# Traces of nilpotent matrices

I needed the following well-known result in the course I’m teaching:

Lemma. Let be a field and a nilpotent matrix. Then .

The classical proof uses the flag of subspaces

to produce a basis for which is upper triangular. Here is a slick basis-independent commutative algebra proof that shows something better:

Lemma. Let be a commutative ring with nilradical , and let be a nilpotent matrix. Then the characteristic polynomial satisfies

Here we write for the polynomials in of degree smaller than whose coefficients lie in a given ideal .

Note that the formulation should ring a bell: in the previous post we saw that . When is a domain, this reduces to , and the lemma just says that .

This suggests that we shouldn’t work with but with its anadrome (or reciprocal) .

Proof of Lemma. We have to show that . Since is nilpotent, there exists with , so . Thus . Evaluating at shows that the constant coefficient is 1.

# Invertible polynomials

Here is a classic little lemma on polynomials:

Lemma. Let be a commutative ring with nilradical . Then .

That is, a polynomial is invertible if and only if is invertible and are nilpotent. This is the formulation you find in Atiyah–MacDonald, exercise 1.2(i). Note that when is a domain (or more generally a reduced ring), this just says that (via the constant polynomials).

Proof. First suppose with and . Multiplying by we may assume ; this does not change the nilpotence assumption on . Then is nilpotent since it is a sum of nilpotent elements in the commutative ring . Thus we get for some , and

shows that . Conversely, let , and suppose for some polynomial . Without loss of generality we may assume . If is a domain, then the leading term of is , so we must have and and are constant. In the general case, we conclude that for any ring homomorphism to a domain . In particular, taking for any prime ideal shows and . Taking the intersection over all prime ideals gives

# Union of hyperplanes over a finite field

The following lemma is a (presumably well-known) result that Raymond Cheng and I happened upon while writing our paper Unbounded negativity on rational surfaces in positive characteristic (arXiv, DOI). Well, Raymond probably knew what he was doing, but to me it was a pleasant surprise.

Lemma. Let be a power of a prime , and let . Then satisfy a linear relation over if and only if

Proof. If for , then for all since . As is a ring homomorphism, we find

so the determinant is zero. Conversely, the union of -rational hyperplanes is a hypersurface of degree (where denotes the dual projective space parametrising hyperplanes in ). Since the determinant above is a polynomial of the same degree that vanishes on all -rational hyperplanes, we conclude that it is the polynomial cutting out , so any for which the determinant vanishes lies on one of the hyperplanes.

Of course when the determinant is zero, one immediately gets a vector in the kernel. There may well be an immediate argument why this vector is proportional to an element of , but the above cleverly circumvents this problem.

For concreteness, we can work out what this determinant is in small cases:

• : a point only satisfies a linear relation over if it is zero.
• : the polynomial cuts out the -rational points of .
• : the polynomial

cuts out the union of -rational lines in . This is the case considered in the paper.

# Algebraic closure of the field of two elements

Recall that the field of two elements is the ring of integers modulo . In other words, it consists of the elements and with addition and the obvious multiplication. Clearly every nonzero element is invertible, so is a field.

Lemma. The field is algebraically closed.

Proof. We need to show that every non-constant polynomial has a root. Suppose does not have a root, so that and . Then , so is the constant polynomial . This contradicts the assumption that is non-constant.

# Internal Hom

This is an introductory post about some easy examples of internal Hom.

Definition. Let be a symmetric monoidal category, i.e. a category with a functor that is associative, unital, and commutative up to natural isomorphism. Then an internal Hom in is a functor

such that is a left adjoint to for any , i.e. there are functorial isomorphisms

Remark. In the easiest examples, we typically think of as ‘upgrading to an object of ‘:

Example. Let be a commutative ring, and let be the category of -modules, with the tensor product. Then with its natural -module structure is an internal Hom, by the usual tensor-Hom adjunction:

The same is true when is the category of -bimodules for a not necessarily commutative ring .

However, we cannot do this for left -modules over a noncommutative ring, because there is no natural -module structure on for left -modules and . In general, the tensor product takes an -bimodule and a -bimodule and produces an -bimodule . Taking gives a way to tensor a right -module with a left -module, but there is no standard way to tensor two left -modules, let alone equip it with the structure of a left -module.

Example. Let . Then is naturally a set, making it into an internal Hom for :

When is the categorical product , the internal (if it exists) is usually called an exponential object, in analogy with the case above.

Example. Another example of exponential objects is from topology. Let be the category of locally compact Hausdorff topological spaces. Then the compact-open topology makes into an internal Hom of topological spaces. (There are mild generalisations of this beyond the compact Hausdorff case, but for an arbitrary topological space the functor does not preserve colimits and hence cannot admit a right adjoint.)

Example. An example of a slightly different nature is chain complexes: let be a commutative ring, and let be the category of cochain complexes

of -modules (meaning each is an -module, and the are -linear maps satisfying ). Homomorphisms are commutative diagrams

and the tensor product is given by the direct sum totalisation of the double complex of componentwise tensor products.

There isn’t a natural way to ‘endow with the structure of a chain complex’, but there is an internal Hom given by

with differentials given by

Then we get for example

since a morphism is given by an element such that , i.e. , meaning that is a morphism of cochain complexes.

Example. The final example for today is presheaves and sheaves. If is a topological space, then the category of abelian sheaves on has an internal Hom given by

with the obvious transition maps for inclusions of open sets. This is usually called the sheaf Hom. A similar statement holds for presheaves.

