Vtyjkyuk

An $n times n$ matrix unit is any matrix which has zeros every, except at one position where it has one. By $E_ij^(n)$ we denote the $n times n$ matrix unit which has its one at the $i$-th row and $j$-th column.

Fix some field $F$, then surely the $n times n$ matrices $M_ntimes n(F)$ could be written in terms of the matrix units. And a more formal way to state that is that if $S$ denotes the semigroup of all matrix units with the zero matrix, then $M_ntimes n(F) cong F_0(S)$, where $F_0(S)$ denotes the contracted semigroup algebra, which is the usual semigroup algebra (i.e. formal sums) with the zero element identified, i.e. $F_0(S)$ has a basis $B$ such that $B cup 0$ is isomorphic with $S$.

This seems quite clear, but now suppose we look at
$$
A = M_n times n(F) oplus M_ntimes n(F)
$$
in the sense that the product of the direct factors is zero, i.e. think as an external direct sum. We can form the semigroup
$$
S = (E_ij^(n), 0), (0, E_ij^(n)) mid 0 le i,j le n cup (0,0)
$$
with the product. Then I think we also have $A cong F_0[S]$? But according to my text this should not be the case, so what am I missing. Why could $A$ be not isomorphic to a semigroup algebra?

Background: The text I am refering to is The algebraic theory of semigroups, Volume I by Clifford/Preston. In this book the following Theorem appears (page 167, Theorem 5.30):

Let $A$ be a semisimple algebra of finite order over a field $F$. If $A = F[S]$ for some (finite) semigroup $S$, then one of the simple components of $A$ is of order $1$ over $F$.

Which clearly contradicts the above situation, where both compoents of $A$ have order $n$ as matrix algebras. Also note that $F_0[S^0] cong F[S]$, where $S^0$ denotes that semigroup $S$ with a zero adjoined (i.e. taking a new element $0 notin S$ and set $0cdot x = x cdot 0 = 0$ for every $x in S$ and $S^0 = S cup0$). Hence every semigroup algebra could be regarded as a contracted semigroup algebra.