Show that $\mathbb{Q}(\sqrt{2})$ is the smallest subfield of $\mathbb{C}$ that contains $\sqrt{2}$ This was an assertion made in our textbook but I have no idea how to show that either statement is true.  Also would like to show that that $\mathbb{Q}(\sqrt{2})$ is strictly larger than $\mathbb{Q}$, which was the second part of the assertion.

Say that $\mathbb Q(\sqrt 2)$ is the smallest subfield of $\mathbb C$ that contains $\sqrt2$ is assume implicitely the ordinary partial order of the set of subsets of $\mathbb C$. Let $K$ be a field such that $K\subset \mathbb Q(\sqrt 2)$.
If $\sqrt2\in K$ then the set $\sqrt2\mathbb Q$ is not a field because is not closed for the multiplication so we need to take the set $\mathbb Q+\sqrt2\mathbb Q$ in order to have a subfield of $K$ generated by its element $\sqrt2$. But this is the definition of $\mathbb Q(\sqrt 2)$ so $K=\mathbb Q(\sqrt 2)$.
