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

Question

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.

Answer 1

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.

Answer 2

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 $f(p) = p$.

Answer 3

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.