• Oct 25th 2006, 03:10 PM
beta12
Question 1)
If alpha is a quadratic integers in Q[sqrt(-d)] , then define a notion of congruence (mod alpha).
Furthermore, define +, -, and X for congruence classes , and show that this notion is well-defined.

Question 2)
Prove that ( the integers of ) Q[sqrt(-5)] do not form a UFD.
Hint: Consider the quadratic integers that divide 6.

Please teach me how to solve these two questions. Thank you very much.
• Oct 25th 2006, 06:29 PM
ThePerfectHacker
Quote:

Originally Posted by beta12
Question 1)
If alpha is a quadratic integers in Q[sqrt(-d)] , then define a notion of congruence (mod alpha).
Furthermore, define +, -, and X for congruence classes , and show that this notion is well-defined.

You can define two quadratic integers congruent when there corresponding parts are congruent modulo $\alpha$ that means,
$x+y\sqrt{d}\equiv z+w\sqrt{d}$ whenever (actually if and only if, since all definitions are biconditional statements)
$x\equiv z\mbox{ and }y\equiv w$ modulo $\alpha$.

Let the ordered pair $(a,b)$ represent the quadratic integer $a+b\sqrt{d}$.

We define $(a,b)\to (c,d)$ when they are congruent (just like the integers).

The relation $\to$ forms an equivalence relation, this is easy to show because it is completely derived from modulo for integers (which does in fact from an equivalence relation).

Now we define $[a,b]$ as the set of all quadratic integers such that a congruent to $(a,b)$.

We define addition as $[a,b]+[c,d]=[a+c,b+d]$ and show it is well-defined (invariant under the representatives). This is easy to show since again it is derived from the fact of well-defineness for integers modulo an integer.
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Again, I never studied quadratic fields, but this looks supprising. Does this have to represent the field of quotients of quadratic integers?
• Oct 25th 2006, 06:35 PM
ThePerfectHacker
Quote:

Originally Posted by beta12

Question 2)
Prove that ( the integers of ) Q[sqrt(-5)] do not form a UFD.
Hint: Consider the quadratic integers that divide 6.
.

You can factor 6 proper and non-trivially as,
$6=(2)(3)$
or,
$6=(1+\sqrt{-5})(1-\sqrt{-5})$
• Oct 26th 2006, 06:19 AM
beta12
Hi Perfecthacker,

I got it. Thank you very much.