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

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?

• $\mathbf{Q}(\sqrt{2}+\sqrt{3}) \not= \{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ because $\sqrt{2}+\sqrt{3}$ does not have degree 2 over $\mathbf{Q}$.
– lhf
Dec 22, 2011 at 14:28
• Hi lhf, then what is it? ? ? ? Dec 22, 2011 at 14:30
• $\alpha=\sqrt{2}+\sqrt{3}$ has degree 4 over $\mathbf{Q}$ and so a basis is $1,\alpha,\alpha^2,\alpha^3$.
– lhf
Dec 22, 2011 at 14:35
• @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.) Dec 22, 2011 at 14:39

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

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

• I like this version, since it's clear how to extend to the situation where more roots are adjoined. May 28, 2013 at 19:23
• 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.
– user284001
Nov 3, 2015 at 9:26
• the best explanation for a general set up Apr 27, 2016 at 14:27
• How does it turn out that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$? How does one prove this? Sep 19, 2018 at 8:16
• @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, $\mathbb Q[\sqrt 2,\sqrt 3] \subseteq \mathbb Q(\sqrt 2,\sqrt 3)$. Sep 19, 2018 at 15:54

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.

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.

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 $\mathbb{Z}[\sqrt{2}+\sqrt{3}]$

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!