# Show that Two Fields are Equal: $\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3})$ ## Problem 215

Show that fields $\Q(\sqrt{2}+\sqrt{3})$ and $\Q(\sqrt{2}, \sqrt{3})$ are equal. Add to solve later

## Proof.

It follows from $\sqrt{2}+\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})$ that we have $\Q(\sqrt{2}+\sqrt{3})\subset \Q(\sqrt{2}, \sqrt{3})$.

To show the reverse inclusion, consider
$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}\in \Q(\sqrt{2}+\sqrt{3}).$ This yields that we have
$\sqrt{6}\in \Q(\sqrt{2}+\sqrt{3}).$

Now we can express $\sqrt{2}$ in terms of elements of $\Q(\sqrt{2}+\sqrt{3})$ as follows. We have
\begin{align*}
\sqrt{6}(\sqrt{2}+\sqrt{3})-2(\sqrt{2}+\sqrt{3})=\sqrt{2}\in \Q(\sqrt{2}+\sqrt{3})
\end{align*}
(Note that the numbers on the left hand side are all in the field $\Q(\sqrt{2}+\sqrt{3})$.)

Hence we also have
$(\sqrt{2}+\sqrt{3})-\sqrt{2}=\sqrt{3}\in \Q(\sqrt{2}+\sqrt{3}).$ Therefore the elements $\sqrt{2}, \sqrt{3}$ are in the field $\Q(\sqrt{2}+\sqrt{3})$, hence
$\Q(\sqrt{2}, \sqrt{3}) \subset \Q(\sqrt{2}+\sqrt{3}).$ Since we showed both inclusions, we have
$\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3}).$ Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Field Theory ##### Galois Group of the Polynomial $x^p-2$.

Let $p \in \Z$ be a prime number. Then describe the elements of the Galois group of the polynomial $x^p-2$....

Close