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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?

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There is no need to deal with embeddings and the like. See my answer below. –  user38268 Nov 19 '12 at 13:55

2 Answers 2

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.

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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.

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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

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