In Milne's Divisors and intersection theory, Example 12.3a) computes the intersection number of the curves $Z_1: Y=X^2$ and $Z_2: Y^2=X^3$ at the intersection point $P=(0,0)$ in the affine plane over a base field $k$.
By definition
$(Z_1 \cdot Z_2)_P = $ $\dim_k O_P/(Y-X^2, Y^2-X^3) = \dim_k k[X,Y]_{(X,Y)}/(Y-X^2, Y-X^3) $
The author states that this dimension is equal to $\dim_k k[X]/(X^4-X^3)$ (so, equal to 3). Why is that true?
I would say at most that
$k[X,Y]_{(X,Y)}/(Y-X^2, Y-X^3) = k[X]_{(X)}/(X^4-X^3) $
but this has dimension greater than 3, since contains for example powers like $1/(X-1)$ which are linearly independent from $1, X, X^2$.
Just reason as follows:
the element $X-1$ is invertible $k[X]_{(X)}$, so the generator $X^4-X^3=X^3(X-1)$ is equal to $X^3$ up to unit.
Conclude that $k[X]_{(X)}/(X^4−X^3)=k[X]_{(X)}/(X^3)=(k[X]/(X^3))_{(X)k[X]/(X^3)}=k[X]/(X^3)$, the last equality being true since $k[X]/(X^3)$ is already a local ring.