Blowing up at a reduced subscheme is geometrically clear to me: it is just like blowing up at a reduced point, and point on the exceptional divisor corresponds to different tangent lines through the base point. However, I always have problem understanding blowing up at a nonreduced subscheme. Is it a geometric way to see it as the reduced cases?

As examples, I would like to ask the following questions:

  1. Consider the blowup of $\mathbb A^2$ at a double point $x={\rm Spec}(k[x,y]/(x^2,y))$. What is the exceptional divisor? Is it isomorphic to $\mathbb P^1$? Is the blowup smooth? (By the comment of mtrying46, the exceptional divisor is $\mathbb P^1$ but the blow up is not smooth)
  2. Consider that first blow up $\mathbb A^2$ at one point and then blow up at one point in the exceptional divisor of the blowup. The composition can be seen as blow up once (at a nonreduced center). Then what is the center?

From the two examples above, it seems that blowup at nonreduced center will either result in singularity or central fiber not being projective space. I would like to know whether this is true in general. That is:

  1. Suppose $Y$ is the blow up of $X$ at $Z\subset X$ and both $X,Y$ are smooth. If we know every fiber of $Z$ is isomorphic to some $\mathbb P^m$, can we conclude that $Z$ is reduced?

Everything is over $\mathbb C$. Any comments will be very helpful!

  • 1
    $\begingroup$ The blow-up of $\mathbb{A}^2$ at $\operatorname{Spec}(k[x,y]/(x^2,y))$ is the closed subscheme of $\mathbb{A}_{x,y}^2\times\mathbb{P}_{u.v}^1$ cut out by the equation $x^2v-yu=0$. The fiber over the origin is then cut out by the additional equations $x=y=0$, so we obtain just $\mathbb{P}^1$. But note that on the open set $v\neq 0$, it is isomorphic to the closed subscheme of $\mathbb{A}^3_{x,y,u}$ cut out by $x^2-yu$, which is singular at the origin as the gradient vanishes. $\endgroup$
    – imtrying46
    Apr 12 at 7:43
  • $\begingroup$ @imtrying46 Thanks for your comment. It looks true but I am a little confused. I can write the blowup as ${\rm Proj}(\oplus_{i\geq 0} (x^2,y)^i)$. Comparing with ${\rm Proj}(k[x,y][u,v])$ with the natural grading given by $u,v$, the ring of the latter has a surjection to the former, hence (?) it seems provide a closed embedding with the equation suggests by your comment. However, (1) why this is OK since Proj is not a functor? (2) is there a easier way to really "see" this instead of this computation? BTW, if you like, you can post it as an answer as this at least solved my first question. $\endgroup$
    – User X
    Apr 12 at 20:33

2 Answers 2


Question 1 hax been handled in the comments and now in another answer - the blowup of $\Bbb A^2$ in $(x^2,y)$ is the closed subscheme of $\Bbb A^2\times\Bbb P^1$ cut out by $x^2v-yu$, which is singular, as on $D(v)$ we have the closed subscheme of $\Bbb A^3$ cut out by $x^2-yu$. (As for dealing with your "proj isn't a functor" concern, you're correct that Proj is not a functor, but we do get some of a map on Proj from a map of graded rings - with a graded ring map $\varphi: A\to B$ we get a map $\operatorname{Proj} B \setminus V(\varphi(A_+)) \to \operatorname{Proj} A$, see for instance Hartshorne exercise II.2.14. If $\varphi$ is surjective, then $\varphi(A_+)=B_+$ and we get an honest map between the two Projs.)

As for question 2, consider the blowup of the ideal $(x^2,xy,y^3)$. This is the subscheme of $\Bbb A^2\times\Bbb P^2$ cut out by the kernel of $k[x,y][a,b,c]\to \bigoplus_{i=0}^\infty (x^2,xy,y^3)^i$ where $a,b,c$ map to the elements $x^2,xy,y^3$ in degree one, respectively. That kernel is $(ya-xb,yb^2-ac,y^2b-xc,xb^3-a^2c)$, and we can consider what it looks like on the three affine patches $D(a)$, $D(b)$, and $D(c)$.

