Is this the right forum for this? I need some help getting started with a couple of proofs. Can anyone get me started in the right direction?

http://i3.photobucket.com/albums/y79/byee614/proofs.jpg

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- Oct 30th 2006, 02:49 PMsplashHelp with some proofs (Chinese remainder theorem)
Is this the right forum for this? I need some help getting started with a couple of proofs. Can anyone get me started in the right direction?

http://i3.photobucket.com/albums/y79/byee614/proofs.jpg - Oct 31st 2006, 06:51 AMThePerfectHacker
I am :confused: confused :confused: by the map,

Because the sets,

are all finite and cannot be expressed as the ideals of .

However,

What you want to show is the isomorphism,

Using the map defined above. Then it would make sense.

Do you understand me?

---

Okay here is the proof on what I said.

First we are given a function,

As,

We need to show that this map is well-defined.

Assume there are two ways to express the same element of

As,

and

Since is in the same coset we can express it as,

for some integer .

Now we show that they map this element into the same ordered coset pair.

Because is an element of and .

Thus the map is well-defined.

We now show that this map is a homomorphism between these two groups.

First if and

Then,

Thus, we need to see whether or not,

That is,

---> True for all.

Thus it is indeed a homomorphism.

Now, we see if this function is injective.

Thus,

So, and for some integers then, for some integer ; note the important fact that otherwise we cannot conclude that has this form. That means that is in the same coset as . Thus, this show the function is indeed injective.

The last thing is whether any element of can be obtained from a map.

Does there exists a such that can be mapped into,

For any ?

Meaning, a such that and . Note the last two equations are saying,

And we can find such an !.

That is the Chinese Remainder Theorem.

If we chose,

Where,

Then, is a simultaneous solution to the congruence.

Thus, we proved isomorphism!

Note, the last two congruences has solutions to them because, . - Oct 31st 2006, 08:41 AMThePerfectHacker
There is a more elegant way to prove this, without solving that congruence via Chinese Remainder Theorem, though I did not find a proper map.

We know that,

Is a one-to-one map.

Then if we can find a map,

That is also one-to-one.

Then, you can conclude that the sets has the same cardinality.

Now there is a theorem, that if a function is injective and maps a**finitie**set into another**finite**set of equal cardinality then that function is also surjective.

So if you can find a one-to-one mapping function backswards then you automatically know that your injective is a bijection which assures us of a solution to those congruences.