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

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

• Does the first field contain an element whose square is $3$? – André Nicolas Jul 11 '16 at 18:30
• No... there is no such element in the first field@Andre – BijanDatta Jul 11 '16 at 18:33
• Well, that settles it, for if an isomorphism takes $a+b\sqrt{2}$ to $\sqrt{3}$, then $(a+b\sqrt{2})^2=3$. – André Nicolas Jul 11 '16 at 18:39

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.

• I cannot understand the statement "an isomorphism would necessarily fix $Q$". May you explain plz. – BijanDatta Jul 11 '16 at 18:46
• @B.R.Datt: Start with the insight that an isomorphism necessarily fixes $1$ as multiplicative identity. It follows that an isomorphism fixes $\mathbb{N}$ and consequently all of $\mathbb{Q}$. – hardmath Jul 11 '16 at 19:40
• Added some cross-references @BijanDatta – Kaj Hansen Apr 11 at 8:47

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.

• What ia the relation between two isomorphic fields and roots of the polynomials? I need more explanation. #DanRust – BijanDatta Jul 11 '16 at 18:44
• I think you mean "By a rationality argument then $a^2 + 3b^2 = 2$ and $2ab = 0$, and so $a$ or $b$ is zero? Of course, the fact that either $a=0$ or $b=0$ follows from $2ab=0$, not $a^2 + 3b^2 = 2$. – MCT Jul 11 '16 at 18:49