HW 3, Exercise 1, part b: Is (f+g)[x, y]=f[x,y]+g[x,y]?

I wanted to point this out and see if this is valid:

Since $2[x, y]=2*1[x, y]=0$ and $\zeta[x,y]=1$, does that mean that $2-\zeta[x, y]=2[x,y]-\zeta[x,y]=0-1=-1$?

More generally, is it true that $(f+g)[x,y]=f[x,y]+g[x,y]$?

I am inclined to think that this is the case, but maybe the way the multiplication is, i.e., $ff'[x,y]=\sum f[x,z]f'[x,y]$, bars us from claiming so.

An example would be $2[x,y]=1[x,y]+1[x,y]$, where $1[x,y]$ is as is in the problem statement. This clearly works, but does it work in general for functions $f, g$?

I'm a math geek. Math is fun!

1. sammonkie

I think so — the functions are in Hom which is a vector space.

2. sammonkie

Yes — they are evaluation maps: $+ =$

3. sammonkie

Yes — they are evaluation maps: $latex \ + \ = \$

4. Wonderful! Well, based on the way the multiplication works on $A(P)$, the above result will very likely be of little consequence to the proof of the exercise, though it is still necessary.