I am 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?

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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, .