# More on algebraic numbers

A complex number is algebraic if it is the root of some polynomial with rational coefficients. is algebraic (e.g. the polynomial ); is algebraic (e.g. the polynomial ); and are not. (A complex number that is not algebraic is called transcendental)

Previously, I wrote some blog posts (see here and here) which sketched a proof of the fact that the sum and product of algebraic numbers is also algebraic (and more). This is not an obvious fact, and to prove this requires some amount of field theory and linear algebra. Nevertheless, the ideas in the proof lead the way to a better understanding of the structure of the algebraic numbers and towards the theorems of Galois theory. In that post, I tried to introduce the minimum algebraic machinery necessary in order to state and prove the main result; I don’t think I entirely succeeded.

However, there is a more direct approach, one which also allows us find a polynomial that has (or ) as a root, for algebraic numbers and . That is the subject of this post. Instead of trying to formally prove the result, I will illustrate the approach for a specific example: showing is algebraic.

This post will assume familiarity with the characteristic polynomial of a matrix, and not much more. (In particular, none of the algebra from the previous posts)

## A case study

Define the set . We will think of this as a four-dimensional vector space, where the scalars are elements of , and the basis is . Every element can be uniquely expressed as , for .

We’re trying to prove is algebraic. Consider the linear transformation on defined as “multiply by “. In other words, consider the linear map which maps . This is definitely a linear map, since it satisfies and . In particular, we should be able to represent it by a matrix.

What is the matrix of ? Well, , , , and . Thus we can represent by the matrix

.

Now, the characteristic polynomial of this matrix, which is defined as , is , which has as a root. Thus is indeed algebraic.

## Why it works

The basic reason is the Cayley-Hamilton theorem. It tells us that should satisfy the characteristic polynomial: is the zero matrix. But the matrix we get when plugging into should correspond to multiplication by ; thus .

Note that I chose randomly. I could have chosen any element of and used this method to find a polynomial with rational coefficients having that element as a root.

At the end of the day, to prove that such a method always works requires the field theory we have glossed over: what is in general, why is it finite-dimensional, etc. This constructive method, which assumes the Cayley-Hamilton theorem, only replaces the non-constructive “linear dependence” argument in Proposition 4 of the original post.

# Two proofs complex matrices have eigenvalues

Today I will briefly discuss two proofs that every matrix over the complex numbers (or more generally, over an algebraically closed field) has an eigenvalue. Notice that this is equivalent to finding a complex number such that has nontrivial kernel. The first proof uses facts about “linear dependence” and the second uses determinants and the characteristic polynomial. The first proof is drawn from Axler’s textbook [1]; the second is the standard proof.

## Proof by linear dependence

Let be a polynomial with complex coefficients. If is a linear map, . We think of this as “ evaluated at ”.

Exercise: Show .

Proof: Pick a random vector . Consider the sequence of vectors This is a set of vectors, so they must be linearly dependent. Thus there exist constants such that .

Define . Then, we can factor By the Exercise, this implies . So, at least one of the maps has a nontrivial kernel, so has an eigenvalue.

## Proof by the characteristic polynomial

Proof: We want to show that there exists some such that has nontrivial kernel: in other words, that is singular. A matrix is singular if and only if its determinant is nonzero. So, let ; this is a polynomial in , called the characteristic polynomial of . Now, every polynomial has a complex root, say . This implies , so has an eigenvalue.

## Thoughts

To me, it seems like the determinant based proof is more straightforward, although it requires more machinery. Also, the determinant based proof is “constructive”, in that we can actually find all the eigenvalues by factoring the characteristic polynomial. On subject of determinant-based vs determinant-free approaches to linear algebra, see Axler’s article “Down With Determinants!” [3].

There is a similar situation for the problem of showing that the sum (or product) of two algebraic numbers is algebraic. Here there is a non-constructive proof using “linear dependence” (which I attempted to describe in a previous post) and a constructive proof using the characteristic polynomial (which will hopefully be the subject of a future blog post). A further advantage of the determinant-based proof is that it can be used more generally to show that the sum and product of integral elements over a ring are integral. In this more general context, we no longer have linear dependence available.

## References

1. Sheldon Axler, Linear algebra done right. Springer 2017
2. Evan Chen, An Infinitely Large Napkin, available online
3. Sheldon Axler. Down with Determinants! The American Mathematical Monthly, 102(2), 139, 1995. doi:10.2307/2975348, available online

# Hilbert’s Basis Theorem

Here is a proof of Hilbert’s Basis Theorem I thought of last night.

Let be a noetherian ring. Consider an ideal in . Let be the ideal in generated by the leading coefficients of the polynomials of degree in . Notice that , since if , , and it has the same leading coefficient. Thus we have an ascending chain , which must terminate, since is noetherian. Suppose it terminates at , so .

Now for each choose a finite set of generators (which we can do since is noetherian). For each generator, choose a representative polynomial in with that leading coefficient. This gives us a finite collection polynomials: define to be the ideal of generated by these polynomials.

Let . I claim . Assume for the sake of contradiction that there is a polynomial of minimal degree (say ) which is in but . If , there is an element of with the same leading coefficient, so is not in but has degree smaller than : contradiction. If is of degree , then there is an element of of which has the same leading coefficient as . Thus is of degree smaller than but is not in : contradiction.

Thus . Since is therefore finitely generated, this proves is noetherian.

# Algebro-geometric proof of Cayley-Hamilton

Here is a sketch of proof of the Cayley-Hamilton theorem via classical algebraic geometry.

The set of n x n matrices over an algebraically closed field can be identified with the affine space . Let be the subset of matrices that satisfy their own characteristic polynomial. We will prove that is in fact all of . Since affine space is irreducible, it suffices to show that is closed and contains a non-empty open set.

Fix a matrix . First, observe that the coefficients of the characteristic polynomial are polynomials in the entries in . In particular, the condition that a matrix satisfy its own characteristic polynomial amounts to a collection of polynomials in the entries of vanishing. This establishes that is closed.

Let be the set of matrices that have distinct eigenvalues. A matrix has distinct eigenvalues if and only if its characteristic polynomial has no double roots when it splits. This occurs if and only if the discriminant of the characteristic polynomial is nonzero. The discriminant is a polynomial in the coefficients of the characteristic polynomial. Thus the condition that a matrix have distinct eigenvalues amounts to a polynomial in the entries of not vanishing. Thus is open.

Finally, we have to show . It is easy to check this for a diagonal matrix. The general result follows from the fact that the determinant and thus the characteristic polynomial is basis-invariant.