# Qudratic field

• Oct 18th 2006, 10:08 AM
beta12
Qudratic field
Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.
• Oct 18th 2006, 04:48 PM
ThePerfectHacker
Quote:

Originally Posted by beta12
Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.

It is false.
---
Here is a theorem from field theory.
Let E be an entension field of field F.
Let 'a' in E be algebraic over F.
Then consider the simple extension F(a) over F.
If 'b' is in F(a) then deg(b,F) divides deg(a,F).

You are working with simple extension field Q(sqrt(2)) over Q.
Trivially, sqrt(2) is in Q(sqrt(2))
We assume it square root is in Q(sqrt(2)).
That is, sqrt(sqrt(2))=4throot(2).
Now, irr(4throot(2),Q)=x^4-2 (use Eisenstein Criterion ).
Thus, deg(4throot(2),Q)=4
But, deg(sqrt(2),Q)=2
And 4 does not divide 2. :eek:
Thus, Q(sqrt(2)) cannot contain its square roots.
• Oct 18th 2006, 05:58 PM
beta12
Hi perfecthacker,

Thank you very much!
=============================================
Consider Q[-1]. Write an equation relating N(alpha) to l alpha l ( the natural absolute value defined for complex numbers). For which Q[sqrt(d)] is this formula correct?
================================================== ========

I am reading the topic from my book is " Quadratic fields" . I found that it is difficult to understand the content of Quadratic field from this book.
Do you know any good web site for " Quadratic fields" ?
• Oct 19th 2006, 06:49 AM
ThePerfectHacker
Quote:

Originally Posted by beta12
For which Q[sqrt(d)] is this formula correct?

I am not sure how the field is defined when d<0, but when d>0 there is no such field Q[sqrt(d)] that contains its square roots. Use the same argument I used before.

If d is a perfect square then Q[sqrt(d)]=Q.
Thus, 2 is in Q[sqrt(d)], yet sqrt(2) is not (irrational).

If d is not a perfect square then Q[sqrt(d)] not = Q.
Also, deg(d,Q)=2 because
f(x) = x^2-d
is the irreducible polynomial over Q.

(You cannot use Eisenstein Criterion here but it happens to be true).

Trivially sqrt(d) is in Q(sqrt(d)).
Its square root is sqrt(sqrt(d))=4throot(d)
And, deg(4throot(d),Q)=Q because
f(x) = x^4-d
is the irreducible polynomial over Q.
(You cannot use Eisenstein Criteroin here either, again it is true.)

But then 4 divides 2 (like in other post) which is impossible.
• Oct 19th 2006, 07:04 AM
beta12
Hi perfecthacker,

Thank you very much.
=================================
Referring to my previous post:

Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.

Can you figure out a counterexample for this question?
• Oct 19th 2006, 07:39 AM
ThePerfectHacker
Quote:

Originally Posted by beta12
Referring to my previous post:

Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.

Can you figure out a counterexample for this question?

I answred it already, it is false.
Unless, you are repeating the same question because I got it wrong did I?
• Oct 19th 2006, 08:06 AM
topsquark
Quote:

Originally Posted by beta12
Hi perfecthacker,

Thank you very much.
=================================
Referring to my previous post:

Consider Q[sqrt(2)].
Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)] ?
Prove if true, and give a counterexample if false.

Can you figure out a counterexample for this question?

TPH was saying that sqrt(sqrt(2)) does not belong to Q[sqrt(2)].

-Dan
• Oct 19th 2006, 07:01 PM
ThePerfectHacker
There is another way to demonstrate that,
$\displaystyle \sqrt[4]{2}\not \in \mathbb{Q}(\sqrt{2})$
Note that simple extensions can be viewed as finite dimensional vector spaces over there subfields. The basis for $\displaystyle \mathbb{Q}(\sqrt{2})$ is,
$\displaystyle B=\{1,\sqrt{2}\}$
That means, any and all linear combinations,
$\displaystyle a+b\sqrt{2}\, a,b\in \mathbb{Q}$ is this field.

Assume that we can express $\displaystyle \sqrt[4]{2}$ as this linear combination that is,
$\displaystyle \sqrt[4]{2}=a+b\sqrt{2}$
Then squaring both sides,
$\displaystyle \sqrt{2}=a^2+2b^2+2ab\sqrt{2}$
Thus,
$\displaystyle 2ab\sqrt{2}-\sqrt{2}=-(a^2+2b^2)$
Thus,
$\displaystyle \sqrt{2}{(2ab-1)}=-(a^2+2b^2)$
Showing that $\displaystyle 2ab-1\not = 0$ is trivial thus,
$\displaystyle \sqrt{2}=\frac{a^2+2b^2}{1-2ab}$
But that is not possible because the number on the right is rational and the one of the left is irrational, contradiction.
• Oct 19th 2006, 11:56 PM
beta12
Quote:

Originally Posted by ThePerfectHacker
There is another way to demonstrate that,
$\displaystyle \sqrt[4]{2}\not \in \mathbb{Q}(\sqrt{2})$
Note that simple extensions can be viewed as finite dimensional vector spaces over there subfields. The basis for $\displaystyle \mathbb{Q}(\sqrt{2})$ is,
$\displaystyle B=\{1,\sqrt{2}\}$
That means, any and all linear combinations,
$\displaystyle a+b\sqrt{2}\, a,b\in \mathbb{Q}$ is this field.

Assume that we can express $\displaystyle \sqrt[4]{2}$ as this linear combination that is,
$\displaystyle \sqrt[4]{2}=a+b\sqrt{2}$
Then squaring both sides,
$\displaystyle \sqrt{2}=a^2+2b^2+2ab\sqrt{2}$
Thus,
$\displaystyle 2ab\sqrt{2}-\sqrt{2}=-(a^2+2b^2)$
Thus,
$\displaystyle \sqrt{2}{(2ab-1)}=-(a^2+2b^2)$
Showing that $\displaystyle 2ab-1\not = 0$ is trivial thus,
$\displaystyle \sqrt{2}=\frac{a^2+2b^2}{1-2ab}$
But that is not possible because the number on the right is rational and the one of the left is irrational, contradiction.

Hi Perfecthacker,

You are the best! This approach is very easy to understand. I got this question fully. Thank you very much.

================
For the below question, can you use a simplier approach to do it? I don't fully understand your answer posted before. I am new to quadratic field . So far the topics which I covered are simple quadratic field knwoledge.