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 *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.

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so

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