# Local structure of finite unramified morphisms

It is well known that a finite étale morphism of schemes is étale locally given by a disjoint union of isomorphisms, i.e. there exists an étale cover such that the pullback is given by . Something similar is true for finite unramified morphisms:

Lemma. Let be a finite unramified¹ morphism of schemes. Then there exists an étale cover such that the pullback is given by , where are closed immersions of finite presentation.

Proof. Let be a point, let be the strict henselisation of at , and let be the base change of along . Then is unramified, so by Tag 04GL it splits as

where is surjective for each and no prime of lies above . But is also finite, so by Tag 00GU the map hits the maximal ideal if . Thus, we conclude that , hence is a product of quotients of .

But is the colimit of for an étale neighbourhood inducing a separable extension . Since is of finite presentation, each of the ideals and the projections are defined over some étale neighbourhood . Then the pullback is given by a finite disjoint union of closed immersions in .

Then might not be a covering, but since was arbitrary we can do this for each point separately and take a disjoint union.

Remark. The number of needed is locally bounded, but if is not quasi-compact it might be infinite. For example, we can take an infinite disjoint union of points, and such that the fibre over for has points.

Remark. In the étale case, we may actually take finite étale, by taking to be the Galois closure of , which exists in reasonable cases². For example, if is normal, we may take to be the integral closure of in the field extension corresponding to the Galois closure of . In general, if is connected it follows from Tag 0BN2 that a suitable component of the -fold fibre product of over is a Galois closure of . If the connected components of are open, apply this construction to each component.

In the unramified case, this is too much to hope for. For example, if , then we may take to be a nontrivial finite étale cover of an elliptic curve . This is finite and unramified, but does not split over any finite étale cover of since there aren’t any. In fact, it cannot split over any connected étale cover whose image contains , since that implies the image only misses finitely many points (as is ample), which is again impossible since .

¹For the purposes of this post, unramified means in the sense of Grothendieck, i.e. including the finite presentation hypothesis. In Raynaud’s work on henselisations, this was weakened to finite type. See Tag 00US for definitions.

²I’m not sure what happens in general.

# Rings that are localisations of each other

This is a post about an answer I gave on MathOverflow in 2016. Most people who have ever clicked on my profile will probably have seen it.

Question. If and are rings that are localisations of each other, are they necessarily isomorphic?

In other words, does the category of rings whose morphisms are localisations form a partial order?

In my previous post, I explained why and are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let be a field, and let . Let

be an infinite-dimensional polynomial ring over , and let

Then is a localisation of , and we can localise further to obtain the ring

isomorphic to by shifting all the indices by 1. To see that and are not isomorphic as rings, note that is closed under addition, and the same is not true in .

Is there a moral to this story? Not sure. Maybe the lesson is to do mathematics your own stupid way, because the weird arguments you come up with yourself may help you solve other problems in the future. The process is more important than the outcome.

# Is the affine line isomorphic to the punctured affine line?

This is the story of Johan Commelin and myself working through the first sections of Hartshorne almost 10 years ago (nothing creates a bond like reading Hartshorne together…). This post is about problem I.1.1(b), which is essentially the following:

Exercise. Let be a field. Show that and are not isomorphic.

In my next post, I will explain why I’m coming back to exactly this problem. There are many ways to solve it, for example:

Solution 1. The -algebra represents the forgetful functor , whereas represents the unit group functor . These functors are not isomorphic, for example because the inclusion induces an isomorphism on unit groups, but not on additive groups.

A less fancy way to say the same thing is that all -algebra maps factor through , while the same evidently does not hold for -algebra maps .

However, we didn’t like this because it only shows that and are not isomorphic as -algebras (rather than as rings). Literal as we were (because we’re undergraduates? Lenstra’s influence?), we thought that this does not answer the question. After finishing all unstarred problems from section I.1 and a few days of being unhappy about this particular problem, we finally came up with:

Solution 2. The set is closed under addition, whereas is not.

This shows more generally that and are never isomorphic as rings for any fields and .

# Rings are abelian

In this post, we prove the following well-known lemma:

Lemma. Let satisfy all axioms of a ring, except possibly the commutativity . Then is abelian.

That is, additive commutativity of a ring is implied by the other axioms.

Proof. By distributivity, we have , so multiplication by is a homomorphism. By our previous post, this implies is abelian.

# Finiteness is not a local property

In this post, we consider the following question:

Question. Let be a Noetherian ring, and and -module. If is a finite -module for all primes , is finite?

That is, is finiteness a local property?

For the statement where local means the property is true on a cover by Zariski opens, see Tag 01XZ. Some properties (e.g. flatness) can also be checked at the level of local rings; however, we show that this is not true for finiteness.

Example 1. Let , and let . Then , because localisation commutes with direct sums and if is prime. Thus, is finitely generated for all primes . Finally, , because is torsion. But is obviously not finitely generated.

Example 2. Again, let , and let be the subgroup of fractions with such that is squarefree. This is a subgroup because can be written with denominator , and that number is squarefree if and are. Clearly is not finitely generated, because the denominators can be arbitrarily large. But , which is finitely generated over . If is a prime, then is the submodule , which is finitely generated over .

Another way to write is .

Remark. The second example shows that over a PID, the property that is free of rank can not be checked at the stalks. Of course it can be if is finitely generated, for then is finite projective [Tag 00NX] of rank , hence free since is a PID.