Let I={ (2x,3x) such that x is in Z mod 6} in R= Z mod 6 x Z mod 6

The book says this isn't an ideal, but I'm not seeing it.

If r=(a,b) and k=(2x,3x) where a,b,x are in Z mod 6 then rk=(a2x, b3x). But couldn't I commute it into (2ax, 3bx) where ax, bx are in Z mod 6? Or am I not allowed to move 2 since it's not explicity in Z mod 6?

EDIT: Nevermind, I figured it out. ax must equal bx, but it isn't so.