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


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