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

Problems and Solutions in Field Theory in Abstract Algebra

Problem 215

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


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
\sqrt{6}(\sqrt{2}+\sqrt{3})-2(\sqrt{2}+\sqrt{3})=\sqrt{2}\in \Q(\sqrt{2}+\sqrt{3})
(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
\[\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3}).\]

LoadingAdd to solve later

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Field Theory
Galois theory problem and solution
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 $x^p-2$....