Is there a name for a regular semigroup with zero in which the product of any two different idempotents is zero? As the title says, my question is: is there a name for a regular semigroup with zero in which the product of any two different idempotents is zero? Note that any such semigroup is necessarily an orthodox semigroup because zero is an idempotent. But there are orthodox semigroups not of this form.
The same question can be asked by dropping "regular", in which case the semigroup is only known to be an E-semigroup. But again in an E-semigroup, the product of two different idempotents is not necessarily zero.
What motivated me to ask this question is that in the Brand semigroup with five elements, (typically denoted by $B_2$), and which has the presentation $\langle a, b\; |\; a^2 = b^2 = 0, aba = a, bab = b \rangle$, you have $(ab)(ba) = (ba)(ab) = 0$. $B_2$ is an inverse semigroup, actually a completely 0-simple inverse semigroup, a class that's well studied. 
EDIT: Embarrassing moment! The example below is not actually motivating this question, because there is (at least one) product of different idempotents in it, namely $(pd)(qd) = pd$ which is not actually zero, neither in the example as I gave it here, nor in my actual application. Two people already upvoted this question (one before I wrote the example), so I'm not sure if should delete the question altogether, but it's possible the class I'm asking about (in the title and in the text above this paragraph) is rather trivially consisting only of inverse semigroups (with zero) or something like that.
I ran however (in an actual Computer Science application, but which is too long to detail here) into the following semigroup that has the same property with respect to its idempotents (i.e. the product of any two different ones is zero) but which is definitely not an inverse semigroup but only orthodox. The basic difference between it and $B_2$ is that instead of two generators there are three, $p$, $q$, and $d$, such that $V(d) = \{p, q\} $, $V(p) = \{d\}$, and $V(q) = \{d\}$ ($V(x)$ denotes the inverses of $x$ per the usual notation), the presentation of the semigroup being (hopefully not missing anything): $$\langle d, p, q\; |\; d^2=p^2=q^2=pq=qp=0, pdp=p, qdq=q, dpd=dqd=d \rangle$$
So the semigroups I ask a name for do seem to exist in a somewhat non-trivial way. If you "smash together" $p$ and $q$, which technically means taking the quotient by the minimum inverse semigroup congruence, you get $B_2$ from this orthodox semigroup, but that's not really exciting.
 A: For what it's worth here is a characterisation of the finite semigroups that are regular with zero in which the product of any two different idempotents equals zero. 
If $S$ and $T$ are semigroups with zero, then the 0-union of $S$ and $T$ is the semigroup formed by taking the union of $S$ and $T$ but identifying their zero elements, and with multiplication extending that of $S$ and $T$ by $st=ts=0$ for all $s\in S$ and $t\in T$. 
A finite Brandt semigroup $S$ is a finite 0-simple inverse semigroup (a semigroup where there are no proper non-zero ideals, and where every element $x\in S$ has a unique $x^{-1}\in S$ such that $xx^{-1}x=x$ and $x^{-1}xx^{-1}=x^{-1}$). 
A finite regular semigroup $S$ with zero has the property that the product of any two distinct idempotents equals zero if and only if it is a 0-union of Brandt semigroups. 
For the converse, let $B_i$, $i\in \{1,\ldots, n\}$ for some $n\in \mathbb{N}$ be Brandt semigroups, and let $S$ denote the 0-union of the $B_i$. Then $S$ is an inverse semigroup, and hence regular. The $\mathscr{J}$-classes of $S$ are simply $\{0\}$, and $B_i\setminus\{0\}$ for all $i$. Suppose that $e\in B_i$ and $f\in B_j$ are idempotents, and $e\not=f$. If $i\not=j$, then $ef=0$ by definition of the multiplication in the 0-union. If $i=j$, then $ef$ is the product of distinct idempotents in a $\mathscr{J}$-class of a finite inverse semigroup, and so  $ef\not\in B_i$. Thus $ef=0$, as required. 
Let $S$ be a finite regular semigroup with the property that the product of any two distinct idempotents equals $0$.
We start by showing that $S$ is an inverse semigroup. It suffices to show that every $\mathscr{L}$-  and $\mathscr{R}$-class of $S$ contains exactly one idempotent. Since $S$ is regular, every $\mathscr{L}$- and $\mathscr{R}$-class contains at least one idempotent. 
Suppose that $e\in S$ is a non-zero idempotent. If there exists $f\in S$ such that $e\mathscr{L}f$, then there exists $x\in S$ such that $xe=f$. Hence $x\cdot ef=f^2=f$ and so $ef\mathscr{L}f$, and, in particular, $ef\not=0$. This contradicts the assumption, and hence 
every $\mathscr{L}$-class of $S$ contains exactly one idempotent. A dual argument shows that every $\mathscr{R}$-class of $S$ contains exactly one idempotent, and so
$S$ is an inverse semigroup. 
Suppose  there are non-zero elements $x,y\in S$ such that $S^1xS^1\subsetneq S^1yS^1$. Let $J_x$ and $J_y$ denote the $\mathscr{J}$-classes of $x$ and $y$ in $S$, respectively. Since $S$ is regular, there exist idempotents $e\in J_y$ and $f\in J_x$, and there exist $u,v\in S$ such that $f=uev$. 
Hence $fv^{-1}\mathscr{R}f$ and $u^{-1}f\mathscr{L}f$, and so, since $\mathscr{L}$ is a right congruence, $u^{-1}fv^{-1}\mathscr{L}fv^{-1}\mathscr{R}f$, i.e. $u^{-1}fv^{-1}\mathscr{D}f$. In particular, $u^{-1}fv^{-1}\not=0$. But $u^{-1}fv^{-1}=(u^{-1}u)e(vv^{-1})$ is a product of idempotents, which does not equal $0$. It follows that 
$u^{-1}u=vv^{-1}=e$, which implies that $u^{-1}fv^{-1}=e$ and so $e\mathscr{J}f$, a contradiction. 
It follows that the union of any $\mathscr{J}$-class and $\{0\}$ is a subsemigroup of $S$, which is $0$-simple and inverse, i.e. a Brandt semigroup. From the previous paragraph, it follows that $S$ is the 0-union of these Brandt semigroups, which completes the proof. 
A: If $S$ is regular and the product of distinct idempotents is $0$ then idempotents commute and you have an inverse semigroup (with a 'Kronecker' semilattice).
