UDC 519.48

MATHEMATICS

Academician of the Academy of Sciences of the MSSR V. A. ANDRUNAKIEVICH, Yu. M. RYABUKHIN

DIRECT SUMS OF DIVISION ALGEBRAS

The classical theorem of Weierstrass—Dedekind is well known: Any nonzero finite-dimensional associative-commutative algebra without nilpotent elements over the field \(K\) of complex numbers is a direct sum of fields isomorphic to \(K\).

In the present note it is proved that in this theorem one may replace the requirement of associativity-commutativity by the considerably weaker condition of associativity at zero, i.e.

\[ x(yz)=0 \Longleftrightarrow (xy)z=0 \]

for arbitrary elements \(x,y,z\) of the algebra under consideration. Moreover, instead of the field of complex numbers one may take any other algebraically closed field \(\Phi\).

Everywhere below, by an algebra we shall always mean a not necessarily associative finite-dimensional algebra \(R\ne 0\) over an arbitrary algebraically closed field \(\Phi\).

Lemma 1. Let \(A\) and \(B\) be \(n\times n\) matrices over the field \(\Phi\), and suppose that the matrix \(A\) is nonsingular. There exists an element \(\lambda\in\Phi\) such that the matrix \(\lambda A+B\) is singular.

Indeed, consider the polynomial*

\[ f(x)=|xA+B|= \begin{vmatrix} a_{11}x+b_{11} & \ldots & a_{1n}x+b_{1n}\\ \cdot & \cdot & \cdot\\ a_{n1}x+b_{1n} & \ldots & a_{nn}x+b_{nn} \end{vmatrix}. \]

It is clear that

\[ f(x)=|A|x^n+\ldots+|B|. \]

But the field \(\Phi\) is algebraically closed. Therefore, from \(|A|\ne 0\) it follows that there exists \(\lambda\in\Phi\) such that \(f(\lambda)=0\). This element \(\lambda\in\Phi\) will be the required one.

Lemma 2. Let the algebra \(R\) have no zero divisors. Then for any nonzero \(a\in R\) the right multiplication \(\rho(a)\), \(\forall x\in R,\ x\rho(a)=xa\), is a nonsingular linear transformation of the linear space \(R\) over \(\Phi\).

Let \(\mathfrak{B}_R=\{e_i\mid 1\le i\le n\}\) be a basis of the space \(R\). Since for arbitrary \(\alpha_i\in\Phi\) the equality

\[ (\Sigma \alpha_i e_i)\rho(a)=\Sigma \alpha_i e_i a=(\Sigma \alpha_i e_i)a \]

holds, and \(R\) is an algebra without zero divisors, it follows that \(\{e_i a\mid 1\le i\le n\}\) is also a basis of the space \(R\). Hence the linear transformation \(\rho(a)\) is invertible, and therefore nonsingular.

Lemma 3. The algebra \(R\) is a division algebra if and only if \(\dim R=1\) and \(R\approx\Phi\).

Recall that an algebra \(R\ne 0\) is called a division algebra if and only if, for any nonzero \(a,b\in R\), the equations \(xa=b\) and \(ay=b\) are solvable in \(R\), and uniquely so. It is well known (see \((^1)\), p. 277) that division algebras are precisely algebras without zero divisors (in the finite-dimensional case!).

* By \(|A|\) is denoted the determinant of the matrix \(A\). The matrix \(A\) is nonsingular if \(|A|\ne 0\).

Let \(R\) be a division algebra, \(\mathfrak{B}_R=\{e_i\mid 1\le i\le n\}\) a basis of the space \(R\). Suppose that \(\dim R=n\ge 2\). Consider the right multiplications \(\rho(e_i)\) and the corresponding matrices \(A_i\).

By Lemma 2 all matrices \(A_i\) are nonsingular, since they correspond to the nonsingular linear transformations \(\rho(e_i)\). Moreover, if among the elements \(\alpha_i\in\Phi\) there are some different from zero, then the matrix \(\sum \alpha_i A_i\) is nonsingular as well, since it corresponds to right multiplication by the element

\[ a=\sum_{i=1}^{n}\alpha_i e_i\ne 0. \]

Using Lemma 1, we find an \(\alpha_1=\lambda\in\Phi\) such that the matrix \(\alpha_1A_1+A_2\) is singular. Taking \(\alpha_2=1\) and \(\alpha_i=0\) for all remaining \(i\), we obtain, by what was said above, that the matrix \(\sum_{i=1}^{n}\alpha_i A_i=\alpha_1A_1+A_2\) is nonsingular—a contradiction.

Consequently, \(\dim R=n=1\), i.e. \(R=\Phi e_1\). But \(R\) is an algebra without zero divisors. Therefore \(e_1^2=\alpha e_1\) for some nonzero \(\alpha\in\Phi\). Denoting \(f=\alpha^{-1}e_1\), we obtain \(f^2=f\), and therefore the map \(\gamma f\to\gamma\) is an isomorphism of the algebra \(R\) onto the field \(\Phi\). The lemma is proved.

Lemma 4. Let the following identities hold in the algebra \(R\):

\[ x^2=0\Longleftrightarrow x=0, \tag{*} \]

\[ x(yz)=0\Longleftrightarrow (xy)z=0. \tag{**} \]

Then, for any element \(a\in R\), the sets

\[ \{x\in R\mid xa=0\},\qquad \{x\in R\mid ax=0\}, \]

\[ (0:a)_R=\{x\in R\mid xa=ax=0\} \]

coincide. Moreover, the annihilator \((0:a)_R\) of the element \(a\) in the algebra \(R\) is an ideal.

