Prove that Q[√d] = {a + b√d | a, b ϵ Q} is a field where d is a positive integer.

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- Apr 6th 2010, 09:34 PMpila0688Prove that Q[√d] = {a + b√d | a, b ϵ Q} is a field where d is a positive integer.
Prove that Q[√d] = {a + b√d | a, b ϵ Q} is a field where d is a positive integer.

Help - Apr 6th 2010, 09:50 PMhamidr
I hope this helps: in order to prove a ring is field we have to prove that the ring has multiplicative inverse, that is it is a unit.

q(rood (d)) is the elements that contains Q and rood (d)

ie: all the elements of the form a + b*sqrt(d) .now you have to see if there exist another element in this form such that (a+ b*root(d))* (c+ d*root(d))=1

I recomend that you divide 1/(c+d*root(d)) and see if you can find an element in Q(root(d))

Hint: use the fact (a-b)*(a+b)=(a^2)-(b^2) (Clapping) - Apr 6th 2010, 10:02 PMDrexel28
Well, he first has to show it's a ring. But, a field is a commutative division ring and so he must not only show that every non-zero element of $\displaystyle \mathbb{Q}[\sqrt{d}]$ has a multiplicative inverse but that the multiplication is commutative (albeit this is trivial)

- Apr 6th 2010, 10:13 PMhamidr
- Apr 6th 2010, 10:18 PMDrexel28
What makes you think that $\displaystyle \mathbb{Q}[\sqrt{2}]\cong\mathbb{C}$? Define $\displaystyle \eta:\mathbb{Q}^2\to\mathbb{Q}[\sqrt{2}]$ by $\displaystyle (p,q)\mapsto p+\sqrt{d}q$. Clearly this is a surjection and so since $\displaystyle \mathbb{Q}^2$ is countable it follows from a basic exercise that so is $\displaystyle \mathbb{Q}[\sqrt{d}]$. So, since $\displaystyle \mathbb{C}$ which is equipotent to $\displaystyle \mathbb{R}$ is uncountable they can't be isomorphic.

- Apr 6th 2010, 10:28 PMBruno J.
- Apr 6th 2010, 10:33 PMDrexel28
- Apr 6th 2010, 10:39 PMhamidr
it seems that I am so lost in ring isomorphisms.

how is Q(p,q) isomorphic to Q[root(d)] ? (Headbang) we dont know what root(d) is, what if it could be expressed as complex? ie d= -1 then we would have a+bi

though I believe you are right since I am so lost in showing isomorphism relations. - Apr 6th 2010, 10:42 PMDrexel28
My point was more elementary than that. I didn't even show that they aren't ring isomorphic I showed that they can't be equipotent (have the same cardinal numbers). So, they can't be isomorphic (for they'd have to be equipotent).

Also, think about what you're saying. If $\displaystyle \mathbb{Q}[\sqrt{2}],\mathbb{Q}[\sqrt{1}]\cong\mathbb{C}$ then $\displaystyle \mathbb{Q}[\sqrt{1}]\cong\mathbb{Q}[\sqrt{2}]$. Stand back and take a look that statement. - Apr 6th 2010, 11:12 PMhamidr