Must the product of a nil ideal with a minimal right ideal be 0? Let $R$ be a ring (not necessarily with identity or commutative). Suppose that $K$ is a a nil ideal, and let $M$ be a minimal right ideal of $R$. Must it then follow that $MK = 0$?
As $M$ is minimal we must have either $MK = 0$ or $MK = M$, but I can't see any reason why this second case cannot occur. If it did we must necessarily have that $K \supset M$, as $MK \subset M$, $MK \subset K$, so $MK \subset M \cap K \subset M$. So $M$ is nil too. But I can't see how to go from here. I also can't find a counterexample.
The context in which I need this uses this to conclude that $l(K) \neq 0$ (the left annihilator). Even if what I wrote above is false, does this still hold?
Thanks in advance!
 A: After asking around, I have the following elementary solution.
Suppose that $MK = M$. Then for some $m \in M$, $mK \neq 0$, and being a right ideal must then be equal to $M$ as $M$ is minimal. But then for some $k \in K$, $mk = m$. $K$ being nil means that $k^n = 0$ for some $n \in \mathbb{N}$. But then $m = mk = mk^2 = ... = mk^n = 0$, and $mK = 0$. Thus we must have that $MK = 0$. 
A: It looks like the Dorroh extension allows you to lasso this problem into the realm of rings with identity to solve, after all.
The Dorroh extension of $R$, if you don't already know, is the ring $R^1=\mathbb Z\times R$ with pointwise addition and multiplication given by $(n, a)(m,b)=(nm, nb+ma+ab)$.  It is easy to check that a right ideal of $R$ is a right ideal of $R^1$, and so on with left ideals and ideals.  
Also importantly, $K$ remains a nil ideal in $R^1$. It's a basic fact (an easy exercise) to show that $K$ is contained in the Jacobson radical of $R^1$.
The last question is whether or not $M$ remains simple as an $R^1$ module. In principle, it could have picked up submodules since $R^1$ is bigger than $R$. Now since it is simple as a right $R$ module, given any nonzero $x,y\in M$, there exists $r\in R$ and $n\in \mathbb Z$ such that $xn+xr=y$ (this says that every nonzero element of $M$ generates $M$, equivalent to simplicity.)
But if you check the multiplication rules, $(n, r)(0,x)=(0,y)$, so it turns out elements of $M$ generate $M$ with $R^1$ just like they do with $R$, so it is minimal as a right $R^1$ ideal too.
A basic characterization of the Jacobson radical is that it annihilates simple modules, so we'd have that $MK=\{0\}$.

This could perhaps be significantly reduced if you had a proof on hand that the Jacobson radical in a ring without identity always annihilates minimal right ideals. I'm simply not positive about this fact, since the situation is a little more complicated than in rings with identity. The only thing I'm pretty sure about is that a nil ideal is still contained in the Jacobson radical of a ring without identity.
Another useful thing would be a Nakayama lemma for rings without identity, which I'm not sure is available and I couldn't find in the sea of references to the ordinary Nakayama lemma.  If such a lemma exists, it would look like $MK=M \implies M=\{0\}$, eliminating the case for you.
