If i want to find out if {(n,n):n in Z} is an ideal of ZxZ, is there a test or prperties i can apply to prove it is an ideal? My book mentions something about finding a homomorphism? i dont know...
If i want to find out if {(n,n):n in Z} is an ideal of ZxZ, is there a test or prperties i can apply to prove it is an ideal? My book mentions something about finding a homomorphism? i dont know...
two magic steps to proving J is an ideal of a ring R:
let a be any element of R, x,y be any elements of J
1) show x - y is in J
2) show ax and xa are in J
that's it!
let's apply these to the set J = {(n,n): n in Z}.
let a = (b,c), x = (k,k), y = (m,m).
x - y = (k,k) - (m,m) = (k-m,k-m). check.
ax = (b,c)(k,k) = (bk,ck)...uh oh, it appears we DON'T have an ideal (because b might not equal c, so bk might not equal ck).
let's try "the homomorphism approach": J is an ideal of R, iff J = ker(φ) for some (ring) homomorphism φ:R→S
if that were so, we must have φ((k,k)) = 0_{S}, for every k in Z, and φ((k,m)) ≠ 0_{S}, if k ≠ m
(we want the kernel to be J, and ONLY J).
now note that if b ≠ c, then bk ≠ ck, but (since φ is a ring homomorphism):
φ((bk,ck)) = φ((b,c))(k,k)) = φ((b,c))φ((k,k)) = φ((b,c))0_{S} = 0_{S}.
but (bk,ck) isn't in J, so any ring homomorphism that "kills" J, kills MORE than J: J is not an ideal.
the way i think of ideals is this: ideals are characterized by a property that "absorbs" all of R.
one example is "even-ness" in the integers. the even numbers form an additive subgroup of Z (this is #1 above, we use the "one-step subgroup test" on (J,+)). and no matter what integer we multiply an even integer by, the result is always even. "even-ness" is "contagious" when you multiply.
the properties of ideals "generalize" the following properties of 0:
0 - 0 = 0
a0 = 0a = 0
so an ideal can be thought of as: "making a bigger 0-type of thingy" (which is exactly what we DO when we form the quotient ring R/J, we let J play the part of "0". and if it's going to play that part, then goshdagnabbit, it'd better have the right properties!).
the "typical" example is Z/nZ, where we let "multiples of n" play 0. so the integers get "reset" every n spaces, and go around in a circle, instead of a line. and yes, nZ is an ideal of Z (the ideal generated by n, in fact, so Z/nZ = Z/(n) = Z_{n}. it's all very cozy).
would it be similar to solve if {(n,m); n+m is even} is an ideal?
so we have a=(b,c), x=(k,r), y=(m,s)
1.) x-y= (k,r)-(m,s)= (k-m, r-s)
Now wouldn't this not be an ideal since k and r have to be either both even or both odd and m and s have to be either both even or both odd, so for example k-m and r-s have the chance of being even + odd which would not equal even
you're on the right track, but you made a wrong turn somewhere. if J = {(n,m) in Z x Z: n+m = 2u, for some u in Z}, and if x,y are in J, then:
x = (k,r) where k+r = 2u
y = (m,s) where m+s = 2v.
therefore: x-y = (k,r) - (m,s) = (k-m,r-s), and k-m + r-s = (k+r) - (m+s) = 2u - 2v = 2(u-v), hence x-y is in J.
so condition #1 IS satisfied.
let's look at condition #2. let a = (b,c) (where b and c might be any integers).
then ax = (b,c)(k,r) = (bk,cr). can we say that bk + cr is even? well, no.
suppose (k,r) = (1,3), which is clearly in J, and suppose a = (2,1). then (2,1)(1,3) = (2,3), and 2+3 = 5 is odd. so we don't have an ideal, but not for the reason you gave.
it's usually condition #2 that is the "tricky" part.