What is the difference of saying finding a basis for $F$ as a vector space and as a $\mathbb Q$-vector space?

Question

Here is the question I am trying to solve:

Let $F = \mathbb Q (\sqrt{2}, \sqrt{3})$ be the smallest subfield of $\mathbb C$ that contains $\sqrt{2}$ and $\sqrt{3}$.

$(a)$ Find a basis for $F$ as a $\mathbb Q$-vector space.

$(b)$ Determine all of the ring automorphisms of $F.$

Here is a trial for the solution of $(a)$:

Since $\mathbb Q(\sqrt{2}) = \{ a + b \sqrt{2}: a,b \in \mathbb Q\},$ so $B_1 = \{1, \sqrt{2}\}$ is a basis of $\mathbb Q(\sqrt{2})$ and since $\mathbb Q(\sqrt{3}) = \{ a + b \sqrt{3}: a,b \in \mathbb Q\},$ so $B_2 = \{1, \sqrt{3}\}$ is a basis of $\mathbb Q(\sqrt{3}).$

Therefore, a basis for $F = \mathbb Q(\sqrt{2}, \sqrt{3})$ is given by:

$B = \{uv: u \in B_1, v \in B_2 \}= \{1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}$ is a basis of $\mathbb Q(\sqrt{2}, \sqrt{3})$ by the bases $B_1$ & $B_2$ given above.

And we can write the following:

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

My questions are:

1- Is the above solution correct?

2- What theorem allows us to say that the basis is the product of the bases as mentioned above?

Could anyone help me answer these questions please?

Answer 1

This is correct, with one small gap.

Saying e.g. "a $\mathbf Q$-vector space" rather than "a vector space" is that in the former case we make explicit that the scalars are elements in $\mathbf Q$.

There is a theorem, which is often called the multiplicativity of degrees, which guarantees the following. If $K\subseteq F\subseteq E$ is a tower of fields and $\{\alpha_i\}$ is a basis for $F$ over $K$, and $\{\beta_j\}$ for $E$ over $F$. Then any $x\in E$ can be written

$\displaystyle x=\sum_j b_j \beta_j$

where $b_j\in F$ and each of the $b_j$ can be written

$\displaystyle b_j=\sum_i a_{ji}\alpha_i$

where $a_{ji}\in K$ so

$\displaystyle x=\sum_j\sum_i a_{ji}\alpha_i\beta_j$

This shows that any element of $E$ can be written as a linear combination of $\{\alpha_i \beta_j\}$ with coefficients in $K$, so this set is spanning.

Now we show linear independence. Assume

$\displaystyle 0=\sum_j\sum_i c_{ji}\alpha_i\beta_j$.

Then

$\displaystyle 0=\sum_j(\sum_i c_{ji}\alpha_i)\beta_j \Rightarrow 0=\sum_i c_{ji}\alpha_i$.

due to linear independence of the $\beta:\mathrm s$. Similarly we show that

$0=\sum_i c_{ji}\alpha_i\Rightarrow c_{ji}=0$.

Thus the product elements of the bases thus form a basis.

So, the small gap in your solution above is that it is not obvious (though it is true) that $\{1, \sqrt 3\}$ is a basis over $\mathbf Q(\sqrt 2)$ just because it is linearly independent over $\mathbf Q$. This is needed in the product step. To show this I would assume there is a relation $\sqrt 3= a+b\sqrt 2$ with $a,b\in \mathbf Q$ and derive a contradiction.