Prove that both rings are fields but that they are not isomorphic.

• Mar 29th 2011, 03:01 PM
joestevens
Prove that both rings are fields but that they are not isomorphic.
Let $\mathbb{Q}[\sqrt{n}] = \{a + b\sqrt{n}\}$
Prove that both $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{5}]$ are fields but that they are not isomorphic.

Note: rings are commutative and associative with unity.

To prove they are fields, just find the general multiplicative inverse.
How do I prove they are not isomorphic?
• Mar 29th 2011, 04:29 PM
hatsoff
You should have a theorem available to show that $\mathbb{Q}[\sqrt{2}]\cong\mathbb{Q}/(x^2-2)$ and $\mathbb{Q}[\sqrt{5}]\cong\mathbb{Q}/(x^2-5)$. Another theorem says that $\mathbb{Q}/(p(x))$ is a field if $p(x)$ is irreducible over $\mathbb{Q}$. So just show that $x^2-2,x^2-5$ are irreducible (using Eisenstein). It follows that both are fields.

Next show that any nonzero homomorphism $\varphi:\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{5}]$ has $\varphi(1)=1$. It follows that $\varphi(\sqrt{2})^2=\varphi(2)=\varphi(1)+\varphi( 1)=2$, which means $\varphi(\sqrt{2})=\sqrt{2}$. But if $\sqrt{2}\in\mathbb{Q}[\sqrt{5}]$, then $\sqrt{2}=a+b\sqrt{5}$. Use this to show that $\sqrt{5}$ is rational, a contradiction. So there are no nonzero homomorphism, i.e. no isomorphisms.
• Mar 29th 2011, 06:47 PM
joestevens
Quote:

Originally Posted by hatsoff
You should have a theorem available to show that $\mathbb{Q}[\sqrt{2}]\cong\mathbb{Q}/(x^2-2)$ and $\mathbb{Q}[\sqrt{5}]\cong\mathbb{Q}/(x^2-5)$. Another theorem says that $\mathbb{Q}/(p(x))$ is a field if $p(x)$ is irreducible over $\mathbb{Q}$. So just show that $x^2-2,x^2-5$ are irreducible (using Eisenstein). It follows that both are fields.

I don't think I have not covered the theorem. I have seen a few examples for which my professor said this is true.
Is it sufficient to just find the general multiplicative inverse?

Quote:

Originally Posted by hatsoff
Next show that any nonzero homomorphism $\varphi:\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{5}]$ has $\varphi(1)=1$. It follows that $\varphi(\sqrt{2})^2=\varphi(2)=\varphi(1)+\varphi( 1)=2$, which means $\varphi(\sqrt{2})=\sqrt{2}$. But if $\sqrt{2}\in\mathbb{Q}[\sqrt{5}]$, then $\sqrt{2}=a+b\sqrt{5}$. Use this to show that $\sqrt{5}$ is rational, a contradiction. So there are no nonzero homomorphism, i.e. no isomorphisms.

Thanks.
• Mar 29th 2011, 06:50 PM
TheChaz
Quote:

Originally Posted by hatsoff
...But if $\sqrt{2}\in\mathbb{Q}[\sqrt{5}]$, then $\sqrt{2}=a+b\sqrt{5}$. Use this to show that $\sqrt{5}$ is rational, a contradiction. So there are no nonzero homomorphism, i.e. no isomorphisms.

Joe, I would just use this part.
• Mar 30th 2011, 02:28 AM
Swlabr
Quote:

Originally Posted by joestevens
I don't think I have not covered the theorem. I have seen a few examples for which my professor said this is true.
Is it sufficient to just find the general multiplicative inverse?

Yes, this is sufficient (as long as you point out that multiplication is commutative). To do this, you solve $(a+b\sqrt{2})(c+d\sqrt{2})=1$ for c and d, and similarly for $\sqrt{5}$.

Personally, I feel that this way is much easier than the way hatsoff gave, although that way is neater...
• Mar 30th 2011, 07:49 AM
joestevens
For my current problem set, all rings are considered associative and commutative with unity.

I agree that his was is neater....
• Mar 30th 2011, 08:04 AM
Swlabr
Quote:

Originally Posted by joestevens
For my current problem set, all rings are considered associative and commutative with unity.

I agree that his was is neater....

The fact that they are assumed it is beside the point. I mean, you should be able to spot right off that $\mathbb{Q}[\sqrt{n}]$ is commutative with unity! To attack the problem properly, you should say that this is so. It doesn't require proof, just a line...

Also, all rings are associative...
• Mar 30th 2011, 08:10 AM
joestevens
Quote:

Originally Posted by Swlabr
The fact that they are assumed it is beside the point. I mean, you should be able to spot right off that $\mathbb{Q}[\sqrt{n}]$ is commutative with unity! To attack the problem properly, you should say that this is so. It doesn't require proof, just a line...

Right. Got it.

Quote:

Originally Posted by Swlabr
Also, all rings are associative...

The definition I was given does not require associativity to be a ring.
See: Wikipedia: Nonassociative ring
• Mar 30th 2011, 08:28 AM
Swlabr
Quote:

Originally Posted by joestevens
The definition I was given does not require associativity to be a ring.
See: Wikipedia: Nonassociative ring

That is...odd. The only `real' benefit of this for an undergrad which I can see is that Lie algebras are nonassociative rings, so you can slip in a section about this. However, both Lie algebras and (associative) rings are rich enough areas that rolling them into one is a bit over the top...