Let $X \subset \mathbb{P}^n$ be a quasiprojective scheme over a field $k$ and let $Z \subset \mathbb{P}^n$ be a closed subscheme. For a closed point $P$ in $X$, let $Y = \operatorname{Spec}(\mathcal{O}_{X,P}/\mathfrak{m}^2_{X,P})$.
I want to prove that $Z \cap Y = \operatorname{Spec}(\mathcal{O}_{X,P}/(\mathfrak{m}^2_{X,P}, \mathcal{I}_{Z,P}))$, where $\mathcal{I}_{Z}$ is the ideal sheaf of $Z$. Is this actually true? Without loss of generality assume $P$ is only in one of the standard open affines, say $D_{+}(x_0)$. Then $Z \times_{\mathbb{P}^n} Y = (Z \cap D_{+} (x_0)) \times_{D_+ (x_0)} Y$. Now we can explicity compute the intersection as we are working with affine schemes and you get the result. Is this the right method?
Best Answer
Your proposed method will work. Let us go through the steps in excruciating detail:
Lemma. Suppose $X\to Z$ and $Y\to Z$ are morphisms of schemes. If we can write $X\to Z$ as a composite $X\to X'\to Z$, then we have $X\times_Z Y \cong X\times_{X'} (X'\times_Z Y )$.
Proof. The fiber product of schemes is associative, and the condition on writing the morphism $X\to Z$ as the composite $X\to X'\to Z$ implies that $X\times_{X'} X'\cong X$. (In fact, this lemma is completely general in any category where the above fiber products exist.) $\blacksquare$
Now I claim $Y=\operatorname{Spec} \mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2\to \Bbb P^n$ can be written $\operatorname{Spec} \mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2\to \Bbb A^n\to \Bbb P^n$ where $\Bbb A^n$ is one of the standard affine opens $D(x_i)$. On the topological side, this is clear: $Y$ is a point, and the composite map on topological spaces is just the composition of immersions. On the sheaf side, we are to show that $\mathcal{O}_{\Bbb P^n}\to i_*\mathcal{O}_Y$ can be written as the composition $\mathcal{O}_{\Bbb P^n}\to i'_*\mathcal{O}_{\Bbb A^n}\to i''_*\mathcal{O}_Y$. This is straightforwards: let $y\in P$ be the underlying point of $i(Y)$. Then $i_*\mathcal{O}_Y(U)$ is $0$ if $y\notin U$ and $\mathcal{O}_Y(Y)$ if $y\in U$. So the map $\mathcal{O}_{\Bbb P^n}(U)\to i_*\mathcal{O}_Y(U)$ is the zero map if $y\notin U$ and can be written as the composition $\mathcal{O}_{\Bbb P^n}(U)\to \mathcal{O}_{\Bbb P^n,y}\to i_*\mathcal{O}_{Y,y} = i_*\mathcal{O}_{Y}(U)$. Once we notice that the same logic holds with the map between $Y$ and $\Bbb A^n$ and combine this with the fact that an open immersion like $\Bbb A^n\to \Bbb P^n$ is a local isomorphism, we get the result.
Next, we claim $(\Bbb A^n\times_{\Bbb P^n} Z)$ is the closed subscheme of $\Bbb A^n$ with structure sheaf $\mathcal{O}_Z|_{Z\cap \Bbb A^n}$. As closed and open immersions are both stable under base change, we see that $(\Bbb A^n\times_{\Bbb P^n} Z)$ is a closed subscheme of $\Bbb A^n$ and an open subscheme of $Z$. So as a topological space it's just $Z\cap \Bbb A^n$, and as it's an open subscheme of $Z$, it's structure sheaf is just $\mathcal{O}_Z|_{\Bbb A^n\cap Z}$. This implies that it's cut out by $\mathcal{I}_Z|_{\Bbb A^n}$ by uniqueness of the kernel. In particular, letting $I=\mathcal{I}_Z(\Bbb A^n)$, we have that $Z\times_{\Bbb P^n} \Bbb A^n=\operatorname{Spec} k[x_1,\cdots,x_n]/I$ is affine.
Now we are in the situation of computing a fiber product of affines: by the lemma above we get $$Y\times_{\Bbb P^n} Z\cong Y\times_{\Bbb A^n} (\Bbb A^n\times_{\Bbb P^n}) Z\cong\operatorname{Spec} \mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2 \times_{\Bbb A^n} \operatorname{Spec} k[x_1,\cdots,x_n]/I.$$
As the fiber product of $\operatorname{Spec} A\to \operatorname{Spec} R$ and $\operatorname{Spec} B\to \operatorname{Spec} R$ is given by $\operatorname{Spec} A\otimes_RB$, we see that our fiber product is given by $\operatorname{Spec} (\mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2 \otimes_{\Bbb k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/I)$. Next, recalling that localization commutes with quotients (because it's exact), localization may be described as a tensor product, the associativity of tensor product, $\mathcal{O}_{X,p}=k[x_1,\cdots,x_n]_{\mathfrak{m}_p}$, and $R/I\otimes_R R/J\cong R/(I,J)$ for any ring $R$ with ideals $I,J$, we may make the following manipulations:
$$\mathcal{O}_{X,p}/\mathfrak{m}_{X,p}^2 \otimes_{\Bbb k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/I \cong (k[x_1,\cdots,x_n]/\mathfrak{m}_p^2)_{\mathfrak{m}_p}\otimes_{k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/I$$
$$ \cong k[x_1,\cdots,x_n]_{\mathfrak{m}_p}\otimes_{k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/\mathfrak{m}_p^2 \otimes_{k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/I$$
$$ \cong k[x_1,\cdots,x_n]_{\mathfrak{m}_p}\otimes_{k[x_1,\cdots,x_n]} k[x_1,\cdots,x_n]/(\mathfrak{m}_p^2,I)$$
$$ \cong k[x_1,\cdots,x_n]_{\mathfrak{m}_p}/(\mathfrak{m}_p^2,I)_{\mathfrak{m}_p} \cong \mathcal{O}_{X,p}/(\mathfrak{m}_{X,p}^2,\mathcal{I}_p) $$
So we are done.
As you become more proficient in algebraic geometry, there's a lot of this that will become second nature and that you won't need to write out so many details for. Basically, each paragraph above will reduce to a sentence (more or less).