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

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

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

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

How to conclude from there?

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9  
$\mathbf{Q}(\sqrt{2}+\sqrt{3}) \not= \{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ because $\sqrt{2}+\sqrt{3}$ does not have degree 2 over $\mathbf{Q}$. –  lhf Dec 22 '11 at 14:28
    
Hi lhf, then what is it? ? ? ? –  Tashi Dec 22 '11 at 14:30
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$\alpha=\sqrt{2}+\sqrt{3}$ has degree 4 over $\mathbf{Q}$ and so a basis is $1,\alpha,\alpha^2,\alpha^3$. –  lhf Dec 22 '11 at 14:35
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@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.) –  joriki Dec 22 '11 at 14:39

5 Answers 5

up vote 47 down vote accepted

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ 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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2  
Thank you, I can understand this answer! –  Tashi Dec 22 '11 at 14:46
    
@Tashi You're welcome : ) –  Matt N. Dec 22 '11 at 14:50
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@Tashi With only slightly more effort, you can understand my answer too (which I have elaborated). That will enable you to understand not only this special case, but much more general cases - many of which arise frequently in practice. –  Bill Dubuque Dec 22 '11 at 15:06

HINT $\ $ If a field F has two F-linear independent combinations of $\rm\ \sqrt{a},\ \sqrt{b}\ $ then you can solve for $\rm\ \sqrt{a},\ \sqrt{b}\ $ in 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 F and $\rm\ F(\sqrt{a}, \sqrt{b}),$ e.g. see PlanetMath's proof.

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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Thank you very much. –  Tashi Dec 22 '11 at 14:46
    
@Matt My answer is excerpted from one of my old sci.math posts. [1] was a link to PlanetMath's proof of the Primitive Element Theorem. But PlanetMath seems to have problems currently (or the link has rotted) –  Bill Dubuque Dec 22 '11 at 15:18
    
Thanks Bill! Now that my comment is obsolete, I'll delete it. –  Matt N. Dec 22 '11 at 15:19
    
I can't seem to access PlanetMath either. Some alternative references: Theorem 3.6 of this handout of Keith Conrad's and Theorem 5.1 of Milne's notes. Oddly enough, they both use Tashi's question as a first example. –  Dylan Moreland Dec 22 '11 at 15:22
    
@BillDubuque In your last sentence, did you mean to write: "...where the determinant of the linear system is non-zero..."? (rather than "invertible") –  Matt N. May 21 '12 at 10:25

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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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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I like this version, since it's clear how to extend to the situation where more roots are adjoined. –  Mark McClure May 28 '13 at 19:23

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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