Abstract Nonsense

Crushing one theorem at a time

Some Facts About The Ring of Algebraic Integers

Point of post: In this post we’ll discuss some very basic facts concerning the ring of algebraic integers that shall become useful in other posts.


We’ve previously discussed the set of all algebraic numbers at least to the extent to show that the set of all of them is countable. In this post we shall restrict our attention to the algebraic integers, which are basically the result of considering algebraic numbers which are roots of monic polynomials. Our goal in this post is to give some alternate characterization of the algebraic integers in terms of integral matrices, show they are a subring of \mathbb{R}, and show that an eigenvalue of a matrix with algebraic integer entries must be itself an algebraic integer.

Definitions and Basics

Let \mathbb{Z}[x] denote, as usual, the polynomial ring over \mathbb{Z}. We call a real number a an algebraic integer if there exists some monic p\in\mathbb{Z}[x] such that p(a)=0. We denote the set of all algebraic integers by \mathbb{A}.

Next, as usual, let \text{Mat}_n\left(\mathbb{Z}\right) denote the set of all n\times n matrices with integer entries. Define then \displaystyle \text{Mat}\left(\mathbb{Z}\right) to be equal to \displaystyle \bigcup_{n\in\mathbb{N}}\text{Mat}_n\left(\mathbb{Z}\right). We call an element of \text{Mat}\left(\mathbb{Z}\right) an integral matrix.

Our first theorem is an interesting one. Indeed, for a matrix A\in\text{Mat}\left(\mathbb{Z}\right) let \sigma\left(A\right) denote it’s spectrum then:

Theorem: \displaystyle \mathbb{A}=\bigcup_{A\in\text{Mat}\left(\mathbb{Z}\right)}\sigma\left(A\right)

Proof: We merely note that if \lambda\in\sigma(A) for some A\in\text{Mat}\left(\mathbb{Z}\right) then p_A(x)\in\mathbb{Z}[x] is monic (the characteristic polynomial of A) and p_A(\lambda)=0. Thus, \lambda\in\mathbb{A}.

Conversely, let \lambda\in\mathbb{A}, then by definition there exists some monic p\in\mathbb{Z}[x] such that p(\lambda)=0. Let M_p denote the companion matrix of p. Then, by common knowledge we have that p_{M_p}(x)=p(x) and so in particular p_{M_p}(\lambda)=p(\lambda)=0 so that \lambda\in\sigma\left(M_p\right). Since M_p\in\text{Mat}\left(\mathbb{Z}\right) the conclusion follows. \blacksquare

With this it’s not difficult to prove that \mathbb{A} is a subring of \mathbb{R}. Indeed

Theorem: \mathbb{A} is a subring of \mathbb{R}.

Proof: Let a,b\in\mathbb{A}. Then, by our previous theorem there exists A,B\in\text{Mat}\left(\mathbb{Z}\right) where A,B are of size m,n respectively such that a\in\sigma\left(A\right) and b\in\sigma\left(B\right). But, this then implies (by basic matrix analysis) that -a\in \sigma\left(-A\right), a+b\in\sigma\left(I_m\otimes A+B\otimes I_n\right), and ab\in\sigma\left(A\otimes B\right) where \otimes denotes the Kronecker product. The conclusion then follows from the previous theorem. \blacksquare

Our final goal in this very short introduction to the ring of algebraic integers is to show that eigenvalues of a matrix whose entires are algebraic integers is itself an algebraic integer. But first, a lemma

Lemma: Let a_1,\cdots,a_k\in\mathbb{A}. Then, there exists some m\in\mathbb{N}, some v\in\mathbb{C}^m, and A_1,cdots,A_k\in\text{Mat}_m\left(\mathbb{Z}\right) such that A_j v=a_j v for every j=1,\cdots,k.

