55 606
Assignments Done
97,9%
Successfully Done
In November 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Abstract Algebra Question for john.george.milnor

Question #23888
For any field k of characteristic p, let G = SL2(Fp) act on the polynomial ring A = k[x, y] by linear changes of the variables {x, y}, and let Vd ⊆ A (d ≥ 0) be the kG-submodule of homogeneous polynomials of degree d in A. It is known (and thus you may assume) that V0, . . . , Vp−1 are a complete set of simple modules over kG.
If {d1, . . . , dn} is any partition of p, show that the tensor product Vd1 ⊗k • • •⊗k Vdn (under the diagonal G-action) is not semisimple over kG, unless p = 2and n = 1.
Expert's answer
Let d1 + · · · + dn= p, where each di ≥ 1. Then f1 ⊗· · ·⊗fn → f1 · · · fn (fi ∈ Vdi ) defines a kG-surjection Vi1 ⊗· · ·⊗Vin → Vp. If p > 2, then Vi1 ⊗· · ·⊗Vin cannot be semisimple over kG, since Vpis not. If p = 2, this argument no longer works. However, if n> 1, the conclusion remains valid; namely, V1 ⊗ V1 is still not semisimple. In fact, for p =2 ,G = SL2(Fp) is the group S3, and V1 is easilyseen to be isomorphic to the module V. Thus, the desired conclusion nowfollows immediately.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question