# Thread: ring isomorphism proof,need help for my homework

1. ## ring isomorphism proof,need help for my homework

How can i prove that $\displaystyle Q[_{\sqrt{2}}]$ is a ring isomorphism to $\displaystyle Q[_{\sqrt{3}}]$

2. Originally Posted by canuhelp
How can i prove that $\displaystyle Q[_{\sqrt{2}}]$ is a ring isomorphism to $\displaystyle Q[_{\sqrt{3}}]$
Hi canuhelp.

Try the mapping $\displaystyle f\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3),$ $\displaystyle f\left(a+b\sqrt2\right)\ \to\ a+b\sqrt3.$

3. Originally Posted by canuhelp
How can i prove that $\displaystyle Q[_{\sqrt{2}}]$ is a ring isomorphism to $\displaystyle Q[_{\sqrt{3}}]$
Originally Posted by TheAbstractionist
Hi canuhelp.

Try the mapping $\displaystyle f\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3),$ $\displaystyle f\left(a+b\sqrt2\right)\ \to\ a+b\sqrt3.$
I disagree with TheAbstractionist because $\displaystyle \mathbb{Q}(\sqrt{2})$ is not isomorphic to $\displaystyle \mathbb{Q}(\sqrt{3})$ as fields (so they cannot be isomorphic as fields). Now since $\displaystyle \sqrt{2},\sqrt{3}$ are algebraic integers it means $\displaystyle \mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2}),\mathbb{Q}[\sqrt{3}] = \mathbb{Q}(\sqrt{3})$. Thus, asking for a ring isomorphism between $\displaystyle \mathbb{Q}[\sqrt{2}]$ and $\displaystyle \mathbb{Q}[\sqrt{3}]$ is asking for an isomorphism between $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ which does not exist.

(They happen to be isomorphic as vector spaces, maybe that is what the poster was really asking).

4. Construct a map $\displaystyle \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}]$ such that: $\displaystyle \phi(\sqrt{2})=\sqrt{3}$ and $\displaystyle \phi(q)=q$, for all $\displaystyle q\in\mathbb{Q}$.
Then check that $\displaystyle \phi$ is bijective and is a homomorphism, to show that $\displaystyle \phi$ is an isomorphism.

5. Hi ThePerfectHacker.

Thanks for pointing this out. I wasn’t thinking clearly when I made my post.

Originally Posted by ThePerfectHacker
(They happen to be isomorphic as vector spaces, maybe that is what the poster was really asking).
That was probably what I had in mind.

6. i tried but i am not sure it is true i wish somebody will check the solution.

Let $\displaystyle \phi:Q[\sqrt{2}]->Q[\sqrt{3}]$ be a supposed ring isomorphism. Since$\displaystyle \phi(x)=\phi(1*x)=\phi(1)*\phi(x)$and $\displaystyle \phi$ is onto, we must have $\displaystyle \phi(1)=1.$ This implies that $\displaystyle \phi(2)=2.$Let $\displaystyle \phi(\sqrt{2})=a+b*\sqrt{3}$, so 2=$\displaystyle \phi(2)=\phi(\sqrt{2})^2=a^2+3*b^2+2*a*b*\sqrt{3}$. Since the $\displaystyle \sqrt{3}$ is irrational we must have either a=0 or b=0. Then we use either $\displaystyle \sqrt{2}$ or $\displaystyle \sqrt{2/3}$ irrational to conclude it is impossible to solve this equation with rational numbers.

7. Originally Posted by canuhelp
and a friend give an answer what do you think about it
http://www.mathhelpforum.com/math-he...-homework.html
Originally Posted by vemrygh
Construct a map $\displaystyle \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}]$ such that: $\displaystyle \phi(\sqrt{2})=\sqrt{3}$ and $\displaystyle \phi(q)=q$, for all $\displaystyle q\in\mathbb{Q}$.
Then check that $\displaystyle \phi$ is bijective and is a homomorphism, to show that $\displaystyle \phi$ is an isomorphism.
By the suggestion above:

$\displaystyle \phi(2) = 2$ and $\displaystyle \phi(2) = (\phi(\sqrt{2}))^2 = 3$

Thus the suggestion is not well defined!!

canuhelp, did you read post #3?

Also, I dont think its a good idea to post the same question in multiple places. It is against forum rules.

8. dear Isomorphism canyou give us a complete explanation about this question as thinking this question to find a RİNG isomorphism beetween
$\displaystyle Q[\sqrt{2}]->Q[\sqrt{3}]$

9. Originally Posted by vemrygh
Construct a map $\displaystyle \phi:\mathbb{Q}[\sqrt{2}]\rightarrow\mathbb{Q}[\sqrt{3}]$ such that: $\displaystyle \phi(\sqrt{2})=\sqrt{3}$ and $\displaystyle \phi(q)=q$, for all $\displaystyle q\in\mathbb{Q}$.
Then check that $\displaystyle \phi$ is bijective and is a homomorphism, to show that $\displaystyle \phi$ is an isomorphism.

That doesn't work as then $\displaystyle 2$ maps to $\displaystyle 3$ and so $\displaystyle 1$ maps to $\displaystyle 3/2$, which is absurd:

$\displaystyle \sqrt{3} = (\sqrt{2})\phi = (\sqrt{2}) \phi*(1) \phi = \sqrt{3}*3/2$. Thus $\displaystyle 3/2 = 1$.

10. Originally Posted by canuhelp
How can i prove that $\displaystyle Q[_{\sqrt{2}}]$ is a ring isomorphism to $\displaystyle Q[_{\sqrt{3}}]$

What does the question ask you? "Give an isomorphism..." or "Is there an isomorphism..."?

I suspect it is the latter; the solution boils down to the question of "is $\displaystyle \sqrt{3/2}$ a rational number", which i don't know off the top of my head.

Hint: Assume there exists an isomorphism. Some element maps to $\displaystyle \sqrt{3}$ - what must it look like?

EDIT: I am entirely confused - I thought I was only the second person to post in this thread, after vemrygh. Thus my repetition of other peoples posts. I'm sure I read the entire thread first though!