Quadratic Fields and Norms

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July 14th 2010, 07:02 AM
Samson

Quadratic Fields and Norms

Hello All,

I had a few more questions about Quadratic Fields and Norms from the example section of my reading. (By the way, I am very appreciative of all of the help I've received so far!) Here are my questions:
Note: Q[x] denotes the Quadratic Field of x, N(y) denotes the Norm of y.

1. There exists integers 'd' for which the field Q[Sqrt(d)] has elements 'a' with negative norm N(a), assuming that d is not a perfect square. Does anyone know what these 'd' is and how to prove this? This seems really general...

2. Examining Q[Sqrt(-1)], an equation can be written relating N(a) to |a| where | | means the natural absolute value defined for complex numbers. What is this equation and for which Q[Sqrt(d)] would this formula be correct? (Undefined previously posted a relation in another problem, please see
http://www.mathhelpforum.com/math-he...-q-fields.html )

Thank you everyone! I really appreciate the help!
July 18th 2010, 06:54 PM
Samson

Useful update!!!

Hello All,

I found the following link and I think it may be helpful. I've been trying to apply it the best way I can to this, but maybe somebody else might be able to make sense of it in relation to my questions!

Link:
PlanetMath: quadratic imaginary norm-Euclidean number fields

I really appreciate the help!
July 21st 2010, 02:48 AM
undefined

Quote:

Originally Posted by Samson

Hello All,

I had a few more questions about Quadratic Fields and Norms from the example section of my reading. (By the way, I am very appreciative of all of the help I've received so far!) Here are my questions:
Note: Q[x] denotes the Quadratic Field of x, N(y) denotes the Norm of y.

1. There exists integers 'd' for which the field Q[Sqrt(d)] has elements 'a' with negative norm N(a), assuming that d is not a perfect square. Does anyone know what these 'd' is and how to prove this? This seems really general...

2. Examining Q[Sqrt(-1)], an equation can be written relating N(a) to |a| where | | means the natural absolute value defined for complex numbers. What is this equation and for which Q[Sqrt(d)] would this formula be correct? (Undefined previously posted a relation in another problem, please see
http://www.mathhelpforum.com/math-he...-q-fields.html )

Thank you everyone! I really appreciate the help!

1. Using alpha and D in place of 'a' and 'd' (which notation I've seen is pretty common), it seems from grade school algebraic manipulation that we have this as long as D > 0.

So in general, $N(\alpha)=\alpha\overline{\alpha} = (a+b\sqrt{D})(a-b\sqrt{D}) = a^2 - b^2D$ .

Suppose D > 0. Set $a^2 - b^2D < 0 \implies a^2 < b^2D$ then we can choose for example b = a, and we are done.

Suppose D < 0. Let's just make sure the above result holds with complex numbers. (I'm not sure this is necessary, but it doesn't hurt to check.) $N(\alpha)=\alpha\overline{\alpha} = (a+b\sqrt{D})(a-b\sqrt{D}) =$ $(a+bi\sqrt{|D|})(a-bi\sqrt{|D|}) = a^2 - b^2i^2\sqrt{|D|}^2 = a^2 + b^2|D| = a^2 - b^2D$ . So we see there is no way to make this negative when D < 0.

2. Again write out from definitions and see if there is an answer just using algebra taught in grade school. $N(\alpha)=a^2 - b^2D=a^2-(b^2)(-1) = a^2+b^2$ . Recall the norm for complex numbers, $|\alpha| = \sqrt{a^2+b^2}$ which is just the distance from the origin to that point in the complex plane. So we see that $N(\alpha) = |\alpha|^2$ and $\sqrt{N(\alpha)} = |\alpha|$ .

So when in doubt, go back to definitions and see if something relatively simple doesn't pop out.
July 21st 2010, 02:56 AM
Samson

Quote:

Originally Posted by undefined

1. Using alpha and D in place of 'a' and 'd' (which notation I've seen is pretty common), it seems from grade school algebraic manipulation that we have this as long as D > 0.

So in general, $N(\alpha)=\alpha\overline{\alpha} = (a+b\sqrt{D})(a-b\sqrt{D}) = a^2 - b^2D$ .

Suppose D > 0. Set $a^2 - b^2D < 0 \implies a^2 < b^2D$ then we can choose for example b = a, and we are done.

Suppose D < 0. Let's just make sure the above result holds with complex numbers. (I'm not sure this is necessary, but it doesn't hurt to check.) $N(\alpha)=\alpha\overline{\alpha} = (a+b\sqrt{D})(a-b\sqrt{D}) =$ $(a+bi\sqrt{|D|})(a-bi\sqrt{|D|}) = a^2 - b^2i^2\sqrt{|D|}^2 = a^2 + b^2|D| = a^2 - b^2D$ . So we see there is no way to make this negative when D < 0.

2. Again write out from definitions and see if there is an answer just using algebra taught in grade school. $N(\alpha)=a^2 - b^2D=a^2-(b^2)(-1) = a^2+b^2$ . Recall the norm for complex numbers, $|\alpha| = \sqrt{a^2+b^2}$ which is just the distance from the origin to that point in the complex plane. So we see that $N(\alpha) = |\alpha|^2$ and $\sqrt{N(\alpha)} = |\alpha|$ .

So when in doubt, go back to definitions and see if something relatively simple doesn't pop out.

Wow, I totally had been over thinking this! But thank you, I really appreciate the help!