Is $\phi: a + b \sqrt{2} \rightarrow a + b\sqrt{3}$ a $\mathbb Q(\sqrt{2}) \rightarrow \mathbb Q(\sqrt{3})$ field isomorphism?

Question

Some book I'm reading says so.

But I'm a bit lost: for field isomorphism, it's required that $\phi(uv) = \phi(u) \phi(v)$.

Choosing $u = v = a + b \sqrt{2}$, we have

$$LHS = \phi((a + b \sqrt{2})(a+b\sqrt{2})) = \phi((a^2 + 2 b^2) + 2 ab \sqrt{2}) = (a^2 + 2b^2) + 2ab \sqrt{3}$$

$$RHS = \phi(a + b\sqrt{2}) \phi(a + b\sqrt{2}) = (a + b\sqrt{3})(a + b\sqrt{3}) = (a^2 + 3b^2) + 2ab \sqrt{3} $$

Seems they don't tally?

The question is : are you sure that they are isomorphic? because they are not isomorphic maybe this is what you need to prove. — Elaqqad, Mar 08 '15 at 16:36
What is the exact wording in that book. (Because what you said, as we have seen in the comments, is false.) — GEdgar, Mar 08 '15 at 16:48

Elaqqad · Accepted Answer · 2015-03-08T18:57:37.300

4

The two fields are not isomorphic: assume the existence of an isomorphisme $\varphi$ and let $\varphi(\sqrt{2})=a+\sqrt3b$, because $\sqrt{2}×\sqrt{2}=1+1$. Hence $\varphi(\sqrt2)^2=\varphi(1)+\varphi(1)=2$. Hence $(a+\sqrt{3}b)^2=2$, so : $$a^2+3b^2+2ab\sqrt{3}=2+0\sqrt3 $$

and using the uniqueness of the writing $a+b\sqrt{3}$ we conclude that $ab=0$ and $a^2+3b^2=2$ hence either $2$ or $\frac{2}{3}$ is a square in $\mathbb{Q}$ which is a contradiction.

edited Mar 08 '15 at 18:57

answered Mar 08 '15 at 16:42

Elaqqad

13,725

1

Is it ok now or I need to prove that $2$ is not a square in $\mathbb{Q}$ – Elaqqad Mar 08 '15 at 17:37
@Elaqqad Maybe you should prove the associativity of addition in $\mathbf{Q}(\sqrt{3})$ as you use it. ;-) ;-) ;-) – Olórin Mar 08 '15 at 18:13
@Steven This Lemma implies $,\sqrt{2}\not\in\Bbb Q(\sqrt{3}),$ by reducing it to the equivalent statement that $,\sqrt2,\sqrt3,\sqrt6\not\in\Bbb Q,,$ i.e. all are all irrational. – Bill Dubuque Mar 08 '15 at 18:19
@Elaqqad This version looks really solid to me. – Steven Stadnicki Mar 08 '15 at 18:56

score 3 · Answer 2 · answered Mar 08 '15 at 19:12

3

You need no tricky computation. If $\phi$ is an ring homomorphism, then $\phi(2)=2$, but $$ \phi(0+1\sqrt{2})=\sqrt{3} $$ and $$ 2=\phi(2)=\phi(\sqrt{2})^2=3 $$

answered Mar 08 '15 at 19:12

egreg

238,574

how can you prove that $\phi(\sqrt 2)=\sqrt 3$ because i don't see any justification for your second equality – Elaqqad Mar 23 '15 at 20:35
@Elaqqad By definition of $\varphi$: $\varphi(a+b\sqrt{2})=a+b\sqrt{3}$, take $a=0$, $b=1$. – egreg Mar 23 '15 at 21:05
@edred The definition of a field isoùorphism is the following: $$\phi(uv)=\phi(u)\phi(v)\ \phi(u+v)=\phi(u)+\phi(v)\ \phi(0)=0;\phi(1)=1 $$ but you can justify your equality because the fields are of the form $\mathbb{Q}(\alpha)$ still the question is that how this justification works? – Elaqqad Mar 23 '15 at 21:09
@Elaqqad I took the definition from your question. Now, $\varphi(2)=2$ by definition and $\varphi(\sqrt{2})=\sqrt{2}$ by definition. However, $2=(\sqrt{2})^2$, but $2=\varphi(2)=\varphi((\sqrt(2))^2)$, but $\varphi(\sqrt{2})^2=(\sqrt{3})^2=3$. So this $\varphi$ is surely not a ring homomorphism. – egreg Mar 23 '15 at 21:14
ok you're right I did not read the question very well (it's not mine) because in my answer I proved that there is no isomorphism from the given fields, but the question asks if the given application is an isomorphism – Elaqqad Mar 23 '15 at 21:25
@Elaqqad Sorry for having thought you were the OP; yes, your answer is more general. – egreg Mar 23 '15 at 21:28

Is $\phi: a + b \sqrt{2} \rightarrow a + b\sqrt{3}$ a $\mathbb Q(\sqrt{2}) \rightarrow \mathbb Q(\sqrt{3})$ field isomorphism?

2 Answers2