# Two Quadratic Fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are Not Isomorphic ## Problem 99

Prove that the quadratic fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic. Add to solve later

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## Hint.

Note that any homomorphism between fields over $\Q$ fixes $\Q$ pointwise.

## Proof.

Assume that there is an isomorphism $\phi:\Q(\sqrt{2}) \to \Q(\sqrt{3})$.
Let $\phi(\sqrt{2})=a+b\sqrt{3}\in \Q(\sqrt{3})$, where $a, b \in \Q$.

Then since $\phi$ fixes the elements of $\Q$, we have
\begin{align*}
2&=\phi(2)=\phi((\sqrt{2})^2)=\phi(\sqrt{2})^2\\
&=(a+b\sqrt{3})^2=a^2+3b^2+2ab\sqrt{3}.
\end{align*}
Hence we have $2=a^2+3b^2$ and $2ab=0$.

From the second equality, we must have $a=0$ or $b=0$. If $a=0$, then $2=3b^2$, and $b=\pm \sqrt{2/3}$ but this is not a rational number. Hence $a \neq 0$ and $b=0$.

But in this case, $2=a^2$ and this implies that $a=\pm \sqrt{2}$, which is not a rational number.
This is a contradiction, and we conclude that the fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic. Add to solve later

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###### More in Field Theory ##### Automorphism Group of $\Q(\sqrt{2})$ Over $\Q$.

Determine the automorphism group of $\Q(\sqrt{2})$ over $\Q$.

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