(\sqrt{2}+\sqrt{3})^3-11(\sqrt{2}+\sqrt{3})}{2}$ and so done.

Now suppose wants to show $\mathbb{Z}[\sqrt{2},\sqrt{3}]\neq \mathbb{Z}[\sqrt{2}+\sqrt{3}]$ then, note minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Z}$ is of degree $4$.

So we can write $\mathbb{Z}[\sqrt{2}+\sqrt{3}]=\{a_1+a_2x+a_3x^2+a_4x^3|x=\sqrt{2}+\sqrt{3},a_i\in \mathbb{Z}\}$ now simple case chase shows $\sqrt{2}$ not in $\mathbb{Z}[\sqrt{2}+\sqrt{3}]$

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Another solution:

Clearly it has to be a subfield of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$. Thus its degree has to be one of these numbers: $1,2,4$. We just have to rule out it is of degree $2$. If it is, then there is a minimal polynomial of degree two s.t $\sqrt{2}+\sqrt{3}$ is its root. However, if there was such polynomial, say $x^2+bx+c$ then $5+c+2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$ which is impossible as $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$ and the equation implies it is linearly dependent!

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