# Homomorphism on rings

1. Dec 12, 2007

### chocok

can someone please explain what these mappings really means? like what is being mapped and mapped to..?? i get confused by the direct sum & product that gets mapped..

Z $$\oplus$$ Z ->Z
Z -> ZxZ

2. Dec 12, 2007

### ircdan

the first is a mapping from the direct sum Z circle(+) Z to Z, the second is a mapping from Z to the direct product Z x Z, i'm not sure where the confusion is, does this help?

3. Dec 12, 2007

### chocok

thanks! can u tell me if i'm understanding correctly?

so for Z $$\oplus$$ Z -> Z (say Z is the integer field),
can i have some maps like this (4 maps): map((a,b)) = $$\pm a \pm b$$
i mean as long as the value on the right side stays in Z?

and for Z $$\rightarrow$$ZxZ, we are mapping from an elt to an ordered pair like
map(a)= (ca,ca) where c in Z?

4. Dec 12, 2007

### ircdan

Z is not a field

so you mean can you have the following maps
f:Z circle(+) Z ->Z given by f(a, b) = a + b
f:Z circle(+) Z ->Z given by f(a, b) = a - b
f:Z circle(+) Z ->Z given by f(a, b) = -a + b
f:Z circle(+) Z ->Z given by f(a, b) = -a - b
?
they are all well defined as is the bottom one in your post

That doesn't mean they are ring homomorphisms though(looking at the 3rd one)
f(a, b)f(c, d) = (a - b)(c - d) = ac - bc -ac + bd
f((a, b)(c, d)) = f(ac, bd) = ac - bd

Last edited: Dec 12, 2007