Intersection of a closed subscheme with a point 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?
 A: 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).
