My setting is that of Hartshorne chapter I ("classical varieties" over an algebraically closed field, which are irreducible [not schemes]).
Background
I've been working on problem 7.7(a) and I've hit a bit of a wall with part (a):
Problem 7.7. Let $Y$ be a variety of dimension $r$ and degree $d>1$ in $\Bbb P^n$. Let $P\in Y$ be a nonsingular point. Define $X$ to be the closure of the union of all lines $PQ$, where $Q\in Y$, $Q\neq P$.
(a) Show that $X$ is a variety of dimension $r+1$.
Clearly $X$ contains $Y$, so $X$ must be of dimension at least $r$. On the other hand, we can define a map $f$ from $(Y\setminus\{P\})\times\Bbb P^1$ to $X$ by sending $(Q,[a:b])\mapsto aQ+bP$ which surjects on to the collection of points in $X$ which are on a line $PQ$ with $Q\in Y$ and $Q\neq P$. Since $(Y\setminus\{P\})\times\Bbb P^1$ is irreducible, this shows that $X$ is irreducible, and the map is dominant, so we get an extension of function fields $k(X)\to K(Y\times\Bbb P^1)$. This shows that $\dim X \leq r+1$. So all I have to do finish the problem is to show that $X$ contains one point which isn't in $Y$. This is turning out to be harder than I expected!
Here's how I'd want to solve this problem: Pick a hyperplane $H$ through $P$ which is transverse to $Y$ (we can do this because $r<n$ by the degree assumption plus the fact that $P$ is a smooth point). Then the algebraic set $Y\cap H$ is smooth at $P$ (I can prove this via the Jacobian criteria), so there's a unique irreducible component $Z\subset Y\cap H$ through $P$. If I could prove that the intersection multiplicity along $Z$ is equal to one, I'd win: either $Y\cap H$ is reducible and I can connect $P$ with a point $Q$ on some other irreducible component and find a point in $X\setminus Y$ on the line $PQ$, or $Y\cap H=Z$, a variety of dimension $r-1$ and degree $d$, and eventually I can get it down to the case of a curve and I understand what to do in that case.
Question
Suppose $Y\subset \Bbb P^n$ is a variety of dimension $r$ and degree $d>1$, and $P$ a smooth point on $Y$. Suppose $H$ is a hyperplane through $P$ so that $Y\cap H$ is smooth at $P$ and $Y\cap H$ is irreducible. Why should $i(Y,H;Y\cap H)=1$?
My thoughts
The problem is that I feel very unsure about why or how I should be able to get this result. I'm supposed to compute the length of $(S/(I_Y+h))_\mathfrak{p}$ over the local ring $S_\mathfrak{p}$, where $\mathfrak{p}$ is the homogeneous prime ideal corresponding to the irreducible subvariety $Y\cap H$ and $h$ is a generator for the ideal of $H$. I can show that this the same as what you'd want in the affine case - that is, if we work in some standard affine patch $U$ containing $P$, I can show that the intersection multiplicity is the same as the length of $(k[Y\cap U]/(\widetilde{h}))_{\mathfrak{p}'}$ over $k[Y\cap U]_{\mathfrak{p}'}$ where $\mathfrak{p}'$ is the ideal corresponding to $Y\cap H\cap U$, but I don't see how I'm supposed to conclude this is $1$ if I know smoothness at $P$.
Edit: I've made a little progress - I can show that $f$ is in the maximal ideal of $P$ in $Y$ but not it's square, so $f$ is in the ideal of $Y\cap H$ but not it's square. I still don't see how this gives that the intersection multiplicity is one - I would need to know something along the lines of $\mathcal{O}_{Y,Y\cap H}$ having principal maximal ideal, which would mean it's a DVR? Still pretty stuck, though.
Or, if you think I'm barking up the wrong tree, is there some other way I'm supposed to finish problem 7.7(a)?
In the language of the post, the local ring of $Y\cap H\cap U$ at $P$ is $\mathcal{O}_{Y,P}/\sqrt{(f)}=k[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/\sqrt{(I_{Y\cap U}+(f))}_{(x_1,\cdots,x_n)}$, which is the reduction of $k[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/(I_{Y\cap U}+(f))_{(x_1,\cdots,x_n)}$. But Hartshorne's theorem I.5.1 says that this last ring is actually a regular local ring by the Jacobian criterion and therefore reduced. So this is an equality, and $\mathcal{O}_{Y\cap U,P}/\sqrt{(f)}=\mathcal{O}_{Y\cap U,P}/(f)$. By the transitivity of localization, we have that $\mathcal{O}_{Y\cap U,Z\cap U}=(\mathcal{O}_{Y\cap U,P})_{I_{Z\cap U,P}}$, so the multiplicity of $k[U]/(I_{Y\cap U}+(f))$ is one. Since this is equal to the multiplicity of the homogeneous version, we have that the intersection multiplicity is one.