Proof: Since a_1,\cdots,a_k\in\mathbb{A} there exists matrices B_j\in\text{Mat}_{n_j}\left(\mathbb{Z}\right) and v_j\in\mathbb{C}^{n_j} for which B_j v_j=a_j v_j. Let then v=v_1\otimes\cdots\otimes v_k where in general  (x_1,\cdots,x_\ell)^{\top}\otimes (y_1,\cdots,y_r)^{\top}=(x_1y_1,x_1y_2,\cdots,x_1y_r,\cdots,x_ny_1,\cdots,x_ny_r)^{\top} (and the case for more than two elements in the tensor product is defined inductively). Then, if A_j=I_{n_1}\otimes\cdots\otimes B_j\otimes\cdots\otimes I_{n_k}  where \otimes here denotes the usual Kronecker product then we note that A_j v=a_j v where we’ve used the common fact that with these definitions

(M_1\otimes\cdots\otimes M_s)(w_1\otimes\cdots\otimes w_s)=(M_1w_1)\otimes\cdots\otimes (M_s w_s)

The conclusion follows. \blacksquare

We now prove the pivotal part of this post. Namely that if M is a matrix with entries in \mathbb{A} then \displaystyle \sigma\left(A\right)\subseteq\mathbb{A}. More formally:

Theorem: Let A\in\text{Mat}_n\left(\mathbb{A}\right) for any n\in\mathbb{N}. Then, \sigma\left(A\right)\subseteq\mathbb{A}.

Proof: Let A=[a_{i,j}] where by assumption a_{i,j}\in\mathbb{A} for every i,j\in[n]. Now, by the previous lemma we may find some m\in\mathbb{N}, a v\in\mathbb{C}^m, and matrices Q_{i,j}\;\; i,j\in[n] such that Q_{i,j}v=a_{i,j}v. Let E_{i,j} be the matrix with a 1 in the (i,j)^{\text{th}} entry and 0 elsewhere. Then, let Q denote the matrix

\displaystyle Q=\sum_{i,j=1}^{n}\left(E_{i,j}\otimes Q_{i,j}\right)


Then, for every \lambda\in\sigma\left(A\right) let w be the associated eigenvector. Then,

\displaystyle \begin{aligned}Q\left(w\otimes v\right) &= \left(\sum_{i,j=1}^{n}\left(E_{i,j}\otimes Q_{i,j}\right)\right)(w\otimes v)\\ &=\sum_{i,j=1}^{n}\left(E_{i,j}(w)\right)\otimes\left(Q_{i,j}(v)\right)\\ &= \sum_{i,j=1}^{n}(0,\cdots,\underbrace{w_j}_{i},\cdots,0)^{\top}\otimes \left(b_{i,j}v\right)\\ &=\sum_{i,j=1}^{n}(0,\cdots,0,\underbrace{w_jb_{i,j}v_1,\cdots,w_jb_{i,j}v_n}_{i^{\text{th}}\text{ block}},0,\cdots,0)^{\top}\\ &= \sum_{j=1}^{n}\left(w_j b_{1,j}v_1,\cdots,w_j b_{1,j}v_m,\cdots,w_ jb_{n,j}v_1,\cdots,w_j b_{n,j}v_m\right)\\ &= \left(v_1\sum_{j=1}^{n}w_1b_{1,j},\cdots,v_m\sum_{j=1}^{n}w_j b_{1,j},\cdots,v_1\sum_{j=1}^{n}w_n b_{n,j},\cdots,v_m\sum_{j=1}^{n}b_{n,j}w_j\right)\\ &= \left(\lambda w_1 v_1,\cdots,\lambda w_1 v_m,\cdots,\lambda w_n v_1,\cdots,\lambda w_n v_m\right)\\ &=\lambda\left(w\otimes v\right)\end{aligned}


and since Q\in\text{Mat}\left(\mathbb{Z}\right) we have that \lambda\in\mathbb{A}. Since \lambda\in\sigma\left(A\right) was arbitrary the conclusion follows. \blacksquare


One last thing to notice is that \mathbb{Q}\cap\mathbb{A}=\mathbb{Z}. More precisely:
Theorem: Let a be rational algebraic integer. Then, a is an integer.

Proof: Suppose that \displaystyle a=\frac{p}{q}\;\; (p,q)=1 and let p(x)=x^n+\cdots+a_1x+a_0 is the polynomial such that p(a)=0. Then, a quick check shows that this implies p^n=q\left(a_{n-1}p^{n-1}q+\cdots+a_0p^n\right) but since (p,q)=1 and q\mid p^n we may conclude that q=1. \blacksquare



1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.


March 3, 2011 - Posted by | Algebra, Linear Algebra, Uncategorized | , , ,


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