id: blowup-scheme-theoretic-viewpoint · parent: _kodaira-network-index.md · depends on: fiber-of-morphism
Yan's own summary — written independently, following the discussion of $\widetilde V$ as a
$V$-scheme, the local-coordinate computation of $B_i\otimes_A\kappa(p)$, and the balancing
relation for $\otimes_A$. Not co-written — this is the reconstruction pass.
The category of $A$-algebras.
An $A$-algebra is a ring homomorphism $\phi:A\to B$.
This is compatible with the traditional definition of an $A$-algebra: An $A$-algebra is an $A$-module with a ring structure. To see this, observe that $a\cdot b:=\phi(a)b$ gives $B$ an $A$-module structure.
A morphism $f:(B,\phi)\to (D,\psi)$ in the category of $A$-algebras is given by a ring homomorphism $f:B\to D$ satisfying $f(\phi(a)b) = \psi(a)f(b)$.
This is nothing but saying that $f$ is a ring homomorphism $B\to D$ that is $A$-linear.
The tensor product in the category of $A$-algebras is the universal object of the pushout diagram $A\to B, A\to C$, denoted by $B\otimes_A C$.
One has $A\otimes_A B \simeq B$ in the category of $A$-algebras.
This follows directly from the universal property of the pushout: $B$ itself, with identity maps, satisfies it — using that a morphism out of $B$ is exactly an $A$-linear ring map (previous point).
Precisely, the isomorphism is given by $x\otimes b \mapsto 1\otimes \phi(x)b \mapsto \phi(x)b$ (via the balancing relation, then dropping the redundant $1\otimes\,$).
Balancing relation: for $A$-modules $M,N$ and $a\in A$, $(a\cdot m)\otimes n = m\otimes(a\cdot n)$ inside $M\otimes_A N$ — lets a scalar be “slid across” the tensor sign. Used above and again below.
Affine base change lemma: Let $\phi:A\to B$ be an $A$-algebra. Let $I$ be an ideal of $A$, then $B\otimes_A (A/I) = B/(IB)$, where $IB$ is the ideal of $B$ generated by $\phi(I)$.
Proof
Let $\Phi:B\to B\otimes_A A/I$, $b\mapsto b\otimes \bar{1}$.
$\Phi$ is constant on cosets of $B/(IB)$: if $b-b'\in IB$ then $\Phi(b) = \Phi(b')$. It follows $\Phi$ gives well-defined ring homomorphism $\bar{\Phi}$ on $B/(IB)$.
By the balancing relation, $\phi(x)\otimes\bar1 = 1\otimes(x\cdot\bar1) = 1\otimes\bar x = 0$ for any $x\in I$ (since $\bar x = 0$ in $A/I$). Writing $b-b'=\sum_k c_k\phi(x_k)$ with $x_k\in I$, $c_k\in B$ (as $IB$ is generated by $\phi(I)$), $\Phi(b-b') = \sum_k c_k\big(\phi(x_k)\otimes\bar1\big) = 0$.
$\bar{\Phi}$ is surjective.
By the balancing relation, $b\otimes\bar a = b\otimes(a\cdot\bar1) = (\phi(a)b)\otimes\bar1 = \Phi(\phi(a)b)$ for any simple tensor $b\otimes\bar a$; since these span $B\otimes_A(A/I)$, $\Phi$ is surjective.
$\bar{\Phi}$ is injective. We must prove $\mathrm{ker}\bar{\Phi}= 0$, or $\mathrm{ker}\Phi = IB$.
The defining exact sequence of $I$: $I\to A\to A/I\to 0$;
Use the fact that $\otimes_A B$ is a right exact functor we get:
$I\otimes_A B\to A\otimes_A B \to A/I \otimes_A B\to 0$;
The second term is isomorphic to $B$ by $x\otimes b \mapsto 1\otimes \phi(x)b\mapsto \phi(x)b$;
It follows that, for $a_1x_1+\cdots+a_nx_n\in I$, where $x_1,\dots,x_n$ are generators of $I$, the first map is $(a_1x_1+\cdots+a_nx_n)\otimes b \mapsto [\phi(a_1)\phi(x_1)+\cdots+\phi(a_n)\phi(x_n)]b\in IB$ on this simple tensor; by additivity the same holds on all of $I\otimes_A B$.