  • On $D(a)$, we have $\operatorname{Spec} k[x,y,b,c]/(y-xb,yb^2-c,y^2b-xc,xb^3-c)$, which is isomorphic to $\operatorname{Spec} k[x,b]$, and the map to $\operatorname{Spec} k[x,y]$ is given by $(x,xb)$.
  • On $D(b)$, we have $\operatorname{Spec} k[x,y,a,c]/(ya-x,y-ac,y^2-xc,x-a^2c)$, which is isomorphic to $\operatorname{Spec} k[a,c]$, and the map to $\operatorname{Spec} k[x,y]$ is given by $(a^2c,ac)$.
  • On $D(c)$, we have $\operatorname{Spec} k[x,y,a,b]/(ya-xb,yb^2-a,y^2b-x,xb^3-a^2)$, which is isomorphic to $\operatorname{Spec} k[y,b]$, and the map to $\operatorname{Spec} k[x,y]$ is given by $(y^2b,y)$.

But these patches and their maps are exactly those of the iterated blowup. With a little more work, one sees that everything glues together the same way, so this is a single ideal which you can blow up which gives you your iterated blowup. (NB: the ideal defining a blowup is not unique.) I don't know a great way to do this algorithmically - I just kind of played around until I found this.

For question three, the answer is no. Consider the blowup of $\operatorname{Spec} k[x,y]$ in the ideal $(x^2,xy,y^2)=(x,y)^2$. This is $\operatorname{Proj}(\bigoplus_{n=0}^\infty (x,y)^{2n})$, and therefore isomorphic to $\operatorname{Proj}(\bigoplus_{n=0}^\infty (x,y)^n)$, as proj is invariant under taking Veronese subrings like this. But $Z=V(x^2,xy,y^2)$ is not reduced.

  • $\begingroup$ Many thanks for your detailed answer! This is very helpful. However, for the question three, I really meant to ask “whether there exists one $Z$ which is reduced” (sorry for not being clear). So, is the answer still negative? $\endgroup$
    – User X
    Apr 30 at 17:40
  • $\begingroup$ That's an interesting question - I'm not immediately sure what the answer is. If I manage to figure something out, I'll edit the post and ping you. $\endgroup$
    – KReiser
    May 1 at 3:29

As written in the comments, one can argue as follows: by definition, the blow-up of $\mathbb{A}^2$ at $\operatorname{Spec}(k[x,y]/(x^2,y))$ is $\operatorname{Proj}(\oplus_{i\geq 0}(x^2,y)^{i})$. As you correctly pointed out, there is a surjective graded morphism $\varphi: k[x,y][u,v]\to \oplus_{i\geq 0}(x^2,y)^{i}$ obtained by sending $u\mapsto x^2$ and $v\mapsto y$. By exercises II.2.14 and II.3.12 in Hartshorne, this induces a closed immersion $\iota:\operatorname{Proj}(\oplus_{i\geq 0}(x^2,y)^{i})\to\operatorname{Proj}(k[x,y][u,v])=\mathbb{A}^2_{x,y}\times\mathbb{P}_{u,v}$. This closed subscheme is determined by the homogeneous ideal $\ker\varphi$, which one can compute to be $(x^2v-yu)$.

Now to compute the fiber over the origin, we have to take the fiber product with $O=\operatorname{Spec}k[x,y]/(x,y)$. On can directly compute that $$O\times D_+(u)=\operatorname{Spec}\left(k[x,y]/(x,y)\otimes k[x,y][v/u]/(x^2(v/u)-y)\right)=\operatorname{Spec}\left(k[x,y][v/u]/(x,y)\right)=\mathbb{A}^1_{v/u}$$ and similarly $O\times D_+(v)=\mathbb{A}^1_{v/u}$. By gluing, we obtain that the fiber is given by $\mathbb{P}^1$.

On the other hand, the blow-up isn't smooth. Indeed, on $D_+(v)$ it is given by $V(x^2-(u/v)y)\subseteq \mathbb{A}^3_{x,y,u/v}$, which isn't smooth at the origin by the Jacobian criterion.

This settles 1, I'll think about 2 and 3!


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