4
$\begingroup$
  1. How can I show that $\mathbb{Q}[x]/(x^2+2)$ and $\mathbb{Q}[x]/(x^2-2)$ are not isomorphic?

If $\alpha$ and $\beta$ are zeros of $x^2+2$ and $x^2-2$ in certain extensions respectively, we have that $\mathbb{Q}[x]/(x^2+2)\cong\mathbb{Q}(\alpha)$ and $\mathbb{Q}[x]/(x^2-2)\cong\mathbb{Q}(\beta)$.

First I thought that if they were isomorphic then there would be a $\mathbb{Q}$-isomophism mapping $\alpha$ to $\beta$ (for some natural reason that I was missing) and then it would be easy to prove that such $\mathbb{Q}$-homomorphism is not possible, but the truth is that this is not always true.

For example $\mathbb{R}[x]/(x^2+1)$ and $\mathbb{R}[x]/(x^2+x+1)$ are isomorphic to $\mathbb{C}$ and there is no $\mathbb{R}$-homomorphism mapping a zero of $x^2+1$ into a zero of $x^2+x+1$. So I run out of ideas.

  1. How can I show that $\mathbb{R}[x]/(x^2+x+1)$ is isomorphic to $\mathbb{C}$?
$\endgroup$
4
$\begingroup$

Hints:

For 1, show that in $\mathbb{Q}(\sqrt{-2})$, $-2$ is a perfect square. In $\mathbb{Q}(\sqrt{2})$, $-2$ is not a perfect square because this field has a real embedding (fixing $\mathbb{Q}$). Any isomorphism between the two fields would exhibit a square root for $-2$ in $\mathbb{Q}(\sqrt{2})$ which is impossible.

For 2, just factor your polynomial over $\mathbb{C}$. You get $(x-\rho)(x-\overline{\rho})$ with $\rho=\frac{-1+i\sqrt{3}}{2}$. This way your field embeds in $\mathbb{C}$, contains $\mathbb{R}$ and contains $i$ (why?). The fundamental theorem of algebra says that $\mathbb{R}(i)=\mathbb{C}$ so the embedding to $\mathbb{C}$ given by $x\mapsto \frac{-1+i\sqrt{3}}{2}$ is an isomorphism.

$\endgroup$
  • $\begingroup$ Is there any way of answer 1) without invoking $\mathbb{R}$ ? $\endgroup$ – Chilote Sep 18 '15 at 2:26
  • 1
    $\begingroup$ @Chilote in a sense, yes. To show that the second field does not have a square root of $-2$, try to write $-2=(a+b\sqrt{2})^2$. From this conclude that either $a$ or $b$ is zero (otherwise $-2$ would be irrational), so $-2=a^2$ or $-2=2b^2$, both of which have no solutions in the integers. $\endgroup$ – guest Sep 18 '15 at 3:54
  • $\begingroup$ I am not sure if this counts, since I did the computations inside a real embedding of your field, but there is no other way: the field is totally real, any way you realize it will be insider $\mathbb{R}$. $\endgroup$ – guest Sep 18 '15 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.