Does there exist any isomorphism between $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 3]$?

Question

Does there exist any isomorphism between the fields $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 3]$ ?

Answer 1

Consider the equation $x^2=2$ which has a solution in the first field. Suppose $f \colon \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{3}]$ is an isomorphism, so $f(x)^2 = f(x^2) = f(2)$. But note that $f(2)=2$, as an isomorphism must fix the multiplicative identity. Hence, we have $y^2=2$ where $y = f(x) \in \mathbb{Q}[\sqrt{3}]$. Then $y^2 = (a+b\sqrt{3})^2 = 2$ and so $a^2+2ab\sqrt{3} +3b^2=2$. By a rationality argument then $a^2+3b^2 =2$ and so $a$ or $b$ is zero. If $b$ is zero then we conclude that $2$ has a rational square root. If $a$ is zero then we conclude that $2/3$ has a rational square root. Both lead to contradictions.

Answer 2

Suppose there were an isomorphism of fields $\phi: \mathbb{Q}[ \sqrt{2}] \rightarrow \mathbb{Q}[\sqrt{3}]$. It is straightforward to show that such an isomorphism would necessarily fix $\mathbb{Q}$ since it would fix $1$ (see here).

Next, we must have $\phi(\sqrt{2}) = a + b\sqrt{3}$ for some $a, b \in \mathbb{Q}$. The multiplicativity of $\phi$ gives us $\phi( \sqrt{2}) \phi(\sqrt{2}) = \phi \Big( \left(\sqrt{2} \right)^2 \Big) = \phi(2) = a^2 + 2ab \sqrt{3} + 3b^2$. Because $\phi$ fixes $\mathbb{Q}$, we arrive at $a^2 + 2ab \sqrt{3} + 3b^2 = 2$, and herein lies a contradiction.

---

$\underline{\textbf{Caution}}$: Although $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are not isomorphic as fields, they are isomorphic as vector spaces since they are both of dimension $2$ over $\mathbb{Q}$. So without context, one needs to be careful with unqualified statements like "$\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are not isomorphic." Click here for further discussion.