**1 Definitions and a positive result **

For a brief history of the problem of how to identify a complete matrix ring,

and of how to present a matrix ring by generators and relations, we refer to

the introduction of [1]. In that paper it was established that a k- algebra R

is an (m+n) ×(m+n) matrix algebra if and only if there are three elements

a, b, c ∈ R satisfying the relations
and . There it

was also shown that the k-algebra presented by the two generators a and b

and (1) be low is nonzero if i/m = j/n. This paper deals further with these

two-element presentations (1), whose theory is by no means as complete as

that of the three-element presentation studied in [1].

Unless otherwise stated k is an arbitrary associative commutative ring with

1. All algebras will here be as sumed associative and with 1.

Let us make

**Definition 1.1** Let i, j, m and n be positive integers. Let R(k; i, j, m,
n)

denote the k-algebra presented by two generators a and b and the relations

We will show that the question whether R(k; i, j, m, n) is nonzero, and

whether it admits a homomorphism to a matrix algebra over k, will only

depend on the ordered pair of positive rational numbers (p, q) = (i/j, m/n).

For a field k we will show that R(k; i, j, m, n) can be mapped into

for some N and hence is non zero if p is arbitrary and q = 1. On the other

hand in the next section we will show that for all k, R(k; i, j, m, n) is
trivial

if q ≠ 1 and p = 1.

We will begin with the following lemma which will be useful in proving the

next theorem.

**Lemma 1.2** Let R be a ring, and d a positive integer, and regard R as

embedded in by the diagonal embedding. Then
every element of R

has a d-th root in .

Proof: A d-th root of x is given by

We can now show that the behavior of R(k; i, j, m, n) only depends on the

ordered pair (i/j, m/n).

**Theorem 1.3** For positive integers c and d we have
the following:

1. R(k; i, j, m, n) is nonzero if and only if R(k; ci, cj, dm, dn) is nonzero.

2. R(k; i, j, m, n) can be mapped into for
some positive integer N,

if and only if R(k; ci, cj, dm, dn) can be mapped into
for some

N'.

Proof: Let c and d be positive integers. It is clear that if R(k; ci, cj, dm, dn)

is nonzero then so is R(k; i, j, m, n), also if R(k; ci, cj, dm, dn) can be
mapped

into for some N then so can R(k; i, j, m, n).

Assume R(k; i, j, m, n) to be nonzero. If we let e be a common multiple

of c and d, then is a nonzero k-algebra in which

the relations defining R(k; ei, ej, em, en) have a solution by Lemma 1.2.

R(k; ei, ej, em, en) is therefore nonzero and hence also R(k; ci, cj, dm, dn).

Let us finally assume that R(k, i, j, m, n) can be mapped into
for

some N. Equivalently there are matrices a and b in
satisfying (1).

Let again e be a common multiple of c and d . Then by Lemma 1.2 we

can find e-th roots of a and b in . Hence R(k; ei, ej, em, en) can be

mapped into , and therefore so can R(k; ci,
cj, dm, dn).

**Definition 1.4** Let be the set of
ordered pairs of positive rational numbers

(i/j, m/n) such that R(k; i, j, m, n) is nonzero, and let
be the

set of (i/j, m/n) such that R(k; i, j, m, n) can be mapped into
for

some positive integer N.

**Theorem 1.3** shows that the sets and
are indeed well
defined.

There are some basic relations worth noticing. We see by left-right symmetry

that for X ∈ { B, C} we have

If there exists a homomorphism of commutative rings k → k'
then

If k and k' contain fields of the same characteristic,
then .

We will prove the next theorem in a few steps .

**Theorem 1.5** If k is a field and p a positive rational number then (p, 1) ∈

Let us first notice a few things that will ease the task. When seeking a map

R(k; i, j, 1, 1) →
for some N it suffices by Theorem 1.3 to consider

the case gcd(i, j) = 1.

**Lemma 1.6** If is the algebraic closure
of the field k, then .

Proof: Clearly if k is a sub field of k'. If k'/k is a
finite field

extension then . Since
where the union is taken over

all finite field extensions k'/k, the lemma follows.

Theorem 1.5 will clearly follow once we have proved the following lemma:

**
**

Lemma 1.7 If k is an algebraically closed field and i, j are relatively prime

positive integers then one can always find a nonzero k-algebra homomorphism

R(k; i, j, 1, 1) → .

Proof: We will consider the following three cases:

Case 1. i + j is even (hence i, j are odd): Here we clearly have a surjection

R(k; i, j, 1, 1) → given by
.

Case 2. char(k) : By left-right symmetry of R(k; i, j, 1, 1) we may

assume j > i to hold. Consider the polynomials
.

Since char(k) does not divide i + j we have that
is separable and

has therefore a root which is not a root of .
Hence in this case there

is an r ∈ k such that and
.

Now for this r ∈ k the k-algebra homomorphism R(k; i, j, 1, 1) →

given by

is clearly well defined.

Case 3. char(k) l i+j and i+j is odd: Let char(k) = p. Since gcd(i, j) = 1,

we have gcd(i, p) = 1 and thus, i is invertible in the prime field of k with an

inverse c. Consider the k-algebra homomorphism R(k; i, j, 1, 1) →

defined as follows:

We easily see that:

and computation using the fact that i+j is odd shows that
we get

1 and . Hence the above homomorphism is well
defined.