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).