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