Page 476 - Advanced Linear Algebra

460 Advanced Linear Algebra

Theorem 18.5 If is a finite-dimensional -algebra, then every element of (
(
-
is algebraic. There are infinite-dimensional algebras in which all elements are
algebraic.
Proof. The first statement has been proved. To prove the second, let us consider
the complex field as a -algebra. The set of algebraic elements of is a
d
r
d
(
field, known as the field of algebraic numbers . These are the complex numbers
(
that are roots of some nonzero polynomial with rational or integral)
coefficients.
To see this, if  ( , then the subalgebra ´ µ is finite-dimensional. Also, ´ µ
r
r
is a field. To prove this, first note that since is a field, the minimal polynomial
d
of any nonzero  ( is irreducible, for if ²%³ ~ ²%³ ²%³ , then

~ ² ³ ² ³ and so one of ² ³ and ² ³ is , which implies that
²%³ ~ ²%³ or ²%³ ~ ²%³. Since ²%³ is irreducible, it has nonzero

constant term and so the inverse of is a polynomial in , that is, c  r ´ . µ

Of course, r is closed under addition and multiplication and so r´ µ ´ µ is a
subfield of .
d
d
Thus, is an algebra over r ´ µ . By similar reasoning, if  ( , then the

minimal polynomial of over ´ µ is irreducible and so c  r ´ µ´ µ . Since
r
r r´ µ´ µ ~ ´ Á µ is the set of all polynomials in the “variables” and , it is
closed under addition and multiplication as well. Hence, r´ Á µ is a finite-
dimensional algebra over r , as well as a subfield of . Now,
d´ µ
dim r r dim r ²´ Á µ³ hr dim r ²´ Á µ³ ~ ²´ µ³
r ´ µ
and so r is finite-dimensional over . Hence, the elements of ´ Á µ are
r
r´ Á µ
c
r
r
r
algebraic over , that is, ´ Á µ ‹ ( . But ´ Á µ contains Á b Á c and
and so ( is a field.
We claim that is not finite-dimensional over . This follows from the fact
(
r
by
that for every prime , the polynomial % c is irreducible over r (

)

Eisenstein's criterion . Hence, if is a complex root of % c , then has

minimal polynomial %c over and so the dimension of ´ µ over is .

r
r
r
(
Hence, cannot be finite-dimensional.
The Spectrum of an Element
Let be an algebra. A nonzero element  ( is a left zero divisor if ~
(
for some £ and a right zero divisor if ~ for some £ . In the
exercises, we ask the reader to show that an algebraic element is a left zero
divisor if and only if it is a right zero divisor.
(
Theorem 18.6 Let be a algebra. An algebraic element  ( is invertible if
and only if it is not a zero divisor.

Proof. If is invertible and ~ , then multiplying by c gives ~ .

Conversely, suppose that is not invertible but ~ implies ~ . Then