1. ## congruence and quadratic integers.

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.

2. 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 $\displaystyle \alpha$ that means,
$\displaystyle x+y\sqrt{d}\equiv z+w\sqrt{d}$ whenever (actually if and only if, since all definitions are biconditional statements)
$\displaystyle x\equiv z\mbox{ and }y\equiv w$ modulo $\displaystyle \alpha$.

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

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

The relation $\displaystyle \to$ 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 $\displaystyle [a,b]$ as the set of all quadratic integers such that a congruent to $\displaystyle (a,b)$.

We define addition as $\displaystyle [a,b]+[c,d]=[a+c,b+d]$ 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?

3. 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,
$\displaystyle 6=(2)(3)$
or,
$\displaystyle 6=(1+\sqrt{-5})(1-\sqrt{-5})$

4. Hi Perfecthacker,

I got it. Thank you very much.