So I began working this out to see how far I could get. It is rather easy to translate the whole problem into a question on some integers. Namely, assume that is an integer, and we are given integers with . Define the quantity as One can show that this is always a non-negative integer. Now I claim that is either 0, or .

Where does this strange expression comes from? Consider the function field over defined by (In fact, one can do this over any field whose characteristic is coprime to , and which has at least elements. Moreover, over an algebraically closed field , any function field extension which is cyclic of order , where is coprime to the characteristic of , can be written in this form, since this is a Kummer extension.) One can show that this equation is irreducibe if and only if , and that the genus of this function field is given by . This also explains why we must have that , since the genus is always a nonnegative integer.

Now we have a struggle: is it easier to show the claim using some Elementary Number Theory, or using some (advanced?) Algebraic Geometry (considering the geometric side) or Algebraic Number Theory (considering the function field side)? I don’t have any idea how to use the latter two to prove this, but I found an elementary proof.

First, note that if , or in case and , at least five of the terms must be since is not divisible by . Therefore, .

Moreover, if , we have , and therefore and . Hence, Therefore, it suffices to consider the case and .

For , note that implies . Hence, there is nothing to show.

For , let me consider three cases.

1. for some prime ;
2. for some prime ;
3. for two distinct prims .

In all three cases, one can show by analyzing several cases that the claim is true. Thus, we are only interested in the cases where no is of the form , , . Therefore, for all . Since not all ‘s cannot be the same – this would contradict – we must have

Let me demonstrate how to do . (In fact, this case suffices if one does not wants the constant , but one is happy with the constant , since .) Assume that for some prime . Since we must have , with in case . In both cases, we have .

In case , implies . Therefore, . In case , implies . Therefore, we also have This yields The latter inequality is true if and only if If we write , this translates to if , , and if , . Hence, the only cases were the above argument does not work are () and (, ).

In case , we must have and . In that case, , whence In case , since we do by assumption , we obtain , whence since .

The cases and are proven analogously, with a few more case distinctions.

So we are left with the case . Here, one can proceed in a similar, painful way. Or one increases the constant to , since we know that not all ‘s can be , whence one is at least . Hence, which yields the claim.

To sum everything up, we showed the following theorem:

Theorem.

Let be an integer, , and satisfy . Then , defined as satisfies , and if it is strictly larger than zero,

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As mentioned, if one invests more work, one can actually show . For my preprint though, this is not worth the trouble.

Unfortunately, the answer turns out to be “no” for the case (as well as ):

Example.

Let . Consider the matrix as well as the sequence Clearly, . Assume that is open in ; then we must have for almost all . But is in Jordan canonical form, and clearly not diagonalizable; but this means that for all . Therefore, is not open in .

But nonetheless, contains a Zariski-open subset of in case is algebraically closed (which implies that lies dense in ). For that recall that is the characteristic polynomial of .

Proposition.

Consider the set Then and is Zariski-open in . In fact, is the complement of a hypersurface in .

Proof.

Note that in case is squarefree, splits into distinct linear factors since is algebraically closed. Hence, has distinct eigenvalues in and therefore one obtains linearly independent eigenvectors of ; i.e., is diagonalizable over . Therefore, .
Now we show that is a hypersurface in , i.e. there exists a polynomial such that . For that, consider the maps defined by for all . Obviously, these must be polynomials. Next, consider the discriminant of ; this is a polynomial expression in the coefficients of , i.e. in , whose value is zero if, and only if, is squarefree. Therefore, . Finally, is a polynomial, whence is Zariski-open in .

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Note that the situation is different over :

Proposition.

In the standard topology,

Proof.

Assume that has at least one eigenvalue with imaginary part . If is a sequence of matrices with , each must have an eigenvalue with . But then, for infinitely many , we must have (since is open), whence we cannot have for infinitely many . Hence, is not in the closure of .
Now assume that has only real eigenvalues. Then there exist some with in Jordan canonical form. By pertubing the diagonal elements of slightly, we can obtain a sequence of matrices with . But then, and for every .
Note that this implies ; moreover, this also implies . Hence, the first two equalities hold. The third equality is standard.

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Also note that for , as the example (which is diagonalizable over , with eigenvalues ) shows. So what about ? In fact, as in the case of , it turns out that is .

Proposition.

We have

For the proof, we need a little lemma.

Lemma.

Let . Then .

Proof.

Let be an arbitrary polynomial. Write , where and is irreducible and monic, . Now the coefficients of all ‘s (except the highest coefficients) are a finite set in of elements, whence there exists sequences with pairwise distinct such that converges to one coefficent of one . In particular, we can construct monic polynomials with , and for every . Even more, we can make sure that every is irreducible; this enforces that is squarefree, i.e. . Therefore, we found a sequence in converging to , whence .

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Proof (Proof of the Proposition).

Let whose characteristic polynomial can be written as , with not necessarily distinct, but monic and irreducible polynomials . There exists a matrix with where is the companion matrix of ; this is a Frobenius normal form of . Now we can find a sequence of squarefree polynomials such that for every , are pairwise coprime, and that . Then set clearly, . Moreover, the characteristic polynomial of is given by , i.e. it is squarefree by choice of the . Therefore, and .

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