# Are the fields $\mathbb{Q}$ and $\mathbb{Q[\sqrt2]}$ isomorphic?

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.

• @JSwanson, Is my approach correct? – Subhasis Biswas Sep 11 '19 at 0:09

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.

• Technically you also need to show that $2$ can't be moved by isomorphism. – Noah Schweber Sep 11 '19 at 1:24
• Yes. Nice point. – José Carlos Santos Sep 11 '19 at 1:29
• 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. – Subhasis Biswas Sep 11 '19 at 5:48

• $$\mathbb Q[\sqrt 2]$$ 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.

• 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 $\mathbb Q(\sqrt 3)$. – lhf Sep 12 '19 at 11:33

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 $$f(p) = p$$.

• 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? – Subhasis Biswas Sep 11 '19 at 5:48