I was reading commutative algebra from Miles Reid where I got stuck in the geometry of nilpotent element on page 29.
The geometric picture of nilpotents is in the spirit of a nonrigorous introduction to calculus, where one may talk about a "number e so small that $e^2 = 0$", and use this to calculate the derivative f'(x) of a polynomial by the formula f(x + e) = f(x) + ef'(x). This is of course nonsense in the context of an element of the real field, but is perfectly sensible and useful in algebra. Thus the element $Y \in k[X, Y]/(Y^2)$ is pictured as the function on the X-axis with values in $k[e]/(e^2 = 0) $ which remembers the $Y$ derivative $(df/dY)(x, 0) $ of $f(X, Y) \in k[X, Y] $ at each point $(x,0)$. Higher order nilpotents can be used to give an algebraic treatment of Taylor expansions of polynomials up to any order.
What does this paragraph even mean? How is the geometry of nilpotent element related to calculus?
Recall that the (average) slope of a function over the interval $[x,x+\Delta x]$ is given by $$ m = \frac{f(x + \Delta x) - f(x)}{\Delta x}, $$ and the derivative of a function is defined by $$ f'(x) = \lim_{\Delta x \to 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}. $$ The reason we need to take a limit is that taking the slope of a function between points $x$ and $x + \Delta x$ doesn't make sense if $\Delta x = 0$; plugging in $\Delta x = 0$ forces us to divide by zero.
The (non-rigorous version of the) idea of $e$ is to play the role of an "infinitely small element", so that we can calculate the derivative as a "slope". That is, $e$ is "so small that" the average slope over $[x,x+e]$ is $f'(x)$, i.e. exactly equal to the slope at $x$. In terms of the formulas above, this amounts to saying that $$ f'(x) = \frac{f(x + e) - f(x)}{e} \implies f(x+e) = f(x) + f'(x) \cdot e. $$