# How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3).$ [duplicate]

How to show that $\mathbb Q(\sqrt 2)$ is not field isomorphic to $\mathbb Q(\sqrt 3)?$

My text provides the hint as: Any isomorphism from $\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ is identity when restricted to $\mathbb Q.$

Ya! I can see that: $\mathbb Q(\sqrt 2)\simeq\mathbb Q[x]/\langle x^2-2\rangle$ and $\mathbb Q(\sqrt 3)\simeq\mathbb Q[x]/\langle x^2-3\rangle.$ Now if $\phi$ be such an isomorphism then $\phi:1+\langle x^2-2\rangle\mapsto1+\langle x^2-3\rangle.$ Hence the restriction of $\phi$ over $\{q+\langle x^2-2\rangle:q\in\mathbb Q\}$ remains the identity.

Now what?

• Find a number that is a square in one field, but not in the other. – Daniel Fischer Apr 18 '14 at 12:32
• $\sqrt{2}$ must map to an element whose square is $2$. (Can you see that?) Now prove such an element does not exist in $Q(\sqrt{3})$. – Magdiragdag Apr 18 '14 at 12:32
• Let $\phi:\sqrt 2\mapsto p.$ Then $2\mapsto p^2$ whence $p^2=2.$ But $x^2-2=0$ has no solution in $\mathbb Q(\sqrt 3)$ since any element of $\mathbb Q(\sqrt 3)$ is of the form $ax+b+\langle x^2-3\rangle$ and $(ax+b+\langle x^2-3\rangle)^2-(2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies(a^2x^2+2abx+b^2-2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies b=0.$ So $p$ is of the form $ax+\langle x^2-3\rangle$ and $a^2x^2+\langle x^2-3\rangle=2+\langle x^2-3\rangle\implies\dfrac{2}{a^2}=3$ a contradiction to $a\in\mathbb Q.$ CORRECT? – user143484 Apr 18 '14 at 12:59
• @user143484 Not sure. I think it's better to think about these things concretely as subfields of $\mathbb{C}$ rather than as quotient rings of polynomial rings, just to avoid mistakes slipping in calculations, e.g., you can't conclude that $b=0$ in your proof. But it can still be salvaged: if $(a+b\sqrt{3})^2=2$, then $(a^2+3b^2)+(2ab)\sqrt{3}=2$, i.e., $a^2+3b^2=2$ and $ab=0$. If $b=0$, then we arrive at a contradiction since $2$ is not the square of a rational number (why?), and if $a=0$, then we also arrive at a contradiction since $\frac{2}{3}$ is not the square of a rational number. – Amitesh Datta Apr 18 '14 at 13:45
• @user143484 As an exercise, I would suggest trying to generalise the above argument to show that: $\mathbb{Q}(\sqrt{p})$ is not isomorphic to $\mathbb{Q}(\sqrt{q})$ if $p$ and $q$ are distinct primes. – Amitesh Datta Apr 18 '14 at 13:50

Consider the polynomial $$p(x)=x^{2}-2$$

Since any isomorphism $$\varphi:\,\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{3})$$ is the identity when restricted to $\mathbb{Q}$ we get that if $q\in\mathbb{Q}(\sqrt{2})$ then $$\varphi(p(q))=\varphi(q^{2})-\varphi(2)=\varphi^{2}(q)-2$$

Since $p$ have a root $r\in\mathbb{Q}(\sqrt{2})$ we get that $$0=\varphi(0)=\varphi(p(r))=\varphi^{2}(r)-2$$

but $\varphi(r)\in\mathbb{Q}(\sqrt{3})$ and there is no element $\alpha$ in $\mathbb{Q}(\sqrt{3})$ that satisfies the equation $\alpha^{2}-2=0$ since this would imply $\alpha=\sqrt{2}\in\mathbb{Q}(\sqrt{3})$.

Note: It is true in general that if $\mathbb{F}_{1}\cong\mathbb{F}_{2}$ and $\sum_{i=0}^{n}a_{n}x^{n}=p(x)\in\mathbb{F}_{1}[x]$ have a root $r$ then the polynomial $\sum_{i=0}^{n}\varphi(a_{n})x^{n}=q\in\mathbb{F}_{2}[x]$ have a root $\varphi(r)\in\mathbb{F}_{2}$.

Isomorphic quadratic number fields have the same discriminant. But $\mathbb{Q}(\sqrt{2})$ has discriminant $8$, and $\mathbb{Q}(\sqrt{3})$ has discriminant $12$, see here.

Let $\phi:\sqrt 2\mapsto p.$ Then $2\mapsto p^2$ whence $p^2=2.$ But $x^2-2=0$ has no solution in $\mathbb Q(\sqrt 3)$ since any element of $\mathbb Q(\sqrt 3)$ is of the form $ax+b+\langle x^2-3\rangle$ and $(ax+b+\langle x^2-3\rangle)^2-(2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies(a^2x^2+2abx+b^2-2+\langle x^2-3\rangle)=\langle x^2-3\rangle$$\implies b=0.$ So $p$ is of the form $ax+\langle x^2-3\rangle$ and $a^2x^2+\langle x^2-3\rangle=2+\langle x^2-3\rangle\implies\dfrac{2}{a^2}=3$ a contradiction to $a\in\mathbb Q$

Well, we use the following facts:

• $\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} : a,b \in \mathbb{Q}\}$
• $\mathbb{Q}(\sqrt{3}) = \{a + b\sqrt{3} : a,b \in \mathbb{Q}\}$
• $\sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$

Suppose they are isomorphic, and let $\phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ be an isomorphism. Then $\phi(a + b\sqrt{2}) = a + b \phi(\sqrt{2})$ (by the fact that $\phi$ restricted to $\mathbb{Q}$ is the identity. But notice $\phi(\sqrt{2})^2 = \phi(2) = 2$, so $\phi(\sqrt{2}) = \pm \sqrt{2} \not\in \mathbb{Q}(\sqrt{3})$.

• Your third fact is almost what the OP is asking... – DonAntonio Apr 18 '14 at 14:18