## Everything is a Normal Cone

Well, really, for intersection theory, it’s true.  We start with $X\subset Y$ a closed subscheme, with normal cone $C$.  We’re going to construct a family of embeddings that deforms $X\subset Y$ to the zero section of $C$.  Then, because intersections should vary nicely in families, we’ll have essentially reduced the problem of doing intersections to the case of normal cones.

Now, look at $Y\times \mathbb{P}^1$.  Set $M$ to be the blowup along $X\times\{\infty\}$.  Now, the normal cone to this $X$ will be $C\oplus 1$, so the exceptional divisor, form the connections we talked about earlier, will be $P(C\oplus 1)$.

Now, also, the blowup of $X\times\mathbb{P}^1$ along $X\times\{\infty\}$ embeds as a closed subscheme of $M$.  By the universal property of the blowup, the image of $X\times\{\infty\}$ is a Cartier divisor.  But then, it already was, and so the blowup is an isomorphism.  So now we have a map $X\times\mathbb{P}^1\to M$.  We can also embed the blowup of $Y$ along $X$ (call it $\tilde{Y}$, by using $Y\times\{\infty\}$.

So now, there are two properties that we want for $M\to \mathbb{P}^1$ to have, in addition to flatness.

1. The fibers of $M$ over finite points of $\mathbb{P}^1$ should just by $Y$, with the usual embedding of $X$.
2. Over $\infty$, we should get the sum of two divisors, $P(C\oplus 1)+\tilde{Y}$, with $X$ embedded by the zero section of $C$ followed by the inclusion of $C$ into $P(C\oplus 1)$, and such that $P(C\oplus 1)\cap \tilde{Y}$ is the exceptional divisor of $\tilde{Y}$.

For the first, we know that we have maps $M\to Y\times \mathbb{P}^1\to\mathbb{P}^1$ which are all flat, so flatness follows.  To get the isomorphism away from $\infty$, we note that $M$ is a blowup along a subvariety contained in $Y\times \{\infty\}$, and so is an isomorphism there, as desired.  The second property is the one that requires some work.

For the second, we have embeddings of the two divisors, so we can just look locally on $Y$.  So we reduce to $Y=\mathrm{Spec}(A)$, and $X$ given by some ideal $I$.  We identify $\mathbb{P}^1\setminus\{0\}$ with $\mathbb{A}^1$, and so get an indeterminate $T$.  The blowup, $M$, will then just be $\mathrm{Proj}(S^*$), with $S^n=(I,T)^n=I^n+I^{n-1}T+\ldots+AT^n+AT^{n+1}+\ldots$.

Now, this is covered by open sets that are the specs of $S^*_{(a)}$ (that is, $s/a^n$ when $s\in S^n$) with $a$ running through a set of generators of $(I,T)$ in $A[T]$.  Now, the exceptional divisor $P(C\oplus 1)$ is given by $a/1$, and $\tilde{Y}$ is given by $T/a$, and then $T=(a/1)(T/a)$, which vanishes precisely at infinity.  Thus, $M_\infty$ decomposes as we’d like.

Now, why do we want to use the embedding of $X$ into the normal cone? Well, for one, there is a retraction $P(N\oplus 1)$ to $X$ when $X$ is regularly embedded, and two, there’s a vector bundle on the normal bundle of rank the codimension of $X$ with a section vanishing precisely on $X$.  This is a lot like the tubular neighborhood construction in topology, which simplifies a lot of problems. 1. Andrea says:
Flatness follows? The first map $M \to Y \times \mathbb{P}^1$ is a blowup, so it is hardly flat.
• L says: