Parameterizing rational points on a singular projective cubic

Let be a singular rational projective cubic. It follows from Bézout’s theorem (applied over ) that there can only be one singularity; if not, the line joining two singularities would intersect with total multiplicity at least , but a cubic and a line intersect with multiplicity at most . Moreover, another application of Bézout’s theorem proves that the singularity must be a double point; if it were higher, once again we’d arrive at a contradiction after joining the singularity and another point of the curve with a line.

Let’s call the singularity . We know that ‘s coordinates are algebraic over , so we can pick a finite Galois extension containing all of them. Now, we consider the action of the Galois group over . Since the Galois group acts on curves, it must preserve its singularities (since a singularity is a common zero of both and its derivatives). Therefore, it must fix , and so we conclude that is a rational point.

Now, any non-tangent line at with rational slope must intersect the cubic exactly one more time (once again by Bézout). Thus, we completely parameterize the cubic’s rational points in terms of its singularity.

A non-free, torsion-free abelian group

In this post we’ll prove that is not free as a -module, although it is torsion-free. Recall that, in the finitely generated case, torsion-free modules are free, by the structure theorem.

For the sake of contradiction, assume that is free. Let be the submodule of consisting of sequences such that the sequence goes to infinity, where is a fixed prime and is the biggest natural such that divides . Since is a PID, every submodule of a free module is free, and so is free. Moreover, we have an injection defined by . Since is uncountably generated, it follows that is, too.

Now we consider the -vector space . Since we supposed was an uncountably generated free -module, should be an uncountably generated vector space (since implies ). On the other hand, every element in can be represented as a sequence that is eventually zero, and there are a countable number of such sequences. So is countable; in particular, it’s countably generated, which contradicts our previous observation. We then conclude that is not free, as we wanted to prove.

Notice that this proves that a direct product of projective modules need not be projective as well, since any projective module over a PID is free.

Finite-dimensional subspaces of normed spaces are complemented

Let be a finite-dimensional subspace of a normed space and pick a basis for . Let be a bounded linear functional that extends the map if and . Such a functional exists by the Hahn-Banach theorem; even more, we can pick one such that its norm is bounded by . We now define as

It’s easy to see that is bounded; in fact , since each is bounded by 1. Therefore, the kernel of this map is closed. Finally, we have a direct sum decomposition , since is a projector. One easily sees that , and so is complemented.

Cyclic modules over local algebras are indecomposable

Let be a local algebra and let be a cyclic -module. Suppose , with . Let be a generator for ; then with , both non-zero. Since is cyclic, there exists some such that . But , and since both and are -modules, this implies and . In particular, is not invertible. Therefore, we must have that is invertible, since is local. But now , and so . This is a contradiction, so we conclude that must be indecomposable.

The representation type of C_p x C_p in the modular case

In this post we’ll prove that has infinite representation type in the modular case. This illustrates the fact that the representation theory of groups in the modular case is a hard problem; we have infinite representation type even for very simple groups.

The group algebra is isomorphic to , and since , we have that and so . We’ll now exhibit an infinite family of non-isomorphic indecomposable modules for this algebra.

Take an -dimensional -vector space; we’ll consider as a -module with and acting in the following way:

where is a -linear morphism such that its matrix is an Jordan block of eigenvalue in some basis. We obviously have , so this endows with a -module structure.

Let’s now describe the endomorphism algebra of this module. If , then where are -linear morphisms. For to be -linear, the following identity must hold:

From this, we deduce that and so . Doing a similar analysis for , we obtain that and so is a -linear morphism iff

where commutes with . An easy calculation shows that any such must be a lower triangular matrix constant in each diagonal in the Jordan basis for . If is invertible (that is, if its main diagonal in the Jordan basis for is not identically zero), then is (one has to check that the obvious -linear inverse is -linear too, but this is easy). If is not invertible, then it is a stricly lower triangular matrix in some basis, and so it is nilpotent, and therefore is too. We have then proved that every -endomorphism of is either invertible or nilpotent. This is a sufficient condition for to be local, and so is indecomposable.

Since we can produce an indecomposable module for every natural , this gives an infinite number of non-isomorphic indecomposable -modules.

Representations of p-groups in the modular case

In this post we’ll prove that if is a -group and , then the only simple -module is the trivial one.

Let be a simple -module and pick . We now define as the subgroup of the additive group of generated by the elements . Since we’re working in characteristic , a subgroup of the additive group of a vector space must be a -group of exponent . Notice that must be finite, since it’s a finitely generated abelian group of finite exponent.

