The question first asks to show that $\sqrt{3}$ has a root in $\mathbb{Q}(\sqrt{3})=\{a+b\sqrt{3}|a,b \in \mathbb{Q}\}$ but not in $\mathbb{Q}(\sqrt{2})$. This is I solved by assuming, towards a contradiction, that $\exists a,b\in \mathbb{Q}$ such that $a+b\sqrt{2} = \sqrt{3}$, so $b = \frac{\sqrt{3} - a}{\sqrt{2}}$. Clearly this is a contradiction since $\frac{\sqrt{3}}{\sqrt{2}}$ is irrational, and $\frac{a}{\sqrt{2}}$ is either zero or irrational, so their difference must be irrational.

Then the queestion asks me to explain why this implies that $\mathbb{Q}(\sqrt{2})$ is not isomorphic to $\mathbb{Q}(\sqrt{3})$. In previous questions, I showed that $\mathbb{Q}(\sqrt{2})$ is isomorphic to $\mathbb{Q}[x]/\mathopen{<}x^{2}-2\mathclose{>}$, where the latter is the set of congruence classes modulo $x^2 - 2$, and the same for root 3. I'm guessing that we come to some contradiction if we assumed that a field not containing $\sqrt{3}$ is isomorphic to $\mathbb{Q}[x]/\mathopen{<}x^{2}-3\mathclose{>}$, but I don't see how.

Any hints or suggestions would be appreciated. Thanks!

  • $\begingroup$ The difference between two irrational numbers is not "clearly irrational". The same goes for the ratio of two irrational numbers. $\endgroup$ – Arthur Jan 23 '18 at 18:39
  • $\begingroup$ Consider using \langle and \rangle to get $\langle$ and $\rangle$ instead of < and > which give $<$ and $>$. For example $\langle x^2-2\rangle$ instead of $<x^2-2>$. $\endgroup$ – Fly by Night Jan 23 '18 at 18:43
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    $\begingroup$ Hint: show that if $f : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ were an isomorphism of fields, then $x = f^{-1}(\sqrt{3})$ would have to satisfy $x^2 = 3$. $\endgroup$ – Daniel Schepler Jan 23 '18 at 18:45
  • $\begingroup$ +1 for sharing your thinking. There are many very closely related threads that have discussed this and more general variants of you question. I picked the oldest variant because it happened to deal with exactly these two fields. Please check out the answers to other threads first. If your question is still unanswered, describe it in more detail. We (=not just me) then reconsider whether your question is sufficiently different. $\endgroup$ – Jyrki Lahtonen Jan 23 '18 at 18:50
  • $\begingroup$ @Arthur I didn't flesh out why root 3 over root 2 is irrational, and I'll admit that I assumed the ratio of a non-zero rational and an irrational is irrational. I see why saying an irrational minus an irrational must be irrational is incorrect, thanks for pointing that out. $\endgroup$ – hiroshin Jan 23 '18 at 18:50