Notation question related to $\mathbb{Q}(\sqrt{3})$.

Question

Someone asked me a question today about the dimensionality of:

$$ \mathbb{Q}(\sqrt{3})=\{ a+b \cdot \sqrt{3} : a,b \in \mathbb{Q}\} $$

I am thinking that they are interpreting it as a vector space when they ask a question about dimensionality. Seeing it has two parameters and we could interpret $a$ and $b \cdot \sqrt{3}$ as two independent vectors (not multiples of each other by irrationality of $\sqrt{3} $ argument).

I would say the answer is $2$.

Am I correct in thinking this, is it me or is this weird notation, has anybody seen this before? Here is the guy's textbook:

I ask this because I found the question interesting, but the most confusing thing one can do is using weird notation and not explaining what you mean.

Answer 1

As a $\Bbb Q$-vector space, ${\Bbb Q}(\sqrt 3) = \{a+b\sqrt 3\mid a,b\in{\Bbb Q}\}$ is isomorphic to ${\Bbb Q}^2$ with the assignment $$a+b\sqrt 3\mapsto{a\choose b},$$ since addition in both spaces are componentwise, in particular, $$(a+b\sqrt 3) + (c+d\sqrt 3) = (a+c) + (b+d)\sqrt 3.$$

Answer 2

It is entirely true that $\Bbb Q(\sqrt3)$ is a $2$-dimensional vector space over $\Bbb Q$, with standard basis $\{1, \sqrt3\}$. It is even very common to think about it like that. In fact, given a field extension $F$ over a field $E$, the dimension of $F$ as a vector space over $E$ is a very importand invariant, called the degree of the extension.

Answer 3

Part of the text is missing from the picture; I'm assuming that's the part o the text that clarifies what they mean : in particular one can see in the picture "$\mathbb{Q}$ with dimension" and so I think the rest of the text says something like "blabla is a vector space over $\mathbb{Q}$ with dimension ". So I don't understand what clarification you need. Let me adress your points about the answer though.

First of all, the textbook is very imprecise when it says "let $\mathbb{Q}$ be a set of rational numbers". In my answer I'll be assuming they meant "the set of rational numbers".

Then, your intuition is correct that there are two parameters, and so the dimension over $\mathbb{Q}$ "should be $2$".

But the notion of dimension is a precise one and so you have to go beyond the intuition and actually prove your answer. The point you were trying to make when you said "we could interpret $a$ and $b\sqrt{3}$ as two independent vectors" can be made precise by stating that $\{1,\sqrt{3} \}$ is a basis of $\mathbb{Q}(\sqrt{3})$.

The proof is easy : it is a generating family, by definition; and it is linearly independent because $\sqrt{3}$ is irrational (you should check the details if this isn't clear).

Thus $\mathbb{Q}(\sqrt{3})$ has a basis with $2$ elements, so it is of dimension $2$.