Cycle of non-equidimensional scheme

Question

In Fulton's intersection theory, example 1.7.1, he mentioned an example that contradicts to the splitting of cycles with respect to irreducible components. Consider the subscheme $X$ in $\mathbb{A}^3$ defined by $(zx,zy)$, and consider the Cartier divisor $E$ defined by $z-x$. Then should the cycle of $E$ should be the y-axis with an additional zero point? Or. it is just the y-axis itself.

Answer 1

Probably this is more than what you were looking for. I hope I haven't made silly blunders.

$X$ has two components (the x-y plane) $X_1=V(z)$ and (the z-axis) $X_2=V(x, y)$ with geometric multiplicities 1 each.

As in the Lemma 1.7.2, the RHS is $1[E\cap X_1]+1[E\cap X_2]=1[(0, 0, 0)]+1[y-axis]$.

Now the LHS: the fundamental cycle $[E]=2[(0, 0, 0)]+1[y-axis]$.

(To see this: $\mathcal{O}(E)=\dfrac{k[x,y,z]}{(xz, yz, x-z)}\simeq \dfrac{k[x, y]}{(x^2, xy)}$.

So $E$ has two irreducible components corresponding to the ideals $(x, z)$ and $(x, y, z)$ in $k[x, y, z]$ (or the ideals $(x)$ and $(x, y)$ in $k[x, y]$ via the above isomorphism.

The geometric multiplicity of $E$ at (the origin) $V((x, y))$ is given by the length $l(\mathcal{O}_{E, V((x, y)})$ as $\mathcal{O}_{E, V((x, y)}$-module. $\mathcal{O}_{E, V((x, y)}\simeq \Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)_{(x, y)}.$ This has a filtration $\Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)_{(x, y)}\supset (x, y) \supset \{0\}$ such that the consecutive quotients are simple $\Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)_{(x, y)}$-modules.

So the length $l(\mathcal{O}_{E, V((x, y)})=2$.)

The other multiplicity can be calculated similarly.

All these equalities are in the graded group $Z_*(X)$.

Thus, the 'LHS' and 'RHS' in Lemma 1.7.2 above don't agree in $Z_{*}(X)$.