Proof that the set $mathbb{Q}left[sqrt2right]$ is an $mathbb{Q}$-vector space

I think it is easiest when you view $mathbb{Q}(sqrt{2})$ like you would think about $mathbb{C}$.

You might know that $mathbb{C}=mathbb{R}^2$ is a vector space.

The elements of $mathbb{C}$ can be written in the form $a+ib$.

The elements of $mathbb{Q}(sqrt{2})$ can be written in the form $a+sqrt{2}b$.

Everything you have to do is to show all the axioms of a vector space. There are no tricks needed.

So we want to show that $mathbb{Q}(sqrt{2})$ is a $mathbb{Q}$-vector space.

There are many axioms to check. For example that $+$ is closed.

$+:mathbb{Q}(sqrt{2})timesmathbb{Q}(sqrt{2})tomathbb{Q}(sqrt{2})$

You would check this like this.

Let $x,yinmathbb{Q}(sqrt{2})$. Then $x=a+bsqrt{2}$ and $y=c+dsqrt{2}$, where $a,b,c,dinmathbb{Q}$

Then is $x+y=(a+bsqrt{2})+(c+dsqrt{2})=(a+c)+sqrt{2}(b+d)inmathbb{Q}(sqrt{2})$ as $a+c$ and $b+d$ are elements of $mathbb{Q$

What should the neutral element of addition look like? How should an inverse element of $xinmathbb{Q}(sqrt{2})$ look like?

This for example you have to figure out. So you have to make a guess first, and then verify that your guess is correct. It is not difficult, and the calculations can be done as shown above.

You have to use that you know that $mathbb{Q}$ is a field. So we know how inverses look in $mathbb{Q}$, we know its algebraic structure (associativ, commutativ and so on). This will be used execsivly in the calculations, as it verifies that our calculations are correct.

Just start, you will eventually figure out that it is indeed pretty easy. If you have any questions feel free to ask.