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.
 A: $\mathbb{Q}(\sqrt{2})$ is by definition the smallest subfield of $\mathbb{C}$ that contains both $\mathbb{Q}$ and $\sqrt{2}$.  The surprising thing (perhaps) is that any element of $\mathbb{Q}(\sqrt{2})$ can be written in the form $a + b\sqrt{2}$, with $a,b \in \mathbb{Q}$.  To prove this, you have to show the following:


*

*That the set of all elements of the form $a + b\sqrt{2}$, with $a,b \in \mathbb{Q}$ is a field (in other words, that it is closed under addition and multiplication, contains both additive and multiplicative identities, contains additive inverses, and contains multiplicative inverses for all nonzero elements);

*That it contains $\mathbb{Q}$

*That it contains $\sqrt{2}$

A: If $\mathbb Q(\sqrt2)$ is defined as the set of number of the form $p+q\sqrt2$ with $p,q\in \mathbb Q$, it is easy to show it is a field, the only tricky part might be multiplicative inverse, you can use the conjugate for that.
On the other hand any field $F$ that contains $\mathbb Q$ and $\sqrt 2$ also contains $q\sqrt2$ with $q$ rational, since $F$ is closed under products and $q$ and $\sqrt2$ belong to $F$. Finally, since $F$ contains every rational $p$ and every number of the form $q\sqrt2$ it contains their sum: $p+q\sqrt2$.
This proves $Q(\sqrt2)\subseteq F$ as desired.

Please note it would be unusual for $\mathbb Q (\sqrt 2)$ to be defined that way.
A: 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 $K=\mathbb Q(\sqrt 2)$.
A: Another method can be:
We know that if $L$ is a finite extension of $K$ and if $K$ is a finite extension of $F$ then $L$ is a finite extension of $F$ and $[L:F]=[L:K]$ $[K:F]$ .
So suppose there is a field $F$ such that  $\mathbb Q\subseteq F\subseteq\mathbb Q$, then $[F:\mathbb Q] | [\mathbb Q:\mathbb Q(\sqrt2)]$ $\implies$ $[F:\mathbb Q]|2$ .
Now if $[F:\mathbb Q]=1$ then $F=\mathbb Q$
If $[F:\mathbb Q]=2$ $\implies$ $[\mathbb Q(\sqrt2):\mathbb Q]=1$ $\implies$ $F=\mathbb Q(\sqrt2)$
So you can see that only subfields of $\mathbb Q(\sqrt2)$ are $\mathbb Q$ and  $\mathbb Q(\sqrt2)$ therefore your result.