Indeed, for any \(x,y\in R\), \(\alpha,\beta\in\Phi\), we obtain

\[ xa=0\Rightarrow a(xa)=0\Rightarrow (ax)a=0\Rightarrow \]

\[ \Rightarrow ((ax)a)x=0\Rightarrow (ax)^2=0\Rightarrow ax=0\Rightarrow xa=0, \]

\[ xa=0\Rightarrow y(xa)=0\Rightarrow (yx)=0, \]

\[ xa=0\Rightarrow ax=0\Rightarrow (ax)y=0\Rightarrow \]

\[ \Rightarrow a(xy)=0\Rightarrow (xy)a=0, \]

\[ \begin{matrix} xa=0\\ ya=0 \end{matrix} \quad\Rightarrow\quad (\alpha x+\beta y)a=\alpha xa+\beta ya=0. \]

Therefore the indicated sets coincide, and the annihilator \((0:a)_R\) is an ideal of the algebra \(R\).

Lemma 5. Let the identities \((*)\) and \((**)\) hold in the algebra \(R\). If \(M\) is a minimal ideal of the algebra \(R\), then the algebra \(M\) has no zero divisors.

Denote by \((x)_R\) the principal ideal of the algebra \(R\) generated by the element \(x\in R\). Let \(M\) be a minimal ideal of the algebra \(R\). Suppose that \(M\) is not an algebra without zero divisors. Then there exist nonzero elements \(a,b\in M\) such that \(ab=0\).

It is clear that \((0:a)_R\cap M\ne 0\) and \((0:b)_R\cap M\ne 0\). By the minimality of \(M\), taking Lemma 4 into account, we obtain

\[ (0:a)_R\cap M=(0:b)_R\cap M=M(a)_R=(b)_R. \]

But then \(a^2=b^2=0\), and therefore \(a=b=0\) by \((*)\). We have obtained a contradiction. Consequently, the algebra \(M\) has no zero divisors. The lemma is proved.

Theorem. Let \(R\) be a nonzero finite-dimensional algebra over an algebraically closed field \(\Phi\). If in the algebra \(R\) the identities ...

properties

\[ x^2=0 \Longleftrightarrow x=0,\qquad x(yz)=0 \Longleftrightarrow (xy)z=0, \]

then \(R\) decomposes into a direct sum of fields isomorphic to \(\Phi\),

\[ R=\sum_{i=1}^{n}\oplus R_i,\qquad R_i\approx \Phi,\qquad n=\dim R. \]

In particular, the algebra \(R\) is associative-commutative.

Proof. Choose in \(R\) some minimal ideal \(M_1\). By Lemmas 3, 5, the algebra \(M_1\) is isomorphic to the field \(\Phi\), and \(\dim M_1=1\). In particular, \(M_1\) is a field and \(M_1=\Phi e_1\), where \(e_1\) is the identity element of the field \(M_1\). We note that for any \(x\in R\),

\[ M_1=Re_1=\Phi e_1, \]

\[ (x-xe_1)e_1=xe_1-(xe_1)e_1=xe_1-xe_1=0, \]

since \(M_1\) is an ideal in \(R\) and \(e_1\) is the identity of the field \(M_1\). This means that \(x-xe_1\in(0:e_1)_R\). But \(x=(x-xe_1)+xe_1\), and therefore the equality

\[ R=M_1\oplus(0:e_1)_R \]

holds, since, obviously, \(\Phi e_1\cap(0:e_1)_R=0\).

If \((0:e_1)_R=0\), then \(R=M_1\approx\Phi\), and there is nothing to prove. If, however, \((0:e_1)_R\ne0\), then choose some minimal ideal \(M_2\subseteq(0:e_1)_R\). As above, \(M_2\) is a field and \(M_2=\Phi e_2\), where \(e_2\) is the identity of the field \(M_2\). But then, repeating almost word for word for \(M_2\) the arguments already given for \(M_1\), we obtain

\[ R=M_1\oplus M_2\oplus(0:(e_1+e_2))_R,\qquad M_1\approx\Phi\approx M_2. \]

Continuing the process, at the \(n\)-th step, where \(n=\dim R\), we obtain

\[ R=M_1\oplus M_2\oplus\ldots\oplus M_n,\qquad M_i\approx\Phi. \]

This completes the proof of the theorem.

It is clear that the theorem of Weierstrass—Dedekind follows from the theorem just proved, since the field \(K\) of complex numbers is algebraically closed.

Let us give an example showing that condition \((**)\) is essential. Let \(R\) be a two-dimensional algebra over \(\Phi\) with basis \(\mathfrak{B}_R=\{e_1,e_2\}\) and defining relations

\[ e_1^2=e_1,\qquad e_2^2=e_1,\qquad e_1e_2=0,\qquad e_2e_1=e_2. \]

It is easy to see that \(R\) is a simple algebra. Moreover, if \(a=\alpha e_1+\beta e_2\), then \(a^2=(\alpha^2+\beta^2)e_1+\beta\alpha e_2\), and therefore

\[ a^2=0\Rightarrow \alpha^2+\beta^2=\beta\alpha=0\Rightarrow \alpha=\beta=0\Rightarrow a=0, \]

i.e., in \(R\) the conditional identity \((*)\) holds. The conditional identity \((**)\) does not hold, since

\[ (e_2e_2)e_2=e_1e_2=0\ne e_2=e_2e_1=e_2(e_2e_2). \]

The example of an algebra with zero multiplication shows that condition \((*)\) is also essential.

Finally, the example of the field of complex numbers, considered as an algebra over the field of real numbers, shows that algebraic closedness of the field \(\Phi\) is necessary for the validity of Lemma 3 and of the theorem.

Institute of Mathematics with Computing Center
Academy of Sciences of the MSSR

Received
13 VIII 1969

REFERENCES

1. A. G. Kurosh, Lectures on General Algebra, Moscow, 1962.