The Problem: Let $A$ be the Weyl algebra, generated by two elements $x,y$ with the relations
$$yx – xy – 1=0$$
(a) If char$k = 0$, what are the finite dimensional representations of $A$? What are the two-sided ideals in $A$?
Hint. For the first question, use the fact that for two square matrices $B, C$, $Tr(BC) = Tr(CB)$. For the second question, show that any nonzero two-sided ideal in $A$ contains a nonzero polynomial in $x$ and use this to characterize this ideal.
For the first part of the question I'm not sure how the hint will help us. I know that a representation is defined to be a vector space $V$ and a homomorphism from $A$ to End$V$. I'm guessing of we suppose that $a,b \in A$ was assigned to a $\phi_1, \phi_2 \in $End $V$, then we first consider the matrix representation of $\phi$'s and use the hint?
For the other part I know the definition of an ideal $I$ is a subspace such that $\forall a \in A$: $aI \subset I$ and $Ia \subset I$. From proposition in this section I know a basis for the Weyl algebra is $\{x^iy^j, i,j \geq 0\}$, but I'm not sure how this would imply that "any nonzero two-sided ideal in $A$ contains a nonzero polynomial in $x$".
Thank you!
Best Answer
In characteristic $k=0$ the Weyl Algebra $A$ has no nontrivial finite dimensional representation. Suppose it did, then given such a representation, elements $x$ and $y$ will act as two matrices $B$ and $C$. Notice that $Tr(BC)=Tr(CB)$, and $[B,C]= BC-CB$. But notice also that $[x,y]=1$ cannot possibly hold since the trace of the commutator is $0$ but the trace of the identity is not $0$
Hint for the computation of the two sided ideals: show that the $A$ is a simple ring. That way the only two sided ideals of $A$ are $(0)$ and $A$ itself.