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Next: 6.3 Orders and Maximal Up: 6 Algebraic Number Fields Previous: 6.1 Algebraic Numbers

6.2 Algebraic Number Fields as Matrix Algebras

Let n be any positive integer and consider a sub-algebra K of the algebra of n×n matrices with rational entries; by this we mean that K contains scalar multiples of the identity matrix and is closed under matrix addition, subtraction and multiplication. To handle division we also insist that non-zero matrices in K are invertible (this actually implies that the inverses are also in K but it is not entirely trivial to prove this). Finally, one knows that the algebra of matrices is not commutative for n $ \geq$ 2. So we put in an additional hypothesis that matrix multiplication between elements of K is commutative.

Now, consider the map $ \alpha$ $ \mapsto$ $ \alpha$ . v where v is any (fixed) non-zero column vector such as the transpose of (1, 0..., 0). When $ \alpha$ and $ \beta$ are an elements of K with $ \alpha$ . v = $ \beta$ . v, we obtain ($ \alpha$ - $ \beta$) . v = 0. But we have assumed that every non-zero element of K is invertible so we must have $ \alpha$ - $ \beta$ = 0. In other words this map is one-to-one on K. Thus K is actually isomorphic to a vector space of rank at most n over the rationals. By a suitable change of basis (and restricting to a submatrix) we may as well assume that the space K . v contains all column vectors or equivalently that K has rank n. Then K . w is the space of all column vectors for any non-zero vector w. We will henceforth make this additional assumption as well.

For any n×n matrix $ \alpha$ we have (the Cayley-Hamilton theorem) that characteristic polynomial ch$\scriptstyle \alpha$(T) of degree n and ch$\scriptstyle \alpha$($ \alpha$) = 0. (In the words of one mathematician khudh kaa nahi satisfy karega to kiska satisfy karega?(Hindi); if it doesn't satisfy its' own then whose will it satisfy?). On the other hand, we have the minimal polynomial $ \min_{\alpha}^{}$(T), which is the polynomial of least degree with rational coefficients that is satisfied by $ \alpha$. If $ \min_{\alpha}^{}$(T) = P(T)Q(T), then P($ \alpha$)Q($ \alpha$) = 0. Since, P($ \alpha$) and Q($ \alpha$) are in K at least one of them must be zero thus one of them must be a constant; in other words the minimal polynomial is irreducible. It also follows as before that it divides the characteristic polynomial. One can show that, under the hypothesis of the previous paragraph (and the fact the we are working over rationals; a perfect field), there is an element $ \alpha$ in K whose characteristic polynomial is irreducible, i. e. its characteristic polynomial equals its minimal polynomial. In particular, the field K has a basis over the field $ \mathbb {Q}$ of rationals of the form 1, $ \alpha$, ..., $ \alpha^{n-1}_{}$.

Proof. [(Sketch of Proof)] Let $ \alpha_{1}^{}$, $ \alpha_{2}^{}$, ..., $ \alpha_{n}^{}$ be a basis of K over the field $ \mathbb {Q}$. Consider the characteristic polynomial of T1$ \alpha_{1}^{}$ + ... + Tn$ \alpha_{n}^{}$ as a function of the variables T1, ..., Tn. The condition that this is reducible will impose certain non-trivial polynomial relations between the Tk's. Thus all we need to do is to find rational numbers rk that do not satisfy these relations. Then the characteristic polynomial of $ \alpha$ = r1$ \alpha_{1}^{}$ + ... + rn$ \alpha_{n}^{}$ will be irreducible (and of degree n. It follows that, the elements 1, $ \alpha$, ..., $ \alpha^{n-1}_{}$ will be independent over $ \mathbb {Q}$. $ \qedsymbol$

To summarise, we will henceforth think of an algebraic number field as a sub-algebra of the ring of n×n matrices which is commutative, with all non-zero elements being invertible. Moreover, this algebra contains an element $ \alpha$ whose characteristic polynomial P(T) is equal to its minimal polynomial. An further extension of the above argument then shows that any invertible matrix g that commutes with every element of K is contained in K; we will use this in later sections.

As an example, let us consider the ``construction'' of the field associated with an irreducible polynomial P(T) = Tn + a1Tn - 1 + ... + an. We consider the matrix

$\displaystyle \alpha_{P}^{}$ = $\displaystyle \begin{pmatrix}
0 & 1 & \cdots & 0 \\
0 & 0 & \cdots & 0   ...
...& \vdots & \ddots & \vdots \\
-a_n & -a_{n-1} & \cdots & -a_1

This has minimal polynomial and characteristic polynomial equal to P(T). The sub-algebra of matrices generated by $ \alpha_{P}^{}$ is the required field $ \mathbb {Q}$($ \alpha_{p}^{}$), sometimes also denoted by $ \mathbb {Q}$[T]/(P(T)) (one uses Euclid's algorithm for polynomials to show that every non-zero element of this is invertible). The above discussion says that any field under consideration is isomorphic to a field of this form for some irreducible polynomial P(T).

next up previous
Next: 6.3 Orders and Maximal Up: 6 Algebraic Number Fields Previous: 6.1 Algebraic Numbers
Kapil Hari Paranjape 2002-10-20