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This is a probably a stupid question, but I can't figure it out myself.

I think that they are not, but I can't prove it formally. One reason that they are probably not isomorphic is that $x^2-2x-1 \in \mathbb{Q}[x] \subset \mathbb{Q}[\sqrt2][x]$ has no roots in $\mathbb{Q}$, but it has its roots in $\mathbb{Q}[\sqrt2]$.

I am not sure whether or not my argument is valid. Any hint/suggestion would be appreciated.

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  • $\begingroup$ @JSwanson, Is my approach correct? $\endgroup$ – Subhasis Biswas Sep 11 '19 at 0:09
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Your proof is correct, but it would be simpler to observe that, in $\mathbb Q$ , there is no element whose square is $2$, whereas in $\mathbb Q\left[\sqrt2\right]$ there is.

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    $\begingroup$ Technically you also need to show that $2$ can't be moved by isomorphism. $\endgroup$ – Noah Schweber Sep 11 '19 at 1:24
  • $\begingroup$ Yes. Nice point. $\endgroup$ – José Carlos Santos Sep 11 '19 at 1:29
  • $\begingroup$ here's what I have done finally : Suppose isomorphism $\phi: \mathbb{Q} \mapsto \mathbb{Q[\sqrt2]}$ exists. Then $1+\sqrt{2}=\phi(x)$ for some $x\in \mathbb{Q}$. Now, $\phi(1)=1 \implies \phi(2)=\phi(1)+\phi(1)=2$. So, $\phi(2x)=\phi(2)\phi(x)=2\phi(x)$. Hence, $\phi(x^2)-\phi(2x)-1=3+2\sqrt 2-2(1+\sqrt2)-1=0\implies \phi(x^2-2x-1)=0 \implies x^2-2x-1=0\in \mathbb{Q}$. But no such $x$ exists. $\endgroup$ – Subhasis Biswas Sep 11 '19 at 5:48
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Here is a roadmap:

  • $\mathbb Q[\sqrt 2]$ is a vector space of dimension $2$ over $\mathbb Q$.

  • A ring homomorphism $f: \mathbb Q \to \mathbb Q[\sqrt 2]$ is a linear transformation over $\mathbb Q$. (*)

  • Therefore, the image of $f$ has dimension at most $1$, and so $f$ cannot be surjective.

(*) see the answer by Charles Hudgins.

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  • $\begingroup$ Proving that two fields are not isomorphic is not that easy in general. Having the same dimension over the base field is a necessary but not sufficient condition. Typically you have to exploit some arithmetic property. Try proving that $\mathbb Q(\sqrt 2)$ is not isomorphic to $\mathbb Q(\sqrt 3)$. $\endgroup$ – lhf Sep 12 '19 at 11:33
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Note that if $L$ and $K$ are fields containing $\mathbb{Q}$, then any field homomorphism $f : L \to K$ must fix $\mathbb{Q}$. To see this, note that $f(0) = 0$ and $f(1) = 1$, so, by induction $$ f(n) = f(1 + \cdots + 1) = f(1) + \cdots + f(1) = 1 + \cdots + 1 = n $$ It is easily checked that we also have $f(-n) = -n$ for $n \in \mathbb{N}$.

Moreover, for $p \in \mathbb{Z}$, $$ 1 = f(1) =f\left(p \cdot \frac{1}{p}\right) = f(p) f\left(\frac{1}{p}\right) = p f\left(\frac{1}{p}\right) $$ which implies $f(1/p) = 1/p$ for all $p \in \mathbb{Z}$ We conclude, therefore, that if $p,q \in \mathbb{Z}$, then $f(p/q) = p/q$.

A field isomorphism $f : \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}$ cannot exist because it could not be injective. We would have to map $\sqrt{2}$ to some $p \in \mathbb{Q}$, but we already have $f(p) = p$.

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  • $\begingroup$ I am trying to write out the proof in a bit different way now: Suppose isomorphism $\phi: \mathbb{Q} \mapsto \mathbb{Q[\sqrt2]}$ exists. Then $1+\sqrt{2}=\phi(x)$ for some $x\in \mathbb{Q}$. Now, $\phi(1)=1 \implies \phi(2)=\phi(1)+\phi(1)=2$. So, $\phi(2x)=\phi(2)\phi(x)=2\phi(x)$. Hence, $\phi(x^2)-\phi(2x)-1=3+2\sqrt 2-2(1+\sqrt2)-1=0\implies \phi(x^2-2x-1)=0 \implies x^2-2x-1=0\in \mathbb{Q}$. But no such $x$ exists. Is it better now? $\endgroup$ – Subhasis Biswas Sep 11 '19 at 5:48

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