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Is $Q(\sqrt5) $ and $Q(\sqrt6)$ isomorphic? Now if I show $$t^2 -5 = (\frac{\sqrt6}{\sqrt5} t)^2 -6=0 $$, does this say they are isomorphic?

user665125
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    Hint: in the first field, the equation $x^2-5=0$ has solutions, in the second it doesn't. –  Mar 11 '20 at 19:15
  • Being Galois, if they are isomorphic then they are equal. But $(3)$ is a prime ideal of $O_{\Bbb{Q}(\sqrt{5})}$ but not of $O_{\Bbb{Q}(\sqrt{6})}$ – reuns Mar 11 '20 at 19:17
  • Is it because the second field has this form $a+b \sqrt6$ where $a,b \in Q$ and this form can produce 5 in $x^2$? Then, we can they are are not isomorphic. Is this correct? – user665125 Mar 11 '20 at 19:21
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    @user665125 You presumably mean "this form cannot produce 5". Note that the rigorous proof will use the fact that any potential isomorphism of those fields will map $0$ into $0$ and $1$ into $1$, thus will map any rational numbers (including $5$) into themselves. –  Mar 11 '20 at 19:32
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    Yes, the pre-image $a+b\sqrt{6}$ of $\sqrt{5}$ will have to satisfy $0=(a+b\sqrt{6})^2-5=(a^2+6b^2-5)+2ab\sqrt{6}$ implies that either $ab=0$ or $\sqrt{6}=\frac{a^2+6b^2-5}{-2ab}$ is rational. If $a=0\neq b$, then $a^2-5=0$ implies that $\sqrt{5}$ is rational. Finally, if $b=0\neq a$, then $6b^2-5=0$ implies that $\sqrt{5/6}$ is rational. All those claims of $\sqrt{5},\sqrt{6},\sqrt{5/6}$ being rational lead to contradiction by the usual descent arguments. –  Mar 11 '20 at 19:34
  • For example, if $p,q$ are relatively prime integers and $6(p/q)^2-5=0$, then $p=5p_1$, with $p_1$ integer. Then $30p_1^2/q^2-1=0$. So, $q=5q_1$, with $q_1$ integer. But then $p,q$ would have $5$ as a common factor contradicting the assumption. –  Mar 11 '20 at 19:37

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