I'm following Eisenbud's description of blowing up: let $X$ be an affine algebraic variety, $R$ the coordinate ring of $X$, and let $a_1,\ldots,a_r$ generate $R$ as a $k$-algebra. Let $Y\subseteq X$ be an affine subvariety corresponding to an ideal $I\subseteq R$ generated by $g_0,\ldots,g_s$.

Let $S=k[x_1,\ldots,x_r,y_0,\ldots,y_s]$ and define $\varphi:S\to R[t]$ by sending $x_i\mapsto a_i$ and $y_i\mapsto g_it$. Then $\ker(\varphi)$ is an ideal homogeneous in the $y_i$, and so corresponds to an algebraic subset $Z\subseteq\mathbb{A}^r\times\mathbb{P}^s$ (the blowup of $Y$ in $X$).

Let $\pi: \mathbb{A}^r\times\mathbb{P}^s\to\mathbb{A}^r$ be projection, and let $E=Z\cap\pi^{-1}(Y)$ (the exceptional set of the blowup).

I am trying to prove the following:

The restriction $\pi|_{Z\setminus E}:Z\setminus E\to X\setminus Y$ is an isomorphism.

I can show that $\pi(Z)=X$, and so the map is surjective, but I am struggling to show injectivity.

Eisenbud also states:

$E$ corresponds to $R[It]/IR[It]$ as a subvariety of $Z$.

However I can't seem to show this either, or use this to help me prove the isomorphism. Since $S/\ker(\varphi)\cong\text{im}(\varphi)=R[It]$, would we have that $E$ corresponds to the ideal $(y_0,\ldots,y_s)\cap\ker(\varphi)$ in $S$?

Any help would be much appreciated.


1 Answer 1


$\newcommand{\IA}{\mathbb{A}}\newcommand{\onto}{\twoheadrightarrow}\newcommand{\IP}{\mathbb{P}}\newcommand{\from}{\colon}\newcommand{\iso}{\cong}$Let $X \subseteq \IA^r$ be an affine variety with coordinate ring $R$. Let $I = \langle g_0, \dots, g_s \rangle \subseteq I$ be an ideal and let $Y = V(I) = \{ x \in X \mid g_0(x) = \dots = g_s(x) = 0\}$ be its zero set in $X$. Let $J$ be the kernel of the morphism of graded $R$-algebras $$R[y_0, \dots, y_s] \onto B_I(R), \qquad Y_j \mapsto g_j \in B_I(R)_1 = I.$$ Note that if $F = \sum f_{i_0, \dots, i_s} Y_0^{i_0} \dots Y_s^{i_s} \in R[Y_0, \dots, Y_s]$ is a homogenous polynomial of degree $d$, then $F \in J$ iff the function $\sum f_{i_0, \dots, i_s} g_0^{i_0} \dots g_s^{i_s} \in I^d \subseteq R$ vanishes identically on $X$. Let $$Z = V_+(J) = \{(x,(b_0 : \ldots : b_s) \mid F(x, b_0, \dots, b_s) = \sum f_{i_0, \dots, i_s}(x) b_0^{i_0} \dots b_s^{i_s} = 0 \text{ for all homogenous } F \in J\} \subseteq X \times \IP^s$$ be the projective zero set of $J$.

Claim: The projection $\pi \from Z \subseteq X \times \IP^s \to X$ induces an isomorphism $Z\setminus \pi^{-1}(Y) \iso X\setminus Y$.

Proof: We construct the inverse map by sending $x \in X \setminus Y$ to $$(x,(g_0(x):\ldots:g_s(x)) \in Z.$$ To see that this is well-defined, note first that because $x \notin Y$, at least one of the $g_j(x)$ does not vanish, so we have an element of $X \times \IP^s$. To see that this is indeed an element of $Z$, let $F \in J$ be of degree $d$ as above. But then $$F(x,g_0(x), \dots, g_s(x)) = \bigl(\sum f_{i_0,\dots, i_s} g_0^{i_0} \dots g_s^{i_s}\bigr)(x) = 0.$$ It is clear that $\pi(x,(g_0(x) : \ldots : g_s(x)) = x$. Conversely, suppose that $(x, (b_0 : \ldots : b_s)) \in Z \setminus \pi^{-1}(Y)$. Note that we have $$ g_j Y_{j'} - g_{j'} Y_j \in J\qquad \text{for all }j, j'.$$ Hence, these functions vanish identically on $Z$. Plugging in $(x,(b_0:\ldots:b_s))$ we find that $$g_j(x)b_{j'} = g_{j'}(x) b_j \qquad \text{ for all } j, j'.$$ This shows that $$(b_0 : \ldots : b_s) = (g_0(x) : \ldots : g_s(x)) \in \IP^s$$ and we see that the function is indeed inverse to $\pi$.

  • $\begingroup$ Many thanks for such a clear explanation, it has been very helpful for me. I was wondering if you would mind explaining Eisenbud's claim I also mentioned that $\pi^{-1}(Y)$ corresponds to $B_IR/IB_IR$? No problem if not or if you would rather not answer a comment, I'm happy to post another question. Thanks again $\endgroup$
    – Dave
    May 13, 2020 at 21:47
  • 1
    $\begingroup$ @Dave I'm glad my answer was helpful to you. Regarding you follow-up question: Let us write $\pi^\ast \colon R \to B_IR$ for the natural map into degree $0$. The ideal $IB_IR$ is generated by $\pi^\ast(g_0), \dots, \pi^\ast(g_s)$. Given $(x,b) \in Z$ you have $(x,b) \in \pi^{-1}(Y) \iff x \in Y \iff g_0(x) = \dots = g_s(x) = 0 \iff \pi^\ast(g_0)(x,b) = \dots = \pi^\ast(g_s)(x,b) = 0$, so $\pi^{-1}(Y)$ is the zero set of functions generating $IB_IR$. In much generality, pulling back closed subsets amounts to pushing forward their ideals. $\endgroup$
    – user787668
    May 14, 2020 at 10:04
  • $\begingroup$ Thank you once again, I'll look into the generalities now I understand this case $\endgroup$
    – Dave
    May 15, 2020 at 19:44

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