Finitely generated field extensions This is a really dumb question, but why is $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2} : a,b \in \Bbb{Q}\}$? I am having trouble writing field extensions in this way.
 A: $\mathbb{Q}(\sqrt{2})$ is by definition the smallest field containing both $\mathbb{Q}$ and $\sqrt{2}$.  Since fields are closed under addition and multiplication, we have the following containment:
$$\{a+b\sqrt2\mid a,b\in\mathbb{Q}\}\subset\mathbb{Q}(\sqrt{2})$$
Now, see if you can show that $\{a+b\sqrt2\mid a,b\in\mathbb{Q}\}$ is actually a field itself by rationalizing denominators.  This will show the other containment, proving that the two sets are equal.
A: $\mathbb Q(\sqrt2)$ can be interpreted in two ways. One is that it is the field $\mathbb Q[x]/(x^2-2)$. This is an abstract construction of a field extension of $\mathbb Q$ that contains a solution to the equation $x^2-2=0$, and moreover it is in a precise sense (up to isomorphism) the smallest such field extension. Less abstractly, $\mathbb Q(\sqrt2)$ may also be interpreted to mean the smallest subfield of $\mathbb C$ which contains $\mathbb Q$ and $\sqrt2$. In either case, the set $\{a+b\sqrt2\mid a,b\in \mathbb Q\}$ is $\mathbb Q(\sqrt2)$.  
A: Your follow-up question about $\mathbb{Q}(\sqrt3,i)$ can be answered by thinking about how $\mathbb{Q}(\sqrt2)$ differs from $\mathbb{Q}$. Ittay's answer should give you a helpful way to think about these problems, by noting that $(x^2-2)$ is irreducible in $\mathbb{Q}$. Now think of $\mathbb{Q}(\sqrt3,i)$ as extending $\mathbb{Q}$ first by $\sqrt3$, then by $i$. Are these extensions linearly independent?
A: I remember struggling with this until I found a helpful reference in Chrystal's Algebra, 5th edition (1904) where he shows that any quadratic irrational expression in $\mathbb{Q}[\sqrt{2}]$ reduces to the form $a+b \sqrt{2}$. Think of all the possible valid operations in $\mathbb{Q}$ together with the additional element $\sqrt{2}$. In general you'll get a polynomial divided by a polynomial (i.e. an irrational function) Chrystal then shows this expression can always be reduced to the form $a+b \sqrt{2}$.
