wanted to know the following

I am interested in computation of degree of the vector space $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ but I am not sure how to do this.

I am trying to show that we have the basis $\{1,\sqrt{2}, \sqrt{3}, \sqrt{6}\}$ over $\mathbb{Q}$ but how do I proceed?

I assumed $$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=0$$ and then I separated the terms but getting into an ugly calculations like I got $$a+2b^2-3c^2-6d^2=(6cd-2ab)\sqrt{2} $$but unable to move further.

So how do I show this?


2 Answers 2


The easiest way to determine the degree of the extension (that is, the dimension of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$) is to use the multiplicativity of the degree: $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}].$$ We know that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Since $\sqrt{3}$ satisfies $x^2-3\in\mathbb{Q}(\sqrt{2})[x]$, this means that $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]\leq 2$, and so is equal to either $1$ or $2$. So really the only question is whether this extension has degree $1$ or $2$.

The extension has degree $1$ if and only if $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$. This amounts to checking if there exists $a,b\in\mathbb{Q}$ such that $(a+b\sqrt{2})^2 = 3$. It is straighforwad to determine the answer with basic algebra. And so this completely gives you the answer.

Trying to show that $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is linearly independent over $\mathbb{Q}$ is much more time consuming, but amounts to essentially the same thing. The equation you got, $$a+2b^2-3c^2-6d^2=(6cd-2ab)\sqrt{2}$$ tells you that you must have $6cd-2ab=0$ and $a+2b^2-3c^2-6d^2=0$, because the left hand side is a rational, while the right hand side would be irrational if $6cd-2ab\neq 0$. So you know that $ab=3cd$, and that should let you proceed through a somewhat laborious and probably annoying process to conclude that you must have $a=b=c=d=0$.

  • $\begingroup$ (Or, to show $\{ 1,\sqrt{2},\sqrt{3},\sqrt{6} \}$ is a basis, just use the lemma that's presumably part of the proof of the degree multiplication formula, that from $\{ 1, \sqrt{2} \}$ being a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$, and $\{ 1, \sqrt{3} \}$ being a basis for $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, you can conclude that $\{ 1\cdot 1, 1 \cdot \sqrt{3}, \sqrt{2} \cdot 1, \sqrt{2} \cdot \sqrt{3} \}$ is a basis for $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$.) $\endgroup$ Dec 9, 2022 at 18:25
  • $\begingroup$ @DanielSchepler: Yes, that's the proof of Dedekind's Product Theorem, but to do that you have to first show that $1.\sqrt{3}$ is a basis over $\mathbb{Q}(\sqrt{2})$, which means proving the degree is $2$. No sense in re-proving the Product Theorem... $\endgroup$ Dec 9, 2022 at 18:27
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    $\begingroup$ Agreed, if the end goal as stated in the question is to find the degree, then using the product formula is the way to go. But then, if you're interested in finding a concrete basis, that comment would give an easier way. (And if you're not interested in finding a concrete basis, then you could skip the entire last paragraph of this answer.) $\endgroup$ Dec 9, 2022 at 18:40
  • $\begingroup$ @ArturoMagidin, I will have to see those Annoying Calculation to see how $a,b,c $ and $d$ are coming out to be $0$. Thanks for the answer. $\endgroup$ Dec 9, 2022 at 18:42
  • $\begingroup$ @DanielSchepler: Oh, I see the point you were trying to make, sorry for missing it. I was thinking as a replacement for the OPs attempt at showing directly that $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ was a basis, ex nihilo, without using any results from this family of results. $\endgroup$ Dec 9, 2022 at 18:42

It is also possible to get $ac=2bd$ as Arturo Magidin got $ab=3cd$. Then, the procedure becomes easier: If one of $a,b,c,d$ is zero, then at least two of them are zero and then all of them are zero. If all of them are non-zero, then we get $\frac{b}{c}=\sqrt{\frac{3}{2}}$ which is a contradiction.


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