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.

Solution 1:

$\mathbb{Q}(\sqrt{2})$ 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 $\sqrt{2}$

Solution 2:

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.

Solution 3:

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 $K=\mathbb Q(\sqrt 2)$.