$IB$ is the image of the first map.
Under $A\otimes_A B\cong B$ and $(A/I)\otimes_A B\cong B\otimes_A(A/I)$ (swap), the second map is exactly $\Phi$.
The exactness implies $IB = \mathrm{ker}\Phi$. This is precisely saying $\bar\Phi$ is injective.
The Spec of an $A$-algebra $\phi:A\to B$ is a scheme $Y=Spec\ B$ over $X = Spec\ A$, via the induced scheme morphism $f:=Spec\,\phi:Y\to X$. For $\mathfrak p\in Spec\ A$, the fiber of $f$ at $\mathfrak p$ is the base change along $Spec\,k(\mathfrak p)\to X$. The $A$-algebra of the fiber is just the extension of scalars $B\otimes_A k(\mathfrak p)$.
In particular, the case of maximal ideal $\mathfrak m$. The fiber is just $B\otimes_A A/\mathfrak m$. By the affine base change lemma, it is $B/\mathfrak m B$.
Consider the case $V=Spec\ k[z_1,\dots,z_n]$. An affine coordinate of blowing up of $0$ is $U_i = Spec\ k[z_i,y_{1/i},\dots,y_{n/i}]$ without $y_{i/i}$. In algebraic terms, we let $A=k[z_1,\dots,z_n]$, $B=k[z_i,y_{1/i},\dots,y_{n/i}]$, the morphism is given by the $A$-algebra $\phi:A\to B$, $z_i\mapsto z_i$, and $z_j\mapsto z_iy_{j/i}$ for $j\ne i$. Let $\mathfrak m$ be the maximal ideal of $0$. Question: what is the fiber of $\mathfrak m$?
The $A$-algebra of the fiber is $B/\mathfrak m B$ by the affine base change lemma.
A general element of $\mathfrak m B$ is $\sum_k b_k\phi(z_k) = b_iz_i+\sum_{j\ne i}b_jz_iy_{j/i} = z_i\big(b_i+\sum_{j\ne i}b_jy_{j/i}\big)$. Since $b_i$ ranges over all of $B$ (independent of the other $b_j$), the bracketed factor sweeps out all of $B$ — it is the unit ideal, not an effective constraint. So $\mathfrak m B = (z_i)$.
Hence $B/\mathfrak m B = k[z_i,y_{1/i},\dots,y_{n/i}]/(z_i) \cong k[y_{1/i},\dots,y_{n/i}]$: a plain polynomial ring, reduced (no nilpotents). Contrast: had $\mathfrak m B$ come out as $(z_i^2)$ instead, the fiber would carry a nonreduced (fat-point) structure — it doesn't, because blow-up makes $\mathfrak m$ invertible (principal, generated by one non-zero-divisor), not merely contained with multiplicity.
Gluing the charts: does $E\cap U_i = Spec\ k[y_{1/i},\dots,y_{n/i}]$ glue with $E\cap U_k$ to give $Proj\ k[y_1,\dots,y_n] = \mathbb P^{n-1}$?
On the overlap, $\phi_i(z_j)=z_iy_{j/i}$ and $\phi_k(z_j)=z_ky_{j/k}$ compute the same $z_j$, so $z_iy_{j/i}=z_ky_{j/k}$.
Since $z_k=z_iy_{k/i}$ (chart $i$'s own relation), substituting and dividing by $z_i$ gives $y_{j/i} = (z_k/z_i)y_{j/k} = y_{k/i}\,y_{j/k}$, i.e. $y_{j/k}=y_{j/i}/y_{k/i}$.
This is exactly the cocycle relation defining how $Proj\ k[y_1,\dots,y_n]$ glues its standard affine charts $D_+(y_i) = Spec\ k[y_1,\dots,y_n]_{(y_i)}$ (coordinatized by $y_j/y_i$). The identity doesn't involve $z_i,z_k$, so it holds on all of $U_i\cap U_k$, including on $E$ where $z_i=z_k=0$.
Hence $E \cong Proj\ k[y_1,\dots,y_n] = \mathbb P^{n-1}$, reduced, glued exactly as computed chart-by-chart in (4).