# Homework Help: Valid mapping?

1. May 12, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Can there be a mapping that may not map any elements from one domain to another?

The reason is that the mapping has a condition. For example, it will only map elements if the one in the domain are related in some way to the element they are mapped to (i.e congruence via a certain ideal). If this specified relation dosen't hold then no mapping will occur. If the relation is specified than the mapping certainly obeys a homomorphism.

For a concrete example, say you map Z[x] to Z via the identity transformation. Then clearly polynomials of degree 1 or more in Z[x] will not be mapped to Z. Are they just left alone? Is this transformation still valid?

Last edited: May 12, 2007
2. May 13, 2007

### matt grime

You cannot map Z[x] to Z via the identity map, since Z[x] is not identical to Z. A function must be defined on every element of the domain. That is why it is called the domain.

3. May 13, 2007

### pivoxa15

Lets get the terminology correct.

Is Z[x] the domain and Z the codomain?

Do you mean a function must be defined on every element of the domain when it is clear what the function will map every element in the domain to?

So in this case it is not clear what the (identity) function will do to degree 1 polynomials and above?

I could do the identity map to Z[x] -> R[x] even though Z[x] and R[x] are not identitcal. It is an injective homomorphism.

4. May 13, 2007

### matt grime

It isn't the identity map. It is (an) inclusion map. The identity is a map from X to X, whatever X might be.

That sentence doesn't make much sense. But I think the answer is: by definition a function is defined at all points of its domain or it isn't a function. Whether or not that definition needs to be made explicit is an entirely different matter.

5. May 13, 2007

### pivoxa15

What would this map p:Z->Z[x] called such that p is a function which multiplies each element in Z by 1. Hence maps each element in Z to the corresponding integer in Z[x].

Would you call it an inclusion map? It is an injective homomorphism.

6. May 13, 2007

### matt grime

The map which 'multiplies each element of Z by 1' is a map from Z to Z. It is not a map from Z to anything else. The map sending z in Z to z as an element of Z[x] is the inclusion map but it is *not* multiplying anything by 1.