Degree of $\mathbb{Q}(\sqrt 2, \sqrt 3)$ over $\mathbb{Q}$ 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?
 A: 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 $a=b=c=d=0$.
A: To show that a set of vectors is a basis for a vector space, you need to show that the vectors are linearly independent and that they span the vector space. In this case, you are trying to show that the set ${1,\sqrt{2}, \sqrt{3}, \sqrt{6}}$ is a basis for the vector space $\mathbb{Q}$
First, show that the vectors are linearly independent. This means that none of the vectors in the set can be written as a linear combination of the other vectors. Since the vectors in the set are all scalars (i.e., they are all real numbers), this means that none of the scalars in the set can be written as a linear combination of the other scalars. For example, we can't have $\sqrt{2}=a+b\sqrt{3}+c\sqrt{6}$ for some scalars $a$, $b$, and $c$.
To show that the vectors span the vector space, show that every element of the vector space can be written as a linear combination of the vectors in the set. In this case, since the vector space is $\mathbb{Q}$, we need to show that every rational number can be written as a linear combination of the scalars in the set. This is true, because every rational number can be written as a ratio of two integers, and every integer can be written as a linear combination of the scalars in the set (e.g., $4=1+1+1+1$).
Therefore, the set ${1,\sqrt{2}, \sqrt{3}, \sqrt{6}}$ is a basis for the vector space $\mathbb{Q}$.
To compute the degree of the vector space $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$, you can use the fact that the degree of a vector space over a field is equal to the number of elements in any basis for the vector space. In this case, we can use the elements ${1, \sqrt{2}, \sqrt{3}, \sqrt{6}}$ as a basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$, since they are linearly independent and span the vector space.
Thous, the degree of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is 4.
A: 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.
