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

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April 6th 2010, 09:34 PM
pila0688

Prove 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
April 6th 2010, 09:50 PM
hamidr

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)
April 6th 2010, 10:02 PM
Drexel28

Quote:

Originally Posted by pila0688

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

Help

Quote:

Originally Posted by hamidr

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)

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 $\mathbb{Q}[\sqrt{d}]$ has a multiplicative inverse but that the multiplication is commutative (albeit this is trivial)
April 6th 2010, 10:13 PM
hamidr

Quote:

Originally Posted by Drexel28

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 $\mathbb{Q}[\sqrt{d}]$ has a multiplicative inverse but that the multiplication is commutative (albeit this is trivial)

yes, However since Q(root(d)) is isomorphic to complex numbers I thought we could assume that it is an integral domain.
April 6th 2010, 10:18 PM
Drexel28

Quote:

Originally Posted by hamidr

yes, However since Q(root(d)) is isomorphic to complex numbers I thought we could assume that it is an integral domain.

What makes you think that $\mathbb{Q}[\sqrt{2}]\cong\mathbb{C}$ ? Define $\eta:\mathbb{Q}^2\to\mathbb{Q}[\sqrt{2}]$ by $(p,q)\mapsto p+\sqrt{d}q$ . Clearly this is a surjection and so since $\mathbb{Q}^2$ is countable it follows from a basic exercise that so is $\mathbb{Q}[\sqrt{d}]$ . So, since $\mathbb{C}$ which is equipotent to $\mathbb{R}$ is uncountable they can't be isomorphic.
April 6th 2010, 10:28 PM
Bruno J.

Quote:

Originally Posted by hamidr

yes, However since Q(root(d)) is isomorphic to complex numbers I thought we could assume that it is an integral domain.

Not only is this false, as Drexel points out, but as false as could be. Even if you meant $\mathbb{Q}[i]$ (which I presume you did), no two quadratic fields $\mathbb{Q}[\sqrt{d}]$ are isomorphic.
April 6th 2010, 10:33 PM
Drexel28

Quote:

Originally Posted by hamidr

yes, However since Q(root(d)) is isomorphic to complex numbers I thought we could assume that it is an integral domain.

Quote:

Originally Posted by Bruno J.

Not only is this false, as Drexel points out, but as false as could be. Even if you meant $\mathbb{Q}[i]$ (which I presume you did), no two quadratic fields $\mathbb{Q}[\sqrt{d}]$ are isomorphic.

Regardless. If you only want it to be an integral domain it shouldn't (and I'm not going to say why!) be hard to show that it is one. In fact, defining a Euclidean valuation on it to make it into a Euclidean domain.
April 6th 2010, 10:39 PM
hamidr

Quote:

Originally Posted by Drexel28

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

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.
April 6th 2010, 10:42 PM
Drexel28

Quote:

Originally Posted by hamidr

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.

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 $\mathbb{Q}[\sqrt{2}],\mathbb{Q}[\sqrt{1}]\cong\mathbb{C}$ then $\mathbb{Q}[\sqrt{1}]\cong\mathbb{Q}[\sqrt{2}]$ . Stand back and take a look that statement.
April 6th 2010, 11:12 PM
hamidr

Quote:

Originally Posted by Drexel28

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 $\mathbb{Q}[\sqrt{2}],\mathbb{Q}[\sqrt{1}]\cong\mathbb{C}$ then $\mathbb{Q}[\sqrt{1}]\cong\mathbb{Q}[\sqrt{2}]$ . Stand back and take a look that statement.

lools now I understand my stupidity lolz, " $\mathbb{Q}[\sqrt{1}]\cong\mathbb{Q}[\sqrt{2}]$ " that was funny. I cannot think how stupid my claim was lol
thanks(Rofl)