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

**joestevens** · March 29th 2011, 02:01 PM

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?

**hatsoff** · March 29th 2011, 03:29 PM

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.

**joestevens** · March 29th 2011, 05:47 PM

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?

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.

**TheChaz** · March 29th 2011, 05:50 PM

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.

**Swlabr** · March 30th 2011, 01:28 AM

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...

**joestevens** · March 30th 2011, 06:49 AM

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

I agree that his was is neater....

**Swlabr** · March 30th 2011, 07:04 AM

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...

**joestevens** · March 30th 2011, 07:10 AM

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.

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

**Swlabr** · March 30th 2011, 07:28 AM

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...

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