# Answer to Question #17182 in Abstract Algebra 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 &minus; yx = 1. Let p(y) &isin; k[y] be a fixed polynomial. Show that R &bull; (x &minus; p(y)) is a maximal left ideal in R, and that the simple R-module V = R/R &bull; (x &minus; p(y)) has R-endomorphism ring equal to k.
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 EndRV 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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