# showing that two simple field extensions of $\mathbb{Q}$ are not isomorphic

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 $\mathbb{C}$?

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.
• Is there any way of answer 1) without invoking $\mathbb{R}$ ? – Chilote Sep 18 '15 at 2:26
• @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. – guest Sep 18 '15 at 3:54
• 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 $\mathbb{R}$. – guest Sep 18 '15 at 3:58