We have an obvious action of over , and we can see that orbits of this action must be of cardinality (allowing to be zero), by the orbit-stabilizer theorem. Since the orbit of zero is obviously trivial, we have that there’s at least another element with trivial orbit. Thus is a -submodule of dimension 1, and so by simplicity. This proves our claim.

Notice that there are non-trivial -modules; for instance, take any non-zero element of and let it act over as a Jordan block of size of eigenvalue , and let all the elements not in act trivially. Then this is a non-trivial representation of . Since a direct sum of trivial representations is itself trivial, we have proved that no non-trivial module can be expressed as a direct sum of simple modules, and so the group algebra is not semisimple. This is a partial converse to Maschke’s theorem (we have only proved it for -groups, although it holds in full generality).

A bijective proof of a classical identity involving the totient function

In this post we’ll give a simple bijective proof of the well-known identity .

If we write down the expressions and then put each one of them in their irreducible form, we have that each divisor of appears exactly times as a denominator. Thus, we have fractions, comprised of fractions with denominator for each , and so , as we wanted.

Two sets are equinumerous iff their symmetric groups are isomorphic

Let and be sets; we’ll prove that iff , following Andreas Blass’ argument here.

Suppose and pick a bijection . Then is an isomorphism between and .

Reciprocally, suppose . In the finite case, we can recover the cardinalities of and easily, so we’ll restrict to the infinite case. We’ll prove that the smallest non-trivial conjugacy class of has cardinality , and since the cardinality of conjugacy classes is an isomorphism invariant, this will in turn prove that . This is easy to check: the set of all transpositions is a conjugacy class and has cardinality . Moreover, if is a non-trivial conjugacy class, we have that has cardinality at least , since we can produce that many elements conjugating a non-trivial element of with transpositions. This proves our claim.

The irreducible representations of the dihedral group

In this post we’ll characterize all of the irreducible representations of over an algebraically closed field . We will suppose so that Maschke’s theorem holds and has finite representation type.

First, let’s recall the following corollary of the Artin-Wedderburn theorem:

Corollary: Let be a finite group such that the group algebra is semisimple. Let be a complete system of representatives for the isomorphism classes of simple -modules and let and , so that there’s an isomorphism of algebras . Then:

(i) (the so-called sum-of-squares formula).
(ii) If , then .
(iii) If is algebraically closed and then .
(iv) If is algebraically closed, the number of simple modules of -dimension 1 is exactly the same as .
(v) (where stands for the set of conjugacy classes of ) and the equality holds if is algebraically closed.

Let’s begin with our characterization. Let be an irreducible -dimensional representation of , and recall that is a presentation for . The equation implies that the matrix representing (which we’ll simply call for the remainder of this post) is diagonalizable, since its minimal polynomial divides , which has simple roots because of our hypothesis on the characteristic of the field. Moreover, every eigenvalue of is an -th root of unity.

Suppose that ; since has an eigenvector , then is a non-trivial subrepresentation of and so must be by irreducibility. In this case, is a matrix such that , so . Both possibilities can occur, and in fact correspond to the trivial and sign representations, respectively.

Now, if , take an eigenvector of with eigenvalue . Then ; and so . One now can easily check that is a subrepresentation, and we see that has dimension at most . Therefore, in this case, the representation is given by

Recall that was an -th root of unity. If is even, then is an -th root of unity, and so picking makes a scalar matrix. Therefore any eigenvalue of gives rise to a -dimensional subrepresentation. This makes sense, since is or , and so we’ll have a different number of irreducible representations of dimension depending on the parity of .

In any other case, that is, , this representation is clearly irreducible since and share no eigenvalues.

It’s easy to see that if denotes the irreducible representation we just described, then . Let’s prove that if and , then . By (i) ; and since (a finite dimensional -algebra which is a domain must be a field; and therefore is a finite extension of , which is algebraically closed) we have that in the odd case and in the even case. An easy counting argument then shows that our representations are pairwise non-isomorphic, as we wanted.

This completely classifies the representations of , as every representation is a direct sum of the irreducible representations we just classified.

Moreover, we can give explicit submodules of isomorphic to each simple module. Given any character , the submodule is isomorphic to the simple module of dimension associated with the character . This construction works in general for any representation induced by a character. Now, for the -dimensional case, we have that the submodule is isomorphic to the simple submodule .

Dense subsets of a metric space sometimes determine its cardinality

Suppose is a metric space and is a dense subset of such that . Then, for every we can produce a sequence of elements in such that . This in turn defines a mapping , which is injective (notice that we are heavily using the axiom of choice here). Since , we have that and so actually determines the cardinality of , which obviously is not the case when is countable.

One thing to keep in mind is that the crucial step in this proof is the characterization of convergence via sequences, which holds for metric spaces, but not in more general contexts.