# Basic definition of Quadratic Fields

1. Jan 20, 2012

### Math Amateur

I am reading Dummit and Foote Chapter 7.

D&F use a quadratic field as an example of a ring. I am trying to get a good understanding of this ring.

D&F define a quadratic field as follows:

Let D be a rational number that is not a perfect square in and define

$$\mathbb{Q} ( \sqrt D ) = \{ \ a + b \sqrt D \ | \ a,b \in \mathbb{Q} \ \}$$

as a subset of $\mathbb{C}$

In this example D&F write ... "... It is easy to show that the assumption that D is not a square implies that every element of $\mathbb{Q} ( \sqrt D )$ may be written uniquely in the form $a + b \sqrt D$."

How do you show this? Further, I am not sure why this assumption is needed?

Is it because we have both positive and negative roots of a square number like 4, but then only consider the principal root $+ \sqrt 3$ of 3? This seems slightly inconsistent!

Also how does the above fit with the idea that D must be not only not a perfect square but squarefree? Is the squarefree condition on D necessary? If so why?

Can someone please clarify this situation for me?

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2. Jan 20, 2012

### Deveno

it's really pretty simple. if D is a square of a rational number, then Q(√D) = Q.

if D isn't square-free, say D = p2m, for some prime p, and integer m, then we get the same extension of Q with Q(√m) as we do for Q(√D) (in other words, we want "the totally irrational part" of the square root, with any rational factors (hence squares) already factored out).

for example, 12 isn't square free, so adjoining √12 = 2√3, gives us the same field as adjoining √3:

a+b√12 <=> a + 2b√3
c+d√3 <=> c + (d/2)√12

on the other hand, if D IS square-free, we get a UNIQUE quadratic extension (different D's, different extensions).

you can visualize such a ring (which is actually a field, since 1/√D = √D/D, and one can use "the conjugate trick" to find an inverse for a + b√D, as long as not both a and b are 0) as a lattice of rational points in the plane, that is, as a vector space it's isomorphic to Q x Q.

3. Jan 20, 2012

### Math Amateur

Thanks!

Thinking through this now!

4. Jan 20, 2012

### Math Amateur

Thanks again

I can see that if D is not a squarefree number then the field specified by $\mathbb{Q} ( \sqrt D )$ is not unique in the sense that $\mathbb{Q} ( \sqrt 12 )$ is the same field as$\mathbb{Q} ( \sqrt 3 )$.

But D&F's statement seemingly refers to consequences within the field if D is not squarefree.

They write:

"It is easy to show that the assumption that D is not a square implies that every element of $\mathbb{Q} ( \sqrt D )$ may be written uniquely in the form $a + b \sqrt D$."

So they are saying that there is a lack of uniqueness within the field.

Do you agree? Can you clarify?

Another thing that bothers me in D&F's statement is that they do not use the term "squarefree" but go for a lesser condition that D is not a square. Can you clarify this also?

5. Jan 20, 2012

### Deveno

if D is not a square, then Q(√D) is bigger than D. argue by contradiction:

suppose Q(√D) = Q. then √D is in Q, hence we have some rational number m (= √D) with m2 = D, so D is a square.

the uniqueness of the extension is what D being square-free entails. even if D is NOT square-free, there is still only one way to write an element of Q(√D):

suppose a+b√D = c+d√D, where √D is not in Q.

then a-c = (d-b)√D.

case 1) d≠ b:

then (a-c)/(d-b) = √D, and thus √D is in Q, contradiction.

case 2) d = b:

then a-c = 0, so that a = c, which shows uniqueness.

6. Jan 20, 2012

### Math Amateur

Thanks.

OK so the sqarefree part is only to get a unique description of the field

Yes, follow you arguments regarding $Q \sqrt D$ being bigger that Q.

I now regard D&F's statement - that got me going on this - as rather misleading!

7. Jan 20, 2012

### morphism

I fail to see what is misleading.

If D is not a square then $\sqrt D$ is irrational. So if $a+b\sqrt D = a'+b' \sqrt D$ with $a,b,a',b' \in \mathbb Q$, then from $a-a' = (b'-b) \sqrt D$ one easily sees that b'=b and hence a'=a (since otherwise $\sqrt D = (a-a')/(b'-b)$ would be rational). It follows that an element x of $\mathbb Q(\sqrt D)$ has a unique expression of the form $x=a+b\sqrt D$ where $a,b \in \mathbb Q$, in the sense that a and b are uniquely determined by x.

If D is a square, say D=E^2 with E rational, then it's easy to see that this fails. For example, we have $a+b\sqrt D = (a-|E|)+(b+1)\sqrt D$ for any $a,b \in \mathbb Q$.

8. Jan 21, 2012

### Math Amateur

Thanks for that help and guidance

Your last two sentences clarified the situation for me!