Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are $\mathbb{Q} + \mathbb{Q}\sqrt 5 $ and $\mathbb{Q} + \mathbb{Q}\sqrt {10} $ isomorphic fields?

The problem that I'm solving says to prove that they are not. I have proven that they are fields, that they are equal to $\mathbb{Q}\left[ {\sqrt 5 } \right]$ and $\mathbb{Q}\left[ {\sqrt {10} } \right]$, respectively, but they have the same Galois group ${\mathbb{Z}_2}$, so I cant use that to obtain contradiction.

One solution that I came up with is not very elegant and goes like this:

If they are isomorphic, then they are also isomorphic as vector fields over $\mathbb{Q}$, one has basis $\left\{ {1,\sqrt 5 } \right\}$ and the other $\left\{ {1,\sqrt {10} } \right\}$.

Suppose that $\varphi $ is an isomorphism between those 2 fields, which implies it is also a bijective linear map. Then, for $a \in \mathbb{Q}$ either $\varphi \left( a \right) = \lambda a$ or $\varphi \left( a \right) = \lambda a\sqrt {10} $ for some $\lambda \ne 0$.

If $\varphi \left( a \right) = \lambda a$, then ${\lambda ^2}aa = \varphi \left( a \right)\varphi \left( a \right) = \varphi \left( {aa} \right) = \lambda aa$ for all $a \in \mathbb{Q}$, we take $a \ne 0$ and obtain $\lambda = 1$. If $\varphi \left( a \right) = \lambda a\sqrt {10} $ then $\lambda aa\sqrt {10} = \varphi \left( {aa} \right) = \varphi \left( a \right)\varphi \left( a \right) = 10{\lambda ^2}aa,\forall a \in \mathbb{Q}$, we take $a \ne 0$ and obtain $\lambda = \frac{{\sqrt {10} }}{{10}}$ so $\varphi \left( a \right) = \frac{{\sqrt {10} }}{{10}}a\sqrt {10} = a,\forall a \in \mathbb{Q}$.

In either case, $\varphi $ is $\mathbb{Q}$-embedding.

For $\varphi $ to be surjective, it is then necessary that $\varphi \left( {\sqrt 5 } \right) = \lambda \sqrt {10} $ fore some $\lambda \ne 0$. We have $5 = \varphi \left( 5 \right) = \varphi \left( {\sqrt 5 \sqrt 5 } \right) = \varphi \left( {\sqrt 5 } \right)\varphi \left( {\sqrt 5 } \right) = 10{\lambda ^2} \Rightarrow \lambda = \pm \frac{{\sqrt 2 }}{2}$.

So, the only 2 possibilities are ${\varphi _{1,2}}:\mathbb{Q}\left[ {\sqrt 5 } \right] \to \mathbb{Q}\left[ {\sqrt {10} } \right]$ given by ${\varphi _{1,2}}\left( {a + b\sqrt 5 } \right) = a \pm b\sqrt 5 $, but $\sqrt 5 \notin \mathbb{Q}\left[ {\sqrt {10} } \right]$. We conclude that there is no such isomorphism.

Is there an easier, more elegant way, using theory of field extensions, embeddings and Galois theory, to see that these 2 fields are not isomorphic?

EDIT: I realized now that if we know that $\varphi $ is $\mathbb{Q}$-embedding, then it must be $\varphi \left( {a + b\sqrt 5 } \right) = a \pm b\sqrt 5 $ because $ \pm \sqrt 5 $ are the only conjugates of $\sqrt 5 $ over $\mathbb{Q}$.

The only question that remains is: is there an easier way to see that $\varphi $ is $\mathbb{Q}$-embedding?

share|cite|improve this question
There is no need to deal with embeddings and the like. See my answer below. – user38268 Nov 19 '12 at 13:55
up vote 4 down vote accepted

You can do it in a similar, but shorter way.

First of all, if there is an isomorphism $\varphi \colon \mathbb{Q}[\sqrt{10}] \to \mathbb{Q}[\sqrt{5}]$, then $\varphi|_{\mathbb{Q}}$ is an identity map. To prove this, first notice that $\varphi(1)=1$, then that $\varphi(n)=n$ for any $n \in \mathbb{Z}$, and then that $\varphi(r)=r$ for any $r \in \mathbb{Q}$.

Then, in $\mathbb{Q}[\sqrt{10}]$ there is an element $\alpha$ such that $\alpha^2=10$. Set $\beta = \varphi(\alpha) \in \mathbb{Q}[\sqrt{5}]$. Then $$ \beta^2 = (\varphi(\alpha))^2 = \varphi(\alpha^2) = \varphi(10) = 10. $$ But it is very easy to show that $\beta^2=10$ is impossible for any $\beta \in \mathbb{Q}[\sqrt{5}]$, so this is a contradition.

share|cite|improve this answer
There are in fact no homomorphisms. – user38268 Nov 19 '12 at 13:55
Thank you both, the fisrt part of this answer also answers my question regarding necessity of $\varphi $ being $\mathbb{Q}$-embedding, so I accepted this answer and up-voted both. – Alen Nov 19 '12 at 13:59

The argument can be cleaned up a lot as follows. There are in fact no $\Bbb{Q}$ - algebra homomorphisms between the two extensions. For if there were, say that $\varphi(\sqrt{5}) = a+b\sqrt{10}$ for $a,b \in \Bbb{Q}$. Then squaring both sides we get that $5 = (a^2 + 2ab\sqrt{10} + 10b^2)$ and so either $a$ or $b$ is zero. If $a = 0$ we get that $10x^2 - 5$ is reducible over $\Bbb{Q}$, a contradiction by Eisenstein's criterion. If $ b= 0$ we also get a contradiction because then $\sqrt{5}$ is rational. It follows that no non-trivial homomorphism can exist.

share|cite|improve this answer
We don't even need to use Eisenstein...$5 = 10b^2$ clearly has no rational solutions since clearing denominators $\sqrt{2}$ would then be rational. Or after clearing denominators see that one side has even power of $5$ whereas other had odd power. – fretty Nov 19 '12 at 13:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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