I'd like to show that for $p,q$ distinct primes, the extensions $\mathbb Q(\sqrt p),\mathbb Q(\sqrt q)$ are not isomorphic. I don't really have knowledge of the "high-level language" of algebraic number theory, so my approach is going to be very bit-wise, if you will.

Here's my try:

If there were such an isomorphism, call it $\varphi$. I know that we must have: $$\left\{\begin{align}&\varphi\mid_{\mathbb Q} = \text{id} \\ &\varphi(\sqrt p) =\pm\sqrt p \end{align}\right.$$

So essentially the existence of $\varphi$ means that for any $a+b'\sqrt p$ there exist $c,d$ such that $\varphi(a+b'\sqrt p) = a+b\sqrt p = c+d\sqrt q$ (where $b$ might be a sign change of $b'$). Take, to simplify $a=0, b=1$. Let $\alpha = \sqrt p - d\sqrt q = c$, from above. We can rearrange to obtain $$\begin{align}&(\alpha-\sqrt p)^2 = d^2q \Rightarrow \\ \Rightarrow& \ \alpha^2 - 2\alpha\sqrt p + p - d^2q = 0 \\ \Rightarrow& \ \sqrt p -\frac{c^2+p-d^2q}{2c} = \sqrt p - \lambda = 0\end{align}$$

But then it has to be that $(x^2-p) \mid (x-\lambda)$, which cannot be.

This proof however doesn't satisfy me for two reasons. First of all, I'd like to have stopped at the polynomial I found in $\alpha$, shown that it was irreducible in $\mathbb Q(\sqrt2)$, and argued in terms of extension degrees. However I don't know how to show that. The second reason is that I'm not using the primality of $p,q$ other than having it imply the irreducibility of $x^2-p$. It seems to me from the way the exercise is worded that if they were compound and $x^2-p,x^2-q$ were still irreducible, we might have a homomorphism.

Can someone clarify?

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    $\begingroup$ You did use primality of $p$ and $q$ when you divided by $c$. $\endgroup$ – Najib Idrissi Dec 15 '14 at 17:22
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    $\begingroup$ Quick alternative idea: we know that there exist rational $a$ and $b$ such that $\varphi (\sqrt{p}) = a+b \sqrt{q}$, and that $\varphi (\sqrt{p})^2 = p$. It seems easy to get a contradiction. $\endgroup$ – D. Thomine Dec 15 '14 at 17:22
  • $\begingroup$ @GPerez, this post should help: math.stackexchange.com/questions/9188/… $\endgroup$ – JeffW89 Dec 15 '14 at 19:52
  • $\begingroup$ @amWhy I have to disagree that this question is a duplicate of a question asked postierorly... $\endgroup$ – GPerez Dec 22 '18 at 19:01

I'm not convinced by your claim that $\phi(\sqrt{p}) = \pm \sqrt{p}$ (is this a typo for $\pm \sqrt{q}$?)

However, we do know that $\phi(\sqrt{p}) = a + b \sqrt{q}$ for some $a , b \in \mathbb{Q}$.

So $p = \phi(p) = \phi(\sqrt{p}^2) = \phi(\sqrt{p})^2 = (a + b \sqrt{q})^2 = a^2 + b^2q + 2ab \sqrt{q}$

So either $a = 0$ or $b=0$. If $b=0$ then we have $p = a^2$, if $a=0$ then we have $p = b^2q$; neither of these can occur as $p$ is prime and distinct from $q$. Contradiction.

With regards to your point about using primality of $p$, we do have $\mathbb{Q}(\sqrt{pa^2}) \cong \mathbb{Q}(\sqrt{p})$ for any integer $a$ and prime $p$, for example.

  • $\begingroup$ Since it must be a $\mathbb Q$-isomorphism, it must map roots of the minimal polynomial to other roots. That was my reasoning for that particular bit. Thanks! $\endgroup$ – GPerez Dec 15 '14 at 17:33
  • $\begingroup$ @both - fair, I got thrown by the difference between making sure $\phi$ was a valid function and therefore couldn't be a homomorphism, and making sure $\phi$ had the homomorphism property and therefore couldn't exist (they have the same effect, but I would almost never take the latter approach) $\endgroup$ – Christopher Dec 15 '14 at 17:38

What would be $\phi(\sqrt p)$? Let $$\phi(\sqrt p)=a+b\sqrt q$$

Then $$p=p\phi(1)=\phi(p)=a^2+qb^2+2ab\sqrt q$$ therefore $$2ab=0$$

We conclude that $a^2=p$ or $qb^2=p$, and both options are contradictions.


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