Prove that $Z_l \left( K \left[ t,t^{-1} \right]\right)=0$. In a studies of $K \left[ t,t^{-1} \right]$, i have a problem to prove $Z_l \left( K \left[ t,t^{-1} \right]\right)=0$ with the following definitions:


*

*$R=K \left[ t,t^{-1} \right]$, 

*the ideal $Z_l \left( R\right)=\{ f\in R : \text{lan}(f)\text{ is left essential ideal of }R \}$ (this is called the left singular ideal),

*$\text{lan}(f)$ is the left annihilator of $R$, which means $\text{lan}(f)=\{g\in R : gf=0 \}$.


Let $f$ be an element in $Z_l \left( R\right)$, i tried to prove that $f=0$.  Since $\text{lan}(f)$ is essential left ideal of $R$ so the intersection of $\text{lan}(f)$ with any ideal of $I$ of $R$ is non-zero. I want to ask about the clear form for any ideal of $R$ to complete the prove, or is there some other way to prove?
 A: I'm guessing that $K$ means a field, since that is the common convention.  That makes $K[t,t^{-1}]$ an integral domain.
It is completely trivial that $Z_l(R)=Z_r(R)=\{0\}$ for any integral domain, and I'm sure you can do it. Just look at what the definition of an annihilator is and think about what that means for elements of the module $_RR$ (and $R_R$ too).

I want to ask about the clear form for any ideal of  to complete the prove, or is there some orther way to prove?

The way I outline above is the direct way: you just show that the only element with essential annihilator is $0$ by noting that every nonzero element's annihilator is zero. (Oops spoilers.)
The way you started is OK in principle, but what you were setting yourself up for was suppose $f\in Z_l(R)\setminus\{0\}$, aiming to find a contradiction. If $x\neq 0$ and its annihilator was essential, the annihilator would also be nonzero, right? so you would have $xy=0$ for $x,y$ nonzero... and surely you can see the contradiction in that...
