TITLE: How do algebraists intuitively picture normal subgroups and ideals?
QUESTION [18 upvotes]: I have developed some intuitive picture to homomorphisms and kernels, but have troubles developing useful insight to normal subgroups and ideals.

REPLY [8 votes]: To elaborate on a very important point briefly touched on in Matt Samuel’s answer:

in algebraic geometry, a ring $R$ is viewed as representing a space $X$, with elements of the ring being some kind of functions on the space — extending the way that e.g. the polynomial ring $\mathbb{Z}[x,y]$ can be viewed as a ring of functions on the 2-dimensional plane.
any subspace $X' \subset X$ of the space should then induce an ideal $I_X \subseteq R$, the ideal of functions that vanish (i.e. are constantly 0) on the subspace $X'$.  So an ideal can represent a subspace; for instance, the ideal $(x-1) \subseteq \mathbb{Z}[x,y]$ represents the line $x = 1$ in the plane, since a polynomial vanishes on that subspace exactly if it’s a multiple of $x-1$.
slightly more generally, the functions that “vanish to higher order” on a subspace also form an ideal.  E.g. $((x-1)^2)$ represent the functions that vanish to second-order on the line $x=1$.  If you haven’t met this idea of higher-order vanishing before, look at (or imagine) a 3-d graph of the function $z = (x-1)^2$ on the $xy$-plane, and see how it’s not just zero on the line $x=1$, it’s flat on that line.  Intuitively, one can think of it as vanishing not just on that line, but on an infinitessimal thickening of that line.  (And $(x-1)^3$ vanishes on a slightly thicker infinitessimal thickening, and so on.)

So intuitively, in the algebraic geometry picture, ideals of a ring correspond to subspaces of a space.