Blowing up an affine variety along an ideal

After having seen this example, we now describe a general blow-up of an affine variety along an ideal.

We start by considering an affine variety and an ideal of the coordinate ring of The zero locus

defines a closed subvariety of and the blow up of along will be a variety isomorphic to everywhere away from . It is defined as follows:

Definition 2.1

Let be a set of generators for . We define the blow-up of at to be

that is, it is the closure inside of the graph of the morphism


We see that this is well-defined, since at any point in at least one of the does not vanish. Note that comes with a morphism given by projection onto the first coordinate. We define

Definition 2.2

The exceptional locus of the blow up of at is defined to be .

Remark 2.1

You might be worried that the above notation doesn't make sense, at least not until we show that does not depend on the choice of generators for . We do this in Proposition 2.1 below. Before that, we make the following observations:

  1. Note that is isomorphic to since there the morphism is invertible with inverse

  1. Let us investigate more closely how the closure in Definition 2.1 is formed. For this purpose, we need to know how to express the Zariski

closed subsets of . We can embed into and there we know from the Segre embedding that the closed subsets are intersections of vanishing sets of polynomials of the form which are bihomogeneous in the and variables, that is, they satisfy

for some and all (the pair is then called the bidegree of ). Thus a basis for the closed subsets of are precisely vanishing sets of polynomials , which are homogeneous only in the -variables.


We now prove the independence of generators. This proof will become obsolete when we give a much more general definition of blow up, but it is useful for our current hands-on approach.

Proposition 2.1

Let be another set of generators for and let us temporarily write

Then we have that


For this proof we may think of the ideal as lying inside and containing the vanishing ideal of , so that and . Thus we have polynomial relations and for some . We define the morphism

We need to check that this is well-defined, i.e. that cannot vanish simultaneously on . Consider the set of polynomials . These are homogeneous in the -variables and vanish on
It follows from our discussion in Remark 2.1 that they must vanish on the whole of . Thus we can't have all vanishing at a point in , since then vanishing of the will imply , a contradiction.

So is well-defined and it is clearly invertible with an analogously constructed inverse morphism. Hence .


We now define one more notion, regarding subvarieties of the one we wish to blow up.

Definition 2.3

Let be a closed subvariety different from . We define the strict transform of in to be .

Remark 2.2

Note that with this definition it is clear that the strict transform of is in fact . This is the crucial functoriality property of blow ups -- we can compute the blow up of a variety by first embedding it into a larger one, blowing that up and taking strict transform. The price we'll pay for writing down a general, coordinate free definition of blowing up is that this functoriality property will not be obvious (though, of course, still true).


Using the theory we've just developed, we will now revisit the example from the beginning.

Example 2.1

The blow up of at the (reduced) origin. The ideal we wish to blow up in is then and we will write as a shorthand for . So we have

Exercise 2.1

If we are working over , show that is topologically (remove a small disc around the origin and identify antipodal points on the resulting boundary). Then we see that in fact is the Mbius bundle and this justifies the common depiction of the blow up we give in Figure (MISSING)


Exercise 2.2

Similarly, show that is topologically , where the bar denotes opposite orientation. To do this, show that the blow up removes a small centered at the origin and then identifies the Hopf fibres on the introduced boundary .

Exercise 2.3

Let denote the exceptional divisor in . Show that the intersection product of with itself equals .