Prove that $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ I am trying to prove $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$.  I have proved that $\{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ is indeed a field with unit.  However is this enough to show that this is indeed equivilant to $\mathbb{Q}(\sqrt{2})$?
 A: $\mathbb{Q} (\sqrt{2})$ is defined to be the intersection of every subfield of $\mathbb{C}$ that contains $\sqrt{2}$. As such it is a subfield of $\mathbb{C}$ and is  contained in every subfield of $\mathbb{C}$ that contains $\sqrt{2}$. So if you have shown that $\{ a+b \sqrt{2} : a,b \in  \mathbb{Q} \}$ is a field (which clearly has $\sqrt{2}$ as one of its elements) you have shown $\mathbb{Q} (\sqrt{2}) \subseteq \{ a+b \sqrt{2} : a,b \in  \mathbb{Q} \}$. To show the inclusion the other way you need to show why $\mathbb{Q} (\sqrt{2})$ must contain every rational as well as $\sqrt{2}$. Then since $\mathbb{Q} (\sqrt{2})$ is a subfield of $\mathbb{C}$ it must contain any sum and product of such elements. 
A: The minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $t^2-2$, so that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. A basis of the $2$-dimensional vector space
$\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $(1,\sqrt{2})$, which is easy to see. Hence every element has the form $a+b\sqrt{2}$. To verify that it is a field has been done on MSE already, see for example here. 
A: You need to show that $\{a+b\sqrt{2}\} \subseteq \mathbb{Q}(\sqrt{2})$, and that $\mathbb{Q}(\sqrt{2}) \subseteq \{a+b\sqrt{2}\}$. 
Then it follows that $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}\}$. So you don't have to prove that $\{a+b\sqrt{2}\}$ has unit and inverse, if it equals a field then it definitely already has those. 
