# Answer on Abstract Algebra Question for Melvin Henriksen

Question #17182

Let k be a field of characteristic zero, and let R be the Weyl algebra A1(k) with generators x, y and relation xy − yx = 1. Let p(y) ∈ k[y] be a fixed polynomial. Show that R • (x − p(y)) is a maximal left ideal in R, and that the simple R-module V = R/R • (x − p(y)) has R-endomorphism ring equal to k.

Expert's answer

We can identify

0 =

so

*V*as an abelian group with*k*[*y*]. In order to use this identificationeffectively, we must describe the action of*y*and*x*on “*V*=*k*[*y*]”. Of course, the*y*action is just left multiplication by*y*. To describe the*x*action, consider any*v*(*y*)*∈**k*[*y*]. Since*x · v*(*y*) =*v*(*y*)*x*+*dv/dy*=*v*(*y*) (*x − p*(*y*)) +*p*(*y*)*v*(*y*)+*dv/dy,*we see that*x − p*(*y*) acts on*V*=*k*[*y*]as differentiation with respect to*y*. To show that*RV*is simple,consider any*v*(*y*)*<>*0, say of degree*m*. Then(*x − p*(*y*))*m · v*(*y*) is a nonzero constant in*k*.This shows that*R · v*(*y*) =*V*, so*V*is simple. Now consider any*f**∈*End*RV*and let*f*(1) =*g*(*y*).Then, in*V*=*k*[*y*]:0 =

*f*((*x − p*(*y*))*·*1) = (*x − p*(*y*))*· g*(*y*) =*dg/dy,*so

*g**∈**k*. But then*f*(*v*(*y*)) =*f*(*v*(*y*)*·*1) =*v*(*y*)*g,*so*f*isjust multiplication by the constant*g**∈**k*. This completes the proof.Need a fast expert's